Rates

#$&*

course Mth 158

documentshort description of contentwhat you'll know when you're doneRatesintroduces the key concept of rate of changethe meaning of average rate of change and how it might be applied

Copy and paste this document into a text editor, insert your responses and submit using theSubmit_Work_Form.

If your solution to stated problem does not match the given solution, you should self-critique per instructions at

http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm

.

Your solution, attempt at solution. If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what you do or do not understand about it. This response should be given, based on the work you did in completing the assignment, before you look at the given solution.

001.Rates

Note that there are 10 questions in this assignment.The questions are of increasing difficulty--the first questions are fairly easy but later questions are very tricky.The main purposes of these exercises are to refine your thinking about rates, and to see how you process challenging information.Most students in most courses would not be expected to answer all these questions correctly; all that's required is that you do your best and follows the recommended procedures for answering and self-critiquing your work.

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Question: If you make $50 in 5 hr, then at what rate are you earning money?

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Your solution:

If you are paid 50 dollars for 5 hours of work then you would receive a pay rate of 10 dollars an hour. For every hour you are paid ten dollars, therefore to receive payment of 50 dollars you would receive 10 dollars 5 times. So the equation would be 50/5=10 or it could be solved by saying 5*10=50.

confidence rating #$&*:ok

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Given Solution:

The rate at which you are earning money is the number of dollars per hour you are earning.You are earning money at the rate of 50 dollars / (5 hours) = 10 dollars / hour.It is very likely that you immediately came up with the $10 / hour becausealmosteveryoneis familiar with the concept of the pay rate, the number of dollars per hour.Note carefully that the pay rate is found by dividing the quantity earned by the time required to earn it.Time rates in general are found by dividing an accumulated quantity by the time required to accumulate it.

You need to make note of anything in the given solution that you didn't understand when you solved the problem.If new ideas have been introduced in the solution, you need to note them.If you notice an error in your own thinking then you need to note that.In your own words, explain anything you didn't already understand and save your response as Notes.

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Self-critique (if necessary):ok

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Question:`q003.If you make $60,000 per year then how much do you make per month?

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Your solution:

If you make 60,000 a year then you would be making 5000 per month. There are 12 months in a year, therefore 60,000/12=5,000. This means you receive 5,000 dollars 12 times. This is represented by the equation 60,000/12=5,000 or 12*5,000=60,000.

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Given Solution:

Most people will very quickly see that we need to divide $60,000 by 12 months, giving us 60,000 dollars / (12 months) = 5000 dollars / month.Note that again we have found a time rate, dividing the accumulated quantity by the time required to accumulate it.

You need to make note of anything in the given solution that you didn't understand when you solved the problem.If new ideas have been introduced in the solution, you need to note them.If you notice an error in your own thinking then you need to note that.In your own words, explain anything you didn't already understand and save your response as Notes.

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Self-critique (if necessary):ok

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Question:`q004.Suppose that the $60,000 is made in a year by a small business.Would be more appropriate to say that the business makes $5000 per month, or that the business makes an average of $5000 per month?

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Your solution:

I would think it more appropriate to consider this an average made per month because the problem and equation in the prior questions seems to neglect the facts that some months are shorter than others limiting the business time per month, also business (or sales made) would differ from month to month. I hardly think that a business would make the exact same in revenue down to the penny amount every month. This equation also does not seem to let us know if the cost of business (what it costs to purchase materials and run machines) is already factored in.

confidence rating #$&*:ok

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Given Solution:

Small businesses do not usually make the same amount of money every month.The amount made depends on the demand for the services or commodities provided by the business, and there are often seasonal fluctuations in addition to other market fluctuations. It is almost certain that a small business making $60,000 per year will make more than $5000 in some months and less than $5000 in others.Therefore it is much more appropriate to say that the business makes and average of $5000 per month.

You need to make note of anything in the given solution that you didn't understand when you solved the problem.If new ideas have been introduced in the solution, you need to note them.If you notice an error in your own thinking then you need to note that.In your own words, explain anything you didn't already understand and save your response as Notes.

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Question:`q005.If you travel 300 miles in 6 hours, at what average rate are you covering distance, and why do we say average rate instead of just plain rate?

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Your solution:

If you travel 300 miles in a time of 6 hours the rate at which you travel is 50 miles per hour. This is solved by taking the overall distance traveled and then dividing that by the amount of time it required to travel such a distance. 300/6=50 meaning for every mile traveled, it took one hour of time. The reason that this is presented in an average is because of factors that are not added in such as traffic that can negatively affect the travel duration. Or another example is wind resistance for airplane travel. Flying into/against the wind would cause the travel time to increase where as traveling away/with the wind will actually shorten the time of travel.

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Given Solution:

The average rate is 50 miles per hour, or 50 miles / hour.This is obtained by dividing the accumulated quantity, the 300 miles, by the time required to accumulate it, obtainingaverate = 300 miles / ( 6 hours) = 50 miles / hour.Note that the rate at which distance is covered is called speed.The car has an average speed of 50 miles/hour. We say 'average rate' in this case because it is almost certain that slight changes in pressure on the accelerator, traffic conditions and other factors ensure that the speed will sometimes be greater than 50 miles/hour and sometimes less than 50 miles/hour; the 50 miles/hour we obtain from the given information is clearly and overall average of the velocities.

You need to make note of anything in the given solution that you didn't understand when you solved the problem.If new ideas have been introduced in the solution, you need to note them.If you notice an error in your own thinking then you need to note that.In your own words, explain anything you didn't already understand and save your response as Notes.

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Self-critique (if necessary):ok

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Question:`q006.If you use 60 gallons of gasoline on a 1200 mile trip, then at what average rate are you using gasoline, with respect to miles traveled?

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Your solution:

This is much like the problems above however, the end result in which we are trying to conclude is different. Instead of mile per gallon we are calculating usage per mile. What I did was I rephrased the problem to say “if you travel 1200 miles on 60 gallons of gas what is the rate of usage?” there for the problem becomes 1200/60=.05. the rate that gas is being used per mile is .05.

confidence rating #$&*:ok

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Given Solution:

The rate of change of one quantity with respect to another is the change in the first quantity, divided by the change in the second.As in previous examples, we found the rate at which money was made with respect to time by dividing the amount of money made by the time required to make it.

By analogy, the rate at which we use fuel with respect to miles traveled is the change in the amount of fuel divided by the number of miles traveled.In this case we use 60 gallons of fuel in 1200 miles, so the average rate it 60 gal / (1200 miles) = .05 gallons / mile.

Note that this question didn't ask for miles per gallon.Miles per gallon is an appropriate and common calculation, but it measures the rate at which miles are covered with respect to the amount of fuel used.Be sure you see the difference.

Note that in this problem we again have here an example of a rate, but unlike previous instances this rate is not calculated with respect to time.This rate is calculated with respect to the amount of fuel used.We divide the accumulated quantity, in this case miles, by the amount of fuel required to cover t miles.Note that again we call the result of this problem an average rate because there are always at least subtle differences in driving conditions that result in more or fewer miles covered with a certain amount of fuel.

It's very important to understand the phrase 'with respect to'. Whether the calculation makes sense or not, it is defined by the order of the terms.

In this case gallons / miletellsyou how many gallons you are burning, on the average, per mile. This concept is not as familiar as miles / gallon, but except for familiarity it's technically no more difficult.

You need to make note of anything in the given solution that you didn't understand when you solved the problem.If new ideas have been introduced in the solution, you need to note them.If you notice an error in your own thinking then you need to note that.In your own words, explain anything you didn't already understand and save your response as Notes.

STUDENT COMMENT

Very Tricky! I thought I had a rhythm going. I understand where I messed up. I am comfortable with the calculations.

INSTRUCTOR RESPONSE

There's nothing wrong with your rhythm.

As I'm sure you understand, there is no intent here to trick, though I know most people will (and do) tend to give the answer you did.

My intent is to make clear the important point that the definition of the terms is unambiguous and must be read carefully, in the right order.

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Question:`q007.The word 'average' generally connotes something like adding two quantities and dividing by 2, or adding several quantities and dividing by the number of quantities we added.Why is it that we are calculating average rates but we aren't adding anything?

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Your solution:

The reason for using the word average is because there are uncalculated circumstances. For example traffic when traveling by car, air resistance when traveling by airplane. These are neglected and therefore giving us an average instead of a precise amount. For example an air plane travels 600 miles in 6 hours is traveling at a rate of 100 miles per hour. This may be true but this would be an example of an object in motion excluding natural factors, like the force required to overcome gravity and airspeed/direction.

confidence rating #$&*:

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Given Solution:

The word 'average' in the context of the dollars / month, miles / gallon types of questions we have been answering was used because we expect that in different months different amounts were earned, or that over different parts of the trip the gas mileage might have varied, but that if we knew all the individual quantities (e.g., the dollars earned each month, the number of gallons used with each mile) and averaged them in the usual manner, we would get the .05 gallons / mile, or the $5000 / month.In a sense we have already added up all the dollars earned in each month, or the miles traveled on each gallon, and we have obtained the total $60,000 or 1200 miles.Thus when we divide by the number of months or the number of gallons, we are in fact calculating an average rate.

You need to make note of anything in the given solution that you didn't understand when you solved the problem.If new ideas have been introduced in the solution, you need to note them.If you notice an error in your own thinking then you need to note that.In your own words, explain anything you didn't already understand and save your response as Notes.

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Question:`q008.In a study of how lifting strength is influenced by various ways of training, a study group was divided into 2 subgroups of equally matched individuals.The first group did 10 pushups per day for a year and the second group did 50 pushups per day for year.At the end of the year to lifting strength of the first group averaged 147 pounds, while that of the second group averaged 162 pounds.At what average rate did lifting strength increase per daily pushup?

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Your solution:

The answer would be the first group increased at a rate of 14.7. I reached this conclusion by taking the average weight and dividing it by the number of pushups. For the second group, the rate of increase was 3.24. I reached this conclusion by using the same process I use for the first group.

confidence rating #$&*:1

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Given Solution:

The second group had 15 pounds more lifting strength as a result of doing 40 more daily pushups than the first.The desired rate is therefore 15 pounds / 40 pushups = .375 pounds / pushup.

You need to make note of anything in the given solution that you didn't understand when you solved the problem.If new ideas have been introduced in the solution, you need to note them.If you notice an error in your own thinking then you need to note that.In your own words, explain anything you didn't already understand and save your response as Notes.

STUDENT COMMENT:

I have a question with respect as to how the question is interpreted. I used the interpretation given in the solution

to question 008 to rephrase the question in 009, but I do not see how this is the correct interpretation of the question as

stated.

INSTRUCTOR RESPONSE:

This exercise is designed to both see what you understand about rates, and to challenge your understanding a bit with concepts that aren't always familiar to students, despite their having completed the necessary prerequisite courses.

The meaning of the rate of change of one quantity with respect to another is of central importance in the application of mathematics. This might well be your first encounter with this particular phrasing, so it might well be unfamiliar to you, but it is important, unambiguous and universal.

You've taken the first step, which is to correctly apply the wordking of the preceding example to the present question.

You'll have ample opportunity in your course to get used to this terminology, and plenty of reinforcement.

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Self-critique (if necessary):

Now that I have seen the answer to this problem, I believe that I interpreted this completely wrong. I think the reason for this is I was trying to relate the wrong facts. Instead of trying to conclude the rate difference between the two groups, I believe that I was trying to find the average rate for each individual group. I did also did not correlate information from both groups, such as how many more pushups the second groups did compared to the first. Nor did I calculate how much more weight the second group lifted than the first.

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Question:`q009.In another part of the study, participants all did 30 pushups per day, but one group did pushups with a 10-pound weight on their shoulders while the other used a 30-pound weight.At the end of the study, the first group had an average lifting strength of 171 pounds, while the second had an average lifting strength of 188 pounds.At what average rate did lifting strength increase with respect to the added shoulder weight?

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Your solution:

Considering this problem is similar to the prior one, i came to the conclusion that the method used to solve it would also be similar. The first operation I performed was subtracting the lifting strength of the weaker group from the lifting strength of the stronger group. This gave a strength difference of 17. Then to find the difference in total weight lifted I subtracted the 10 pound weight of group one from the 30 pound weight of group two. This gave 20, so to find out the actual increase of strength you divide 17 by 20 (17/20=8.5) giving an average increase of .85

confidence rating #$&*:

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Given Solution:

The difference in lifting strength was 17 pounds, as a result of a 20 pound difference in added weight.The average rate at which strength increases with respect added weight would therefore be 17 lifting pounds / (20 added pounds) = .85 lifting pounds / added pound.The strength advantage was .85 lifting pounds per pound of added weight, on the average.

You need to make note of anything in the given solution that you didn't understand when you solved the problem.If new ideas have been introduced in the solution, you need to note them.If you notice an error in your own thinking then you need to note that.In your own words, explain anything you didn't already understand and save your response as Notes.

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Self-critique (if necessary):

After seeing the answer to the first problem, I understand how to do this better.

Self-critiqueRating:ok

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Question:`q010.During a race, a runner passes the 100-meter mark 12 seconds after the start and the 200-meter mark 22 seconds after the start.At what average rate was the runner covering distance between those two positions?

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Your solution:

The average speed the runner was running between the 100 meter marker and the 200 meter marker is 10 meters per second. This is solved by taking the time and distance of the first 100 meter marker and subtracting it from the 200 meter marker. We do this because the problem asked for the average speed between the two points and both of these times are measured from the begging of the race. This gives us 100 meters in 10 seconds. Now to find the rate of travel we simply take the distance traveled and divide it by the time taken to travel the distance (100/10=10) therefore the average speed between these two points is 10 meters per second.

confidence rating #$&*:ok

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Given Solution:

The runner traveled 100 meters between the two positions, and required 10 seconds to do so.The average rate at which the runner was covering distance was therefore 100 meters / (10 seconds) = 10 meters / second.Again this is an average rate; at different positions in his stride the runner would clearly be traveling at slightly different speeds.

You need to make note of anything in the given solution that you didn't understand when you solved the problem.If new ideas have been introduced in the solution, you need to note them.If you notice an error in your own thinking then you need to note that.In your own words, explain anything you didn't already understand and save your response as Notes.

STUDENT QUESTION

Is there a formula for this is it d= r*t or distance equal rate times time??????????????????

INSTRUCTOR RESPONSE

That formula would apply in this specific situation.

The goal is to learn to use the general concept of rate of change. The situation of this problem, and the formula you quote, are just one instance of a general concept that applies far beyond the context of distance and time.

It's fine if the formula helps you understand the general concept of rate. Just be sure you work to understand the broader concept.

Note also that we try to avoid using d for the name of a variable. The letter d will come to have a specific meaning in the context of rates, and to use d as the name of a variable invite confusion.

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Question:`q011.During a race, a runner passes the 100-meter mark moving at 10 meters / second, and the 200-meter mark moving at 9 meters / second.What is your best estimate of how long it takes the runner to cover the intervening 100 meter distance?

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Your solution:

To solve this problem we need to find the average between the two speeds. If the runner was traveling at 10 meters per second for the first 100 meter and the slowed to 9 meters per second for the second 100 meters, we need to find the average. To do this we need to add the speeds (10+9=19) then we divide by the number of values (19/2=9.5). Now that we have the average of the 2 speeds, we need to find the average speed for 100 meters traveled at this distance. (100/9.5=10.5) so the average speed for the next 100 meters would be estimated at 10.5 meters per second.

confidence rating #$&*:ok

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Given Solution:

At 10 meters/sec, the runner would require 10 seconds to travel 100 meters.However the runner seems to be slowing, and will therefore require more than 10 seconds to travel the 100 meters.We don't know what the runner's average speedis,we only know that it goes from 10 m/s to 9 m/s.The simplest estimate we could make would be that the average speed is the average of 10 m/s and 9 m/s, or (10 m/s + 9 m/s )/ 2 = 9.5 m/s.Taking this approximation as the average rate, the time required to travel 100 meters will be (100 meters) / (9.5 m/s) = 10.5 sec,approx..Note that simply averaging the 10 m/s and the 9 m/s might not be the best way to approximate the average rate--for example we if we knew enough about the situation we might expect that this runner would maintain the 10 m/s for most of the remaining 100 meters, and simply tire during the last few seconds.However we were not given this information, and we don't add extraneous assumptions without good cause.So the approximation we used here is pretty close to the best we can do with the given information.

You need to make note of anything in the given solution that you didn't understand when you solved the problem.If new ideas have been introduced in the solution, you need to note them.If you notice an error in your own thinking then you need to note that.In your own words, explain anything you didn't already understand and save your response as Notes.

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Question:`q012.We just averaged two quantities, adding them and dividing by 2, to find an average rate.We didn't do that before.Why we do it now?

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Your solution:

The reason we did this in this certain problem is because we were presented with more than one value of time. Before we had a set distance and speed, so to find the average speed traveled within that distance, we need only to take the distance traveled and divide it by the time taken to travel that distance. However in the problem above we were presented with a whole value split in to 3 values which we are to find the third. Taking the average speed from either the first or second increment of the race only would not give us an accurate answer for the third, because the time it took to run the 100 meter intervals were not the same. And depending on which interval of the race we took our average from the average speed could be higher or lower, giving us sporadic rates of speed. Therefore it is appropriate to take an average of the time values given to find the speed. For example if a car travels a distance of 300 miles and the first 100 miles was traveled 2 hours and the second 100 miles was traveled in 3 hours and the third 100 mile was traveled in1 hour then we would need to take an average of the time values to discover how fast the car was traveling. In this example the average time is 2 hours, where as if we take the average from just the second or third increment of the distance then we will get higher or lower traveling speed.

confidence rating #$&*:ok

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Given Solution:

In previous examples the quantities weren't rates.We were given the amount of change of some accumulating quantity, and the change in time or in some other quantity on which the first was dependent (e.g., dollars and months, miles and gallons).Here we are given 2 rates, 10 m/s and 9 m/s, in a situation where we need an average rate in order to answer a question.Within this context, averaging the 2 rates was an appropriate tactic.

You need to make note of anything in the given solution that you didn't understand when you solved the problem.If new ideas have been introduced in the solution, you need to note them.If you notice an error in your own thinking then you need to note that.In your own words, explain anything you didn't already understand and save your response as Notes.

STUDENT QUESTION:

I thought the change of an accumulating quantity was the rate?

INSTRUCTOR RESPONSE:

Quick response: The rate is not just the change in the accumulating quantity; if we're talking about a 'time rate' it's the change in the accumulating quantity divided by the time interval (or in calculus the limiting value of this ratio as the time interval approaches zero).

More detailed response: If quantity A changes with respect to quantity B, then the average rate of change of A with respect to B (i.e., change in A / change in B) is 'the rate'. If the B quantity is clock time, then 'the rate' tells you 'how fast' the A quantity accumulates. However the rate is not just the change in the quantity A (i.e., the change in the accumulating quantity), but change in A / change in B.

For students having had at least a semester of calculus at some level: Of course the above generalizes into the definition of the derivative. y ' (x) is the instantaneous rate at which the y quantity changes with respect to x. y ' (x) is the rate at which y accumulates with respect to x.

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Rates

#$&*

course Mth 158

documentshort description of contentwhat you'll know when you're doneRatesintroduces the key concept of rate of changethe meaning of average rate of change and how it might be applied

Copy and paste this document into a text editor, insert your responses and submit using theSubmit_Work_Form.

If your solution to stated problem does not match the given solution, you should self-critique per instructions at

http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm

.

Your solution, attempt at solution. If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what you do or do not understand about it. This response should be given, based on the work you did in completing the assignment, before you look at the given solution.

001.Rates

Note that there are 10 questions in this assignment.The questions are of increasing difficulty--the first questions are fairly easy but later questions are very tricky.The main purposes of these exercises are to refine your thinking about rates, and to see how you process challenging information.Most students in most courses would not be expected to answer all these questions correctly; all that's required is that you do your best and follows the recommended procedures for answering and self-critiquing your work.

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Question: If you make $50 in 5 hr, then at what rate are you earning money?

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Your solution:

If you are paid 50 dollars for 5 hours of work then you would receive a pay rate of 10 dollars an hour. For every hour you are paid ten dollars, therefore to receive payment of 50 dollars you would receive 10 dollars 5 times. So the equation would be 50/5=10 or it could be solved by saying 5*10=50.

confidence rating #$&*:ok

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Given Solution:

The rate at which you are earning money is the number of dollars per hour you are earning.You are earning money at the rate of 50 dollars / (5 hours) = 10 dollars / hour.It is very likely that you immediately came up with the $10 / hour becausealmosteveryoneis familiar with the concept of the pay rate, the number of dollars per hour.Note carefully that the pay rate is found by dividing the quantity earned by the time required to earn it.Time rates in general are found by dividing an accumulated quantity by the time required to accumulate it.

You need to make note of anything in the given solution that you didn't understand when you solved the problem.If new ideas have been introduced in the solution, you need to note them.If you notice an error in your own thinking then you need to note that.In your own words, explain anything you didn't already understand and save your response as Notes.

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Self-critique (if necessary):ok

Self-critiqueRating:ok

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Question:`q003.If you make $60,000 per year then how much do you make per month?

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Your solution:

If you make 60,000 a year then you would be making 5000 per month. There are 12 months in a year, therefore 60,000/12=5,000. This means you receive 5,000 dollars 12 times. This is represented by the equation 60,000/12=5,000 or 12*5,000=60,000.

confidence rating #$&*:ok

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Given Solution:

Most people will very quickly see that we need to divide $60,000 by 12 months, giving us 60,000 dollars / (12 months) = 5000 dollars / month.Note that again we have found a time rate, dividing the accumulated quantity by the time required to accumulate it.

You need to make note of anything in the given solution that you didn't understand when you solved the problem.If new ideas have been introduced in the solution, you need to note them.If you notice an error in your own thinking then you need to note that.In your own words, explain anything you didn't already understand and save your response as Notes.

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Self-critique (if necessary):ok

Self-critiqueRating:ok

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Question:`q004.Suppose that the $60,000 is made in a year by a small business.Would be more appropriate to say that the business makes $5000 per month, or that the business makes an average of $5000 per month?

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Your solution:

I would think it more appropriate to consider this an average made per month because the problem and equation in the prior questions seems to neglect the facts that some months are shorter than others limiting the business time per month, also business (or sales made) would differ from month to month. I hardly think that a business would make the exact same in revenue down to the penny amount every month. This equation also does not seem to let us know if the cost of business (what it costs to purchase materials and run machines) is already factored in.

confidence rating #$&*:ok

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Given Solution:

Small businesses do not usually make the same amount of money every month.The amount made depends on the demand for the services or commodities provided by the business, and there are often seasonal fluctuations in addition to other market fluctuations. It is almost certain that a small business making $60,000 per year will make more than $5000 in some months and less than $5000 in others.Therefore it is much more appropriate to say that the business makes and average of $5000 per month.

You need to make note of anything in the given solution that you didn't understand when you solved the problem.If new ideas have been introduced in the solution, you need to note them.If you notice an error in your own thinking then you need to note that.In your own words, explain anything you didn't already understand and save your response as Notes.

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Question:`q005.If you travel 300 miles in 6 hours, at what average rate are you covering distance, and why do we say average rate instead of just plain rate?

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Your solution:

If you travel 300 miles in a time of 6 hours the rate at which you travel is 50 miles per hour. This is solved by taking the overall distance traveled and then dividing that by the amount of time it required to travel such a distance. 300/6=50 meaning for every mile traveled, it took one hour of time. The reason that this is presented in an average is because of factors that are not added in such as traffic that can negatively affect the travel duration. Or another example is wind resistance for airplane travel. Flying into/against the wind would cause the travel time to increase where as traveling away/with the wind will actually shorten the time of travel.

confidence rating #$&*:ok

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Given Solution:

The average rate is 50 miles per hour, or 50 miles / hour.This is obtained by dividing the accumulated quantity, the 300 miles, by the time required to accumulate it, obtainingaverate = 300 miles / ( 6 hours) = 50 miles / hour.Note that the rate at which distance is covered is called speed.The car has an average speed of 50 miles/hour. We say 'average rate' in this case because it is almost certain that slight changes in pressure on the accelerator, traffic conditions and other factors ensure that the speed will sometimes be greater than 50 miles/hour and sometimes less than 50 miles/hour; the 50 miles/hour we obtain from the given information is clearly and overall average of the velocities.

You need to make note of anything in the given solution that you didn't understand when you solved the problem.If new ideas have been introduced in the solution, you need to note them.If you notice an error in your own thinking then you need to note that.In your own words, explain anything you didn't already understand and save your response as Notes.

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Self-critique (if necessary):ok

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Question:`q006.If you use 60 gallons of gasoline on a 1200 mile trip, then at what average rate are you using gasoline, with respect to miles traveled?

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Your solution:

This is much like the problems above however, the end result in which we are trying to conclude is different. Instead of mile per gallon we are calculating usage per mile. What I did was I rephrased the problem to say “if you travel 1200 miles on 60 gallons of gas what is the rate of usage?” there for the problem becomes 1200/60=.05. the rate that gas is being used per mile is .05.

confidence rating #$&*:ok

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Given Solution:

The rate of change of one quantity with respect to another is the change in the first quantity, divided by the change in the second.As in previous examples, we found the rate at which money was made with respect to time by dividing the amount of money made by the time required to make it.

By analogy, the rate at which we use fuel with respect to miles traveled is the change in the amount of fuel divided by the number of miles traveled.In this case we use 60 gallons of fuel in 1200 miles, so the average rate it 60 gal / (1200 miles) = .05 gallons / mile.

Note that this question didn't ask for miles per gallon.Miles per gallon is an appropriate and common calculation, but it measures the rate at which miles are covered with respect to the amount of fuel used.Be sure you see the difference.

Note that in this problem we again have here an example of a rate, but unlike previous instances this rate is not calculated with respect to time.This rate is calculated with respect to the amount of fuel used.We divide the accumulated quantity, in this case miles, by the amount of fuel required to cover t miles.Note that again we call the result of this problem an average rate because there are always at least subtle differences in driving conditions that result in more or fewer miles covered with a certain amount of fuel.

It's very important to understand the phrase 'with respect to'. Whether the calculation makes sense or not, it is defined by the order of the terms.

In this case gallons / miletellsyou how many gallons you are burning, on the average, per mile. This concept is not as familiar as miles / gallon, but except for familiarity it's technically no more difficult.

You need to make note of anything in the given solution that you didn't understand when you solved the problem.If new ideas have been introduced in the solution, you need to note them.If you notice an error in your own thinking then you need to note that.In your own words, explain anything you didn't already understand and save your response as Notes.

STUDENT COMMENT

Very Tricky! I thought I had a rhythm going. I understand where I messed up. I am comfortable with the calculations.

INSTRUCTOR RESPONSE

There's nothing wrong with your rhythm.

As I'm sure you understand, there is no intent here to trick, though I know most people will (and do) tend to give the answer you did.

My intent is to make clear the important point that the definition of the terms is unambiguous and must be read carefully, in the right order.

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Self-critique (if necessary):ok

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Question:`q007.The word 'average' generally connotes something like adding two quantities and dividing by 2, or adding several quantities and dividing by the number of quantities we added.Why is it that we are calculating average rates but we aren't adding anything?

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Your solution:

The reason for using the word average is because there are uncalculated circumstances. For example traffic when traveling by car, air resistance when traveling by airplane. These are neglected and therefore giving us an average instead of a precise amount. For example an air plane travels 600 miles in 6 hours is traveling at a rate of 100 miles per hour. This may be true but this would be an example of an object in motion excluding natural factors, like the force required to overcome gravity and airspeed/direction.

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Given Solution:

The word 'average' in the context of the dollars / month, miles / gallon types of questions we have been answering was used because we expect that in different months different amounts were earned, or that over different parts of the trip the gas mileage might have varied, but that if we knew all the individual quantities (e.g., the dollars earned each month, the number of gallons used with each mile) and averaged them in the usual manner, we would get the .05 gallons / mile, or the $5000 / month.In a sense we have already added up all the dollars earned in each month, or the miles traveled on each gallon, and we have obtained the total $60,000 or 1200 miles.Thus when we divide by the number of months or the number of gallons, we are in fact calculating an average rate.

You need to make note of anything in the given solution that you didn't understand when you solved the problem.If new ideas have been introduced in the solution, you need to note them.If you notice an error in your own thinking then you need to note that.In your own words, explain anything you didn't already understand and save your response as Notes.

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Question:`q008.In a study of how lifting strength is influenced by various ways of training, a study group was divided into 2 subgroups of equally matched individuals.The first group did 10 pushups per day for a year and the second group did 50 pushups per day for year.At the end of the year to lifting strength of the first group averaged 147 pounds, while that of the second group averaged 162 pounds.At what average rate did lifting strength increase per daily pushup?

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Your solution:

The answer would be the first group increased at a rate of 14.7. I reached this conclusion by taking the average weight and dividing it by the number of pushups. For the second group, the rate of increase was 3.24. I reached this conclusion by using the same process I use for the first group.

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Given Solution:

The second group had 15 pounds more lifting strength as a result of doing 40 more daily pushups than the first.The desired rate is therefore 15 pounds / 40 pushups = .375 pounds / pushup.

You need to make note of anything in the given solution that you didn't understand when you solved the problem.If new ideas have been introduced in the solution, you need to note them.If you notice an error in your own thinking then you need to note that.In your own words, explain anything you didn't already understand and save your response as Notes.

STUDENT COMMENT:

I have a question with respect as to how the question is interpreted. I used the interpretation given in the solution

to question 008 to rephrase the question in 009, but I do not see how this is the correct interpretation of the question as

stated.

INSTRUCTOR RESPONSE:

This exercise is designed to both see what you understand about rates, and to challenge your understanding a bit with concepts that aren't always familiar to students, despite their having completed the necessary prerequisite courses.

The meaning of the rate of change of one quantity with respect to another is of central importance in the application of mathematics. This might well be your first encounter with this particular phrasing, so it might well be unfamiliar to you, but it is important, unambiguous and universal.

You've taken the first step, which is to correctly apply the wordking of the preceding example to the present question.

You'll have ample opportunity in your course to get used to this terminology, and plenty of reinforcement.

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Self-critique (if necessary):

Now that I have seen the answer to this problem, I believe that I interpreted this completely wrong. I think the reason for this is I was trying to relate the wrong facts. Instead of trying to conclude the rate difference between the two groups, I believe that I was trying to find the average rate for each individual group. I did also did not correlate information from both groups, such as how many more pushups the second groups did compared to the first. Nor did I calculate how much more weight the second group lifted than the first.

Self-critiqueRating:

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Question:`q009.In another part of the study, participants all did 30 pushups per day, but one group did pushups with a 10-pound weight on their shoulders while the other used a 30-pound weight.At the end of the study, the first group had an average lifting strength of 171 pounds, while the second had an average lifting strength of 188 pounds.At what average rate did lifting strength increase with respect to the added shoulder weight?

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Your solution:

Considering this problem is similar to the prior one, i came to the conclusion that the method used to solve it would also be similar. The first operation I performed was subtracting the lifting strength of the weaker group from the lifting strength of the stronger group. This gave a strength difference of 17. Then to find the difference in total weight lifted I subtracted the 10 pound weight of group one from the 30 pound weight of group two. This gave 20, so to find out the actual increase of strength you divide 17 by 20 (17/20=8.5) giving an average increase of .85

confidence rating #$&*:

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Given Solution:

The difference in lifting strength was 17 pounds, as a result of a 20 pound difference in added weight.The average rate at which strength increases with respect added weight would therefore be 17 lifting pounds / (20 added pounds) = .85 lifting pounds / added pound.The strength advantage was .85 lifting pounds per pound of added weight, on the average.

You need to make note of anything in the given solution that you didn't understand when you solved the problem.If new ideas have been introduced in the solution, you need to note them.If you notice an error in your own thinking then you need to note that.In your own words, explain anything you didn't already understand and save your response as Notes.

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Self-critique (if necessary):

After seeing the answer to the first problem, I understand how to do this better.

Self-critiqueRating:ok

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Question:`q010.During a race, a runner passes the 100-meter mark 12 seconds after the start and the 200-meter mark 22 seconds after the start.At what average rate was the runner covering distance between those two positions?

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Your solution:

The average speed the runner was running between the 100 meter marker and the 200 meter marker is 10 meters per second. This is solved by taking the time and distance of the first 100 meter marker and subtracting it from the 200 meter marker. We do this because the problem asked for the average speed between the two points and both of these times are measured from the begging of the race. This gives us 100 meters in 10 seconds. Now to find the rate of travel we simply take the distance traveled and divide it by the time taken to travel the distance (100/10=10) therefore the average speed between these two points is 10 meters per second.

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Given Solution:

The runner traveled 100 meters between the two positions, and required 10 seconds to do so.The average rate at which the runner was covering distance was therefore 100 meters / (10 seconds) = 10 meters / second.Again this is an average rate; at different positions in his stride the runner would clearly be traveling at slightly different speeds.

You need to make note of anything in the given solution that you didn't understand when you solved the problem.If new ideas have been introduced in the solution, you need to note them.If you notice an error in your own thinking then you need to note that.In your own words, explain anything you didn't already understand and save your response as Notes.

STUDENT QUESTION

Is there a formula for this is it d= r*t or distance equal rate times time??????????????????

INSTRUCTOR RESPONSE

That formula would apply in this specific situation.

The goal is to learn to use the general concept of rate of change. The situation of this problem, and the formula you quote, are just one instance of a general concept that applies far beyond the context of distance and time.

It's fine if the formula helps you understand the general concept of rate. Just be sure you work to understand the broader concept.

Note also that we try to avoid using d for the name of a variable. The letter d will come to have a specific meaning in the context of rates, and to use d as the name of a variable invite confusion.

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Question:`q011.During a race, a runner passes the 100-meter mark moving at 10 meters / second, and the 200-meter mark moving at 9 meters / second.What is your best estimate of how long it takes the runner to cover the intervening 100 meter distance?

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Your solution:

To solve this problem we need to find the average between the two speeds. If the runner was traveling at 10 meters per second for the first 100 meter and the slowed to 9 meters per second for the second 100 meters, we need to find the average. To do this we need to add the speeds (10+9=19) then we divide by the number of values (19/2=9.5). Now that we have the average of the 2 speeds, we need to find the average speed for 100 meters traveled at this distance. (100/9.5=10.5) so the average speed for the next 100 meters would be estimated at 10.5 meters per second.

confidence rating #$&*:ok

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Given Solution:

At 10 meters/sec, the runner would require 10 seconds to travel 100 meters.However the runner seems to be slowing, and will therefore require more than 10 seconds to travel the 100 meters.We don't know what the runner's average speedis,we only know that it goes from 10 m/s to 9 m/s.The simplest estimate we could make would be that the average speed is the average of 10 m/s and 9 m/s, or (10 m/s + 9 m/s )/ 2 = 9.5 m/s.Taking this approximation as the average rate, the time required to travel 100 meters will be (100 meters) / (9.5 m/s) = 10.5 sec,approx..Note that simply averaging the 10 m/s and the 9 m/s might not be the best way to approximate the average rate--for example we if we knew enough about the situation we might expect that this runner would maintain the 10 m/s for most of the remaining 100 meters, and simply tire during the last few seconds.However we were not given this information, and we don't add extraneous assumptions without good cause.So the approximation we used here is pretty close to the best we can do with the given information.

You need to make note of anything in the given solution that you didn't understand when you solved the problem.If new ideas have been introduced in the solution, you need to note them.If you notice an error in your own thinking then you need to note that.In your own words, explain anything you didn't already understand and save your response as Notes.

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Question:`q012.We just averaged two quantities, adding them and dividing by 2, to find an average rate.We didn't do that before.Why we do it now?

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Your solution:

The reason we did this in this certain problem is because we were presented with more than one value of time. Before we had a set distance and speed, so to find the average speed traveled within that distance, we need only to take the distance traveled and divide it by the time taken to travel that distance. However in the problem above we were presented with a whole value split in to 3 values which we are to find the third. Taking the average speed from either the first or second increment of the race only would not give us an accurate answer for the third, because the time it took to run the 100 meter intervals were not the same. And depending on which interval of the race we took our average from the average speed could be higher or lower, giving us sporadic rates of speed. Therefore it is appropriate to take an average of the time values given to find the speed. For example if a car travels a distance of 300 miles and the first 100 miles was traveled 2 hours and the second 100 miles was traveled in 3 hours and the third 100 mile was traveled in1 hour then we would need to take an average of the time values to discover how fast the car was traveling. In this example the average time is 2 hours, where as if we take the average from just the second or third increment of the distance then we will get higher or lower traveling speed.

confidence rating #$&*:ok

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Given Solution:

In previous examples the quantities weren't rates.We were given the amount of change of some accumulating quantity, and the change in time or in some other quantity on which the first was dependent (e.g., dollars and months, miles and gallons).Here we are given 2 rates, 10 m/s and 9 m/s, in a situation where we need an average rate in order to answer a question.Within this context, averaging the 2 rates was an appropriate tactic.

You need to make note of anything in the given solution that you didn't understand when you solved the problem.If new ideas have been introduced in the solution, you need to note them.If you notice an error in your own thinking then you need to note that.In your own words, explain anything you didn't already understand and save your response as Notes.

STUDENT QUESTION:

I thought the change of an accumulating quantity was the rate?

INSTRUCTOR RESPONSE:

Quick response: The rate is not just the change in the accumulating quantity; if we're talking about a 'time rate' it's the change in the accumulating quantity divided by the time interval (or in calculus the limiting value of this ratio as the time interval approaches zero).

More detailed response: If quantity A changes with respect to quantity B, then the average rate of change of A with respect to B (i.e., change in A / change in B) is 'the rate'. If the B quantity is clock time, then 'the rate' tells you 'how fast' the A quantity accumulates. However the rate is not just the change in the quantity A (i.e., the change in the accumulating quantity), but change in A / change in B.

For students having had at least a semester of calculus at some level: Of course the above generalizes into the definition of the derivative. y ' (x) is the instantaneous rate at which the y quantity changes with respect to x. y ' (x) is the rate at which y accumulates with respect to x.

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#*&!

Rates

#$&*

course Mth 158

documentshort description of contentwhat you'll know when you're doneRatesintroduces the key concept of rate of changethe meaning of average rate of change and how it might be applied

Copy and paste this document into a text editor, insert your responses and submit using theSubmit_Work_Form.

If your solution to stated problem does not match the given solution, you should self-critique per instructions at

http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm

.

Your solution, attempt at solution. If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what you do or do not understand about it. This response should be given, based on the work you did in completing the assignment, before you look at the given solution.

001.Rates

Note that there are 10 questions in this assignment.The questions are of increasing difficulty--the first questions are fairly easy but later questions are very tricky.The main purposes of these exercises are to refine your thinking about rates, and to see how you process challenging information.Most students in most courses would not be expected to answer all these questions correctly; all that's required is that you do your best and follows the recommended procedures for answering and self-critiquing your work.

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Question: If you make $50 in 5 hr, then at what rate are you earning money?

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Your solution:

If you are paid 50 dollars for 5 hours of work then you would receive a pay rate of 10 dollars an hour. For every hour you are paid ten dollars, therefore to receive payment of 50 dollars you would receive 10 dollars 5 times. So the equation would be 50/5=10 or it could be solved by saying 5*10=50.

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Given Solution:

The rate at which you are earning money is the number of dollars per hour you are earning.You are earning money at the rate of 50 dollars / (5 hours) = 10 dollars / hour.It is very likely that you immediately came up with the $10 / hour becausealmosteveryoneis familiar with the concept of the pay rate, the number of dollars per hour.Note carefully that the pay rate is found by dividing the quantity earned by the time required to earn it.Time rates in general are found by dividing an accumulated quantity by the time required to accumulate it.

You need to make note of anything in the given solution that you didn't understand when you solved the problem.If new ideas have been introduced in the solution, you need to note them.If you notice an error in your own thinking then you need to note that.In your own words, explain anything you didn't already understand and save your response as Notes.

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Self-critique (if necessary):ok

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Question:`q003.If you make $60,000 per year then how much do you make per month?

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Your solution:

If you make 60,000 a year then you would be making 5000 per month. There are 12 months in a year, therefore 60,000/12=5,000. This means you receive 5,000 dollars 12 times. This is represented by the equation 60,000/12=5,000 or 12*5,000=60,000.

confidence rating #$&*:ok

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Given Solution:

Most people will very quickly see that we need to divide $60,000 by 12 months, giving us 60,000 dollars / (12 months) = 5000 dollars / month.Note that again we have found a time rate, dividing the accumulated quantity by the time required to accumulate it.

You need to make note of anything in the given solution that you didn't understand when you solved the problem.If new ideas have been introduced in the solution, you need to note them.If you notice an error in your own thinking then you need to note that.In your own words, explain anything you didn't already understand and save your response as Notes.

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Self-critique (if necessary):ok

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Question:`q004.Suppose that the $60,000 is made in a year by a small business.Would be more appropriate to say that the business makes $5000 per month, or that the business makes an average of $5000 per month?

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Your solution:

I would think it more appropriate to consider this an average made per month because the problem and equation in the prior questions seems to neglect the facts that some months are shorter than others limiting the business time per month, also business (or sales made) would differ from month to month. I hardly think that a business would make the exact same in revenue down to the penny amount every month. This equation also does not seem to let us know if the cost of business (what it costs to purchase materials and run machines) is already factored in.

confidence rating #$&*:ok

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Given Solution:

Small businesses do not usually make the same amount of money every month.The amount made depends on the demand for the services or commodities provided by the business, and there are often seasonal fluctuations in addition to other market fluctuations. It is almost certain that a small business making $60,000 per year will make more than $5000 in some months and less than $5000 in others.Therefore it is much more appropriate to say that the business makes and average of $5000 per month.

You need to make note of anything in the given solution that you didn't understand when you solved the problem.If new ideas have been introduced in the solution, you need to note them.If you notice an error in your own thinking then you need to note that.In your own words, explain anything you didn't already understand and save your response as Notes.

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Self-critique (if necessary):ok

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Question:`q005.If you travel 300 miles in 6 hours, at what average rate are you covering distance, and why do we say average rate instead of just plain rate?

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Your solution:

If you travel 300 miles in a time of 6 hours the rate at which you travel is 50 miles per hour. This is solved by taking the overall distance traveled and then dividing that by the amount of time it required to travel such a distance. 300/6=50 meaning for every mile traveled, it took one hour of time. The reason that this is presented in an average is because of factors that are not added in such as traffic that can negatively affect the travel duration. Or another example is wind resistance for airplane travel. Flying into/against the wind would cause the travel time to increase where as traveling away/with the wind will actually shorten the time of travel.

confidence rating #$&*:ok

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Given Solution:

The average rate is 50 miles per hour, or 50 miles / hour.This is obtained by dividing the accumulated quantity, the 300 miles, by the time required to accumulate it, obtainingaverate = 300 miles / ( 6 hours) = 50 miles / hour.Note that the rate at which distance is covered is called speed.The car has an average speed of 50 miles/hour. We say 'average rate' in this case because it is almost certain that slight changes in pressure on the accelerator, traffic conditions and other factors ensure that the speed will sometimes be greater than 50 miles/hour and sometimes less than 50 miles/hour; the 50 miles/hour we obtain from the given information is clearly and overall average of the velocities.

You need to make note of anything in the given solution that you didn't understand when you solved the problem.If new ideas have been introduced in the solution, you need to note them.If you notice an error in your own thinking then you need to note that.In your own words, explain anything you didn't already understand and save your response as Notes.

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Question:`q006.If you use 60 gallons of gasoline on a 1200 mile trip, then at what average rate are you using gasoline, with respect to miles traveled?

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Your solution:

This is much like the problems above however, the end result in which we are trying to conclude is different. Instead of mile per gallon we are calculating usage per mile. What I did was I rephrased the problem to say “if you travel 1200 miles on 60 gallons of gas what is the rate of usage?” there for the problem becomes 1200/60=.05. the rate that gas is being used per mile is .05.

confidence rating #$&*:ok

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Given Solution:

The rate of change of one quantity with respect to another is the change in the first quantity, divided by the change in the second.As in previous examples, we found the rate at which money was made with respect to time by dividing the amount of money made by the time required to make it.

By analogy, the rate at which we use fuel with respect to miles traveled is the change in the amount of fuel divided by the number of miles traveled.In this case we use 60 gallons of fuel in 1200 miles, so the average rate it 60 gal / (1200 miles) = .05 gallons / mile.

Note that this question didn't ask for miles per gallon.Miles per gallon is an appropriate and common calculation, but it measures the rate at which miles are covered with respect to the amount of fuel used.Be sure you see the difference.

Note that in this problem we again have here an example of a rate, but unlike previous instances this rate is not calculated with respect to time.This rate is calculated with respect to the amount of fuel used.We divide the accumulated quantity, in this case miles, by the amount of fuel required to cover t miles.Note that again we call the result of this problem an average rate because there are always at least subtle differences in driving conditions that result in more or fewer miles covered with a certain amount of fuel.

It's very important to understand the phrase 'with respect to'. Whether the calculation makes sense or not, it is defined by the order of the terms.

In this case gallons / miletellsyou how many gallons you are burning, on the average, per mile. This concept is not as familiar as miles / gallon, but except for familiarity it's technically no more difficult.

You need to make note of anything in the given solution that you didn't understand when you solved the problem.If new ideas have been introduced in the solution, you need to note them.If you notice an error in your own thinking then you need to note that.In your own words, explain anything you didn't already understand and save your response as Notes.

STUDENT COMMENT

Very Tricky! I thought I had a rhythm going. I understand where I messed up. I am comfortable with the calculations.

INSTRUCTOR RESPONSE

There's nothing wrong with your rhythm.

As I'm sure you understand, there is no intent here to trick, though I know most people will (and do) tend to give the answer you did.

My intent is to make clear the important point that the definition of the terms is unambiguous and must be read carefully, in the right order.

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Self-critique (if necessary):ok

Self-critiqueRating:ok

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Question:`q007.The word 'average' generally connotes something like adding two quantities and dividing by 2, or adding several quantities and dividing by the number of quantities we added.Why is it that we are calculating average rates but we aren't adding anything?

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Your solution:

The reason for using the word average is because there are uncalculated circumstances. For example traffic when traveling by car, air resistance when traveling by airplane. These are neglected and therefore giving us an average instead of a precise amount. For example an air plane travels 600 miles in 6 hours is traveling at a rate of 100 miles per hour. This may be true but this would be an example of an object in motion excluding natural factors, like the force required to overcome gravity and airspeed/direction.

confidence rating #$&*:

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Given Solution:

The word 'average' in the context of the dollars / month, miles / gallon types of questions we have been answering was used because we expect that in different months different amounts were earned, or that over different parts of the trip the gas mileage might have varied, but that if we knew all the individual quantities (e.g., the dollars earned each month, the number of gallons used with each mile) and averaged them in the usual manner, we would get the .05 gallons / mile, or the $5000 / month.In a sense we have already added up all the dollars earned in each month, or the miles traveled on each gallon, and we have obtained the total $60,000 or 1200 miles.Thus when we divide by the number of months or the number of gallons, we are in fact calculating an average rate.

You need to make note of anything in the given solution that you didn't understand when you solved the problem.If new ideas have been introduced in the solution, you need to note them.If you notice an error in your own thinking then you need to note that.In your own words, explain anything you didn't already understand and save your response as Notes.

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Self-critique (if necessary):ok

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Question:`q008.In a study of how lifting strength is influenced by various ways of training, a study group was divided into 2 subgroups of equally matched individuals.The first group did 10 pushups per day for a year and the second group did 50 pushups per day for year.At the end of the year to lifting strength of the first group averaged 147 pounds, while that of the second group averaged 162 pounds.At what average rate did lifting strength increase per daily pushup?

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Your solution:

The answer would be the first group increased at a rate of 14.7. I reached this conclusion by taking the average weight and dividing it by the number of pushups. For the second group, the rate of increase was 3.24. I reached this conclusion by using the same process I use for the first group.

confidence rating #$&*:1

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Given Solution:

The second group had 15 pounds more lifting strength as a result of doing 40 more daily pushups than the first.The desired rate is therefore 15 pounds / 40 pushups = .375 pounds / pushup.

You need to make note of anything in the given solution that you didn't understand when you solved the problem.If new ideas have been introduced in the solution, you need to note them.If you notice an error in your own thinking then you need to note that.In your own words, explain anything you didn't already understand and save your response as Notes.

STUDENT COMMENT:

I have a question with respect as to how the question is interpreted. I used the interpretation given in the solution

to question 008 to rephrase the question in 009, but I do not see how this is the correct interpretation of the question as

stated.

INSTRUCTOR RESPONSE:

This exercise is designed to both see what you understand about rates, and to challenge your understanding a bit with concepts that aren't always familiar to students, despite their having completed the necessary prerequisite courses.

The meaning of the rate of change of one quantity with respect to another is of central importance in the application of mathematics. This might well be your first encounter with this particular phrasing, so it might well be unfamiliar to you, but it is important, unambiguous and universal.

You've taken the first step, which is to correctly apply the wordking of the preceding example to the present question.

You'll have ample opportunity in your course to get used to this terminology, and plenty of reinforcement.

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Self-critique (if necessary):

Now that I have seen the answer to this problem, I believe that I interpreted this completely wrong. I think the reason for this is I was trying to relate the wrong facts. Instead of trying to conclude the rate difference between the two groups, I believe that I was trying to find the average rate for each individual group. I did also did not correlate information from both groups, such as how many more pushups the second groups did compared to the first. Nor did I calculate how much more weight the second group lifted than the first.

Self-critiqueRating:

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Question:`q009.In another part of the study, participants all did 30 pushups per day, but one group did pushups with a 10-pound weight on their shoulders while the other used a 30-pound weight.At the end of the study, the first group had an average lifting strength of 171 pounds, while the second had an average lifting strength of 188 pounds.At what average rate did lifting strength increase with respect to the added shoulder weight?

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Your solution:

Considering this problem is similar to the prior one, i came to the conclusion that the method used to solve it would also be similar. The first operation I performed was subtracting the lifting strength of the weaker group from the lifting strength of the stronger group. This gave a strength difference of 17. Then to find the difference in total weight lifted I subtracted the 10 pound weight of group one from the 30 pound weight of group two. This gave 20, so to find out the actual increase of strength you divide 17 by 20 (17/20=8.5) giving an average increase of .85

confidence rating #$&*:

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Given Solution:

The difference in lifting strength was 17 pounds, as a result of a 20 pound difference in added weight.The average rate at which strength increases with respect added weight would therefore be 17 lifting pounds / (20 added pounds) = .85 lifting pounds / added pound.The strength advantage was .85 lifting pounds per pound of added weight, on the average.

You need to make note of anything in the given solution that you didn't understand when you solved the problem.If new ideas have been introduced in the solution, you need to note them.If you notice an error in your own thinking then you need to note that.In your own words, explain anything you didn't already understand and save your response as Notes.

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Self-critique (if necessary):

After seeing the answer to the first problem, I understand how to do this better.

Self-critiqueRating:ok

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Question:`q010.During a race, a runner passes the 100-meter mark 12 seconds after the start and the 200-meter mark 22 seconds after the start.At what average rate was the runner covering distance between those two positions?

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Your solution:

The average speed the runner was running between the 100 meter marker and the 200 meter marker is 10 meters per second. This is solved by taking the time and distance of the first 100 meter marker and subtracting it from the 200 meter marker. We do this because the problem asked for the average speed between the two points and both of these times are measured from the begging of the race. This gives us 100 meters in 10 seconds. Now to find the rate of travel we simply take the distance traveled and divide it by the time taken to travel the distance (100/10=10) therefore the average speed between these two points is 10 meters per second.

confidence rating #$&*:ok

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Given Solution:

The runner traveled 100 meters between the two positions, and required 10 seconds to do so.The average rate at which the runner was covering distance was therefore 100 meters / (10 seconds) = 10 meters / second.Again this is an average rate; at different positions in his stride the runner would clearly be traveling at slightly different speeds.

You need to make note of anything in the given solution that you didn't understand when you solved the problem.If new ideas have been introduced in the solution, you need to note them.If you notice an error in your own thinking then you need to note that.In your own words, explain anything you didn't already understand and save your response as Notes.

STUDENT QUESTION

Is there a formula for this is it d= r*t or distance equal rate times time??????????????????

INSTRUCTOR RESPONSE

That formula would apply in this specific situation.

The goal is to learn to use the general concept of rate of change. The situation of this problem, and the formula you quote, are just one instance of a general concept that applies far beyond the context of distance and time.

It's fine if the formula helps you understand the general concept of rate. Just be sure you work to understand the broader concept.

Note also that we try to avoid using d for the name of a variable. The letter d will come to have a specific meaning in the context of rates, and to use d as the name of a variable invite confusion.

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Self-critique (if necessary):ok

Self-critiqueRating:ok

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Question:`q011.During a race, a runner passes the 100-meter mark moving at 10 meters / second, and the 200-meter mark moving at 9 meters / second.What is your best estimate of how long it takes the runner to cover the intervening 100 meter distance?

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Your solution:

To solve this problem we need to find the average between the two speeds. If the runner was traveling at 10 meters per second for the first 100 meter and the slowed to 9 meters per second for the second 100 meters, we need to find the average. To do this we need to add the speeds (10+9=19) then we divide by the number of values (19/2=9.5). Now that we have the average of the 2 speeds, we need to find the average speed for 100 meters traveled at this distance. (100/9.5=10.5) so the average speed for the next 100 meters would be estimated at 10.5 meters per second.

confidence rating #$&*:ok

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Given Solution:

At 10 meters/sec, the runner would require 10 seconds to travel 100 meters.However the runner seems to be slowing, and will therefore require more than 10 seconds to travel the 100 meters.We don't know what the runner's average speedis,we only know that it goes from 10 m/s to 9 m/s.The simplest estimate we could make would be that the average speed is the average of 10 m/s and 9 m/s, or (10 m/s + 9 m/s )/ 2 = 9.5 m/s.Taking this approximation as the average rate, the time required to travel 100 meters will be (100 meters) / (9.5 m/s) = 10.5 sec,approx..Note that simply averaging the 10 m/s and the 9 m/s might not be the best way to approximate the average rate--for example we if we knew enough about the situation we might expect that this runner would maintain the 10 m/s for most of the remaining 100 meters, and simply tire during the last few seconds.However we were not given this information, and we don't add extraneous assumptions without good cause.So the approximation we used here is pretty close to the best we can do with the given information.

You need to make note of anything in the given solution that you didn't understand when you solved the problem.If new ideas have been introduced in the solution, you need to note them.If you notice an error in your own thinking then you need to note that.In your own words, explain anything you didn't already understand and save your response as Notes.

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Self-critique (if necessary):ok

Self-critiqueRating:ok

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Question:`q012.We just averaged two quantities, adding them and dividing by 2, to find an average rate.We didn't do that before.Why we do it now?

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Your solution:

The reason we did this in this certain problem is because we were presented with more than one value of time. Before we had a set distance and speed, so to find the average speed traveled within that distance, we need only to take the distance traveled and divide it by the time taken to travel that distance. However in the problem above we were presented with a whole value split in to 3 values which we are to find the third. Taking the average speed from either the first or second increment of the race only would not give us an accurate answer for the third, because the time it took to run the 100 meter intervals were not the same. And depending on which interval of the race we took our average from the average speed could be higher or lower, giving us sporadic rates of speed. Therefore it is appropriate to take an average of the time values given to find the speed. For example if a car travels a distance of 300 miles and the first 100 miles was traveled 2 hours and the second 100 miles was traveled in 3 hours and the third 100 mile was traveled in1 hour then we would need to take an average of the time values to discover how fast the car was traveling. In this example the average time is 2 hours, where as if we take the average from just the second or third increment of the distance then we will get higher or lower traveling speed.

confidence rating #$&*:ok

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Given Solution:

In previous examples the quantities weren't rates.We were given the amount of change of some accumulating quantity, and the change in time or in some other quantity on which the first was dependent (e.g., dollars and months, miles and gallons).Here we are given 2 rates, 10 m/s and 9 m/s, in a situation where we need an average rate in order to answer a question.Within this context, averaging the 2 rates was an appropriate tactic.

You need to make note of anything in the given solution that you didn't understand when you solved the problem.If new ideas have been introduced in the solution, you need to note them.If you notice an error in your own thinking then you need to note that.In your own words, explain anything you didn't already understand and save your response as Notes.

STUDENT QUESTION:

I thought the change of an accumulating quantity was the rate?

INSTRUCTOR RESPONSE:

Quick response: The rate is not just the change in the accumulating quantity; if we're talking about a 'time rate' it's the change in the accumulating quantity divided by the time interval (or in calculus the limiting value of this ratio as the time interval approaches zero).

More detailed response: If quantity A changes with respect to quantity B, then the average rate of change of A with respect to B (i.e., change in A / change in B) is 'the rate'. If the B quantity is clock time, then 'the rate' tells you 'how fast' the A quantity accumulates. However the rate is not just the change in the quantity A (i.e., the change in the accumulating quantity), but change in A / change in B.

For students having had at least a semester of calculus at some level: Of course the above generalizes into the definition of the derivative. y ' (x) is the instantaneous rate at which the y quantity changes with respect to x. y ' (x) is the rate at which y accumulates with respect to x.

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Self-critique (if necessary):ok

Self-critiqueRating:ok

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Self-critique (if necessary):

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Self-critique rating:

#*&!#*&!

Rates

#$&*

course Mth 158

documentshort description of contentwhat you'll know when you're doneRatesintroduces the key concept of rate of changethe meaning of average rate of change and how it might be applied

Copy and paste this document into a text editor, insert your responses and submit using theSubmit_Work_Form.

If your solution to stated problem does not match the given solution, you should self-critique per instructions at

http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm

.

Your solution, attempt at solution. If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what you do or do not understand about it. This response should be given, based on the work you did in completing the assignment, before you look at the given solution.

001.Rates

Note that there are 10 questions in this assignment.The questions are of increasing difficulty--the first questions are fairly easy but later questions are very tricky.The main purposes of these exercises are to refine your thinking about rates, and to see how you process challenging information.Most students in most courses would not be expected to answer all these questions correctly; all that's required is that you do your best and follows the recommended procedures for answering and self-critiquing your work.

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Question: If you make $50 in 5 hr, then at what rate are you earning money?

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Your solution:

If you are paid 50 dollars for 5 hours of work then you would receive a pay rate of 10 dollars an hour. For every hour you are paid ten dollars, therefore to receive payment of 50 dollars you would receive 10 dollars 5 times. So the equation would be 50/5=10 or it could be solved by saying 5*10=50.

confidence rating #$&*:ok

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Given Solution:

The rate at which you are earning money is the number of dollars per hour you are earning.You are earning money at the rate of 50 dollars / (5 hours) = 10 dollars / hour.It is very likely that you immediately came up with the $10 / hour becausealmosteveryoneis familiar with the concept of the pay rate, the number of dollars per hour.Note carefully that the pay rate is found by dividing the quantity earned by the time required to earn it.Time rates in general are found by dividing an accumulated quantity by the time required to accumulate it.

You need to make note of anything in the given solution that you didn't understand when you solved the problem.If new ideas have been introduced in the solution, you need to note them.If you notice an error in your own thinking then you need to note that.In your own words, explain anything you didn't already understand and save your response as Notes.

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Self-critique (if necessary):ok

Self-critiqueRating:ok

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Question:`q003.If you make $60,000 per year then how much do you make per month?

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Your solution:

If you make 60,000 a year then you would be making 5000 per month. There are 12 months in a year, therefore 60,000/12=5,000. This means you receive 5,000 dollars 12 times. This is represented by the equation 60,000/12=5,000 or 12*5,000=60,000.

confidence rating #$&*:ok

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Given Solution:

Most people will very quickly see that we need to divide $60,000 by 12 months, giving us 60,000 dollars / (12 months) = 5000 dollars / month.Note that again we have found a time rate, dividing the accumulated quantity by the time required to accumulate it.

You need to make note of anything in the given solution that you didn't understand when you solved the problem.If new ideas have been introduced in the solution, you need to note them.If you notice an error in your own thinking then you need to note that.In your own words, explain anything you didn't already understand and save your response as Notes.

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Self-critique (if necessary):ok

Self-critiqueRating:ok

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Question:`q004.Suppose that the $60,000 is made in a year by a small business.Would be more appropriate to say that the business makes $5000 per month, or that the business makes an average of $5000 per month?

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Your solution:

I would think it more appropriate to consider this an average made per month because the problem and equation in the prior questions seems to neglect the facts that some months are shorter than others limiting the business time per month, also business (or sales made) would differ from month to month. I hardly think that a business would make the exact same in revenue down to the penny amount every month. This equation also does not seem to let us know if the cost of business (what it costs to purchase materials and run machines) is already factored in.

confidence rating #$&*:ok

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Given Solution:

Small businesses do not usually make the same amount of money every month.The amount made depends on the demand for the services or commodities provided by the business, and there are often seasonal fluctuations in addition to other market fluctuations. It is almost certain that a small business making $60,000 per year will make more than $5000 in some months and less than $5000 in others.Therefore it is much more appropriate to say that the business makes and average of $5000 per month.

You need to make note of anything in the given solution that you didn't understand when you solved the problem.If new ideas have been introduced in the solution, you need to note them.If you notice an error in your own thinking then you need to note that.In your own words, explain anything you didn't already understand and save your response as Notes.

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Self-critique (if necessary):ok

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Question:`q005.If you travel 300 miles in 6 hours, at what average rate are you covering distance, and why do we say average rate instead of just plain rate?

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Your solution:

If you travel 300 miles in a time of 6 hours the rate at which you travel is 50 miles per hour. This is solved by taking the overall distance traveled and then dividing that by the amount of time it required to travel such a distance. 300/6=50 meaning for every mile traveled, it took one hour of time. The reason that this is presented in an average is because of factors that are not added in such as traffic that can negatively affect the travel duration. Or another example is wind resistance for airplane travel. Flying into/against the wind would cause the travel time to increase where as traveling away/with the wind will actually shorten the time of travel.

confidence rating #$&*:ok

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Given Solution:

The average rate is 50 miles per hour, or 50 miles / hour.This is obtained by dividing the accumulated quantity, the 300 miles, by the time required to accumulate it, obtainingaverate = 300 miles / ( 6 hours) = 50 miles / hour.Note that the rate at which distance is covered is called speed.The car has an average speed of 50 miles/hour. We say 'average rate' in this case because it is almost certain that slight changes in pressure on the accelerator, traffic conditions and other factors ensure that the speed will sometimes be greater than 50 miles/hour and sometimes less than 50 miles/hour; the 50 miles/hour we obtain from the given information is clearly and overall average of the velocities.

You need to make note of anything in the given solution that you didn't understand when you solved the problem.If new ideas have been introduced in the solution, you need to note them.If you notice an error in your own thinking then you need to note that.In your own words, explain anything you didn't already understand and save your response as Notes.

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Self-critique (if necessary):ok

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Question:`q006.If you use 60 gallons of gasoline on a 1200 mile trip, then at what average rate are you using gasoline, with respect to miles traveled?

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Your solution:

This is much like the problems above however, the end result in which we are trying to conclude is different. Instead of mile per gallon we are calculating usage per mile. What I did was I rephrased the problem to say “if you travel 1200 miles on 60 gallons of gas what is the rate of usage?” there for the problem becomes 1200/60=.05. the rate that gas is being used per mile is .05.

confidence rating #$&*:ok

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Given Solution:

The rate of change of one quantity with respect to another is the change in the first quantity, divided by the change in the second.As in previous examples, we found the rate at which money was made with respect to time by dividing the amount of money made by the time required to make it.

By analogy, the rate at which we use fuel with respect to miles traveled is the change in the amount of fuel divided by the number of miles traveled.In this case we use 60 gallons of fuel in 1200 miles, so the average rate it 60 gal / (1200 miles) = .05 gallons / mile.

Note that this question didn't ask for miles per gallon.Miles per gallon is an appropriate and common calculation, but it measures the rate at which miles are covered with respect to the amount of fuel used.Be sure you see the difference.

Note that in this problem we again have here an example of a rate, but unlike previous instances this rate is not calculated with respect to time.This rate is calculated with respect to the amount of fuel used.We divide the accumulated quantity, in this case miles, by the amount of fuel required to cover t miles.Note that again we call the result of this problem an average rate because there are always at least subtle differences in driving conditions that result in more or fewer miles covered with a certain amount of fuel.

It's very important to understand the phrase 'with respect to'. Whether the calculation makes sense or not, it is defined by the order of the terms.

In this case gallons / miletellsyou how many gallons you are burning, on the average, per mile. This concept is not as familiar as miles / gallon, but except for familiarity it's technically no more difficult.

You need to make note of anything in the given solution that you didn't understand when you solved the problem.If new ideas have been introduced in the solution, you need to note them.If you notice an error in your own thinking then you need to note that.In your own words, explain anything you didn't already understand and save your response as Notes.

STUDENT COMMENT

Very Tricky! I thought I had a rhythm going. I understand where I messed up. I am comfortable with the calculations.

INSTRUCTOR RESPONSE

There's nothing wrong with your rhythm.

As I'm sure you understand, there is no intent here to trick, though I know most people will (and do) tend to give the answer you did.

My intent is to make clear the important point that the definition of the terms is unambiguous and must be read carefully, in the right order.

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Self-critique (if necessary):ok

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Question:`q007.The word 'average' generally connotes something like adding two quantities and dividing by 2, or adding several quantities and dividing by the number of quantities we added.Why is it that we are calculating average rates but we aren't adding anything?

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Your solution:

The reason for using the word average is because there are uncalculated circumstances. For example traffic when traveling by car, air resistance when traveling by airplane. These are neglected and therefore giving us an average instead of a precise amount. For example an air plane travels 600 miles in 6 hours is traveling at a rate of 100 miles per hour. This may be true but this would be an example of an object in motion excluding natural factors, like the force required to overcome gravity and airspeed/direction.

confidence rating #$&*:

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Given Solution:

The word 'average' in the context of the dollars / month, miles / gallon types of questions we have been answering was used because we expect that in different months different amounts were earned, or that over different parts of the trip the gas mileage might have varied, but that if we knew all the individual quantities (e.g., the dollars earned each month, the number of gallons used with each mile) and averaged them in the usual manner, we would get the .05 gallons / mile, or the $5000 / month.In a sense we have already added up all the dollars earned in each month, or the miles traveled on each gallon, and we have obtained the total $60,000 or 1200 miles.Thus when we divide by the number of months or the number of gallons, we are in fact calculating an average rate.

You need to make note of anything in the given solution that you didn't understand when you solved the problem.If new ideas have been introduced in the solution, you need to note them.If you notice an error in your own thinking then you need to note that.In your own words, explain anything you didn't already understand and save your response as Notes.

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Self-critique (if necessary):ok

Self-critiqueRating:ok

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Question:`q008.In a study of how lifting strength is influenced by various ways of training, a study group was divided into 2 subgroups of equally matched individuals.The first group did 10 pushups per day for a year and the second group did 50 pushups per day for year.At the end of the year to lifting strength of the first group averaged 147 pounds, while that of the second group averaged 162 pounds.At what average rate did lifting strength increase per daily pushup?

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Your solution:

The answer would be the first group increased at a rate of 14.7. I reached this conclusion by taking the average weight and dividing it by the number of pushups. For the second group, the rate of increase was 3.24. I reached this conclusion by using the same process I use for the first group.

confidence rating #$&*:1

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Given Solution:

The second group had 15 pounds more lifting strength as a result of doing 40 more daily pushups than the first.The desired rate is therefore 15 pounds / 40 pushups = .375 pounds / pushup.

You need to make note of anything in the given solution that you didn't understand when you solved the problem.If new ideas have been introduced in the solution, you need to note them.If you notice an error in your own thinking then you need to note that.In your own words, explain anything you didn't already understand and save your response as Notes.

STUDENT COMMENT:

I have a question with respect as to how the question is interpreted. I used the interpretation given in the solution

to question 008 to rephrase the question in 009, but I do not see how this is the correct interpretation of the question as

stated.

INSTRUCTOR RESPONSE:

This exercise is designed to both see what you understand about rates, and to challenge your understanding a bit with concepts that aren't always familiar to students, despite their having completed the necessary prerequisite courses.

The meaning of the rate of change of one quantity with respect to another is of central importance in the application of mathematics. This might well be your first encounter with this particular phrasing, so it might well be unfamiliar to you, but it is important, unambiguous and universal.

You've taken the first step, which is to correctly apply the wordking of the preceding example to the present question.

You'll have ample opportunity in your course to get used to this terminology, and plenty of reinforcement.

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Self-critique (if necessary):

Now that I have seen the answer to this problem, I believe that I interpreted this completely wrong. I think the reason for this is I was trying to relate the wrong facts. Instead of trying to conclude the rate difference between the two groups, I believe that I was trying to find the average rate for each individual group. I did also did not correlate information from both groups, such as how many more pushups the second groups did compared to the first. Nor did I calculate how much more weight the second group lifted than the first.

Self-critiqueRating:

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Question:`q009.In another part of the study, participants all did 30 pushups per day, but one group did pushups with a 10-pound weight on their shoulders while the other used a 30-pound weight.At the end of the study, the first group had an average lifting strength of 171 pounds, while the second had an average lifting strength of 188 pounds.At what average rate did lifting strength increase with respect to the added shoulder weight?

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Your solution:

Considering this problem is similar to the prior one, i came to the conclusion that the method used to solve it would also be similar. The first operation I performed was subtracting the lifting strength of the weaker group from the lifting strength of the stronger group. This gave a strength difference of 17. Then to find the difference in total weight lifted I subtracted the 10 pound weight of group one from the 30 pound weight of group two. This gave 20, so to find out the actual increase of strength you divide 17 by 20 (17/20=8.5) giving an average increase of .85

confidence rating #$&*:

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Given Solution:

The difference in lifting strength was 17 pounds, as a result of a 20 pound difference in added weight.The average rate at which strength increases with respect added weight would therefore be 17 lifting pounds / (20 added pounds) = .85 lifting pounds / added pound.The strength advantage was .85 lifting pounds per pound of added weight, on the average.

You need to make note of anything in the given solution that you didn't understand when you solved the problem.If new ideas have been introduced in the solution, you need to note them.If you notice an error in your own thinking then you need to note that.In your own words, explain anything you didn't already understand and save your response as Notes.

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Self-critique (if necessary):

After seeing the answer to the first problem, I understand how to do this better.

Self-critiqueRating:ok

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Question:`q010.During a race, a runner passes the 100-meter mark 12 seconds after the start and the 200-meter mark 22 seconds after the start.At what average rate was the runner covering distance between those two positions?

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Your solution:

The average speed the runner was running between the 100 meter marker and the 200 meter marker is 10 meters per second. This is solved by taking the time and distance of the first 100 meter marker and subtracting it from the 200 meter marker. We do this because the problem asked for the average speed between the two points and both of these times are measured from the begging of the race. This gives us 100 meters in 10 seconds. Now to find the rate of travel we simply take the distance traveled and divide it by the time taken to travel the distance (100/10=10) therefore the average speed between these two points is 10 meters per second.

confidence rating #$&*:ok

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Given Solution:

The runner traveled 100 meters between the two positions, and required 10 seconds to do so.The average rate at which the runner was covering distance was therefore 100 meters / (10 seconds) = 10 meters / second.Again this is an average rate; at different positions in his stride the runner would clearly be traveling at slightly different speeds.

You need to make note of anything in the given solution that you didn't understand when you solved the problem.If new ideas have been introduced in the solution, you need to note them.If you notice an error in your own thinking then you need to note that.In your own words, explain anything you didn't already understand and save your response as Notes.

STUDENT QUESTION

Is there a formula for this is it d= r*t or distance equal rate times time??????????????????

INSTRUCTOR RESPONSE

That formula would apply in this specific situation.

The goal is to learn to use the general concept of rate of change. The situation of this problem, and the formula you quote, are just one instance of a general concept that applies far beyond the context of distance and time.

It's fine if the formula helps you understand the general concept of rate. Just be sure you work to understand the broader concept.

Note also that we try to avoid using d for the name of a variable. The letter d will come to have a specific meaning in the context of rates, and to use d as the name of a variable invite confusion.

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Self-critique (if necessary):ok

Self-critiqueRating:ok

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Question:`q011.During a race, a runner passes the 100-meter mark moving at 10 meters / second, and the 200-meter mark moving at 9 meters / second.What is your best estimate of how long it takes the runner to cover the intervening 100 meter distance?

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Your solution:

To solve this problem we need to find the average between the two speeds. If the runner was traveling at 10 meters per second for the first 100 meter and the slowed to 9 meters per second for the second 100 meters, we need to find the average. To do this we need to add the speeds (10+9=19) then we divide by the number of values (19/2=9.5). Now that we have the average of the 2 speeds, we need to find the average speed for 100 meters traveled at this distance. (100/9.5=10.5) so the average speed for the next 100 meters would be estimated at 10.5 meters per second.

confidence rating #$&*:ok

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Given Solution:

At 10 meters/sec, the runner would require 10 seconds to travel 100 meters.However the runner seems to be slowing, and will therefore require more than 10 seconds to travel the 100 meters.We don't know what the runner's average speedis,we only know that it goes from 10 m/s to 9 m/s.The simplest estimate we could make would be that the average speed is the average of 10 m/s and 9 m/s, or (10 m/s + 9 m/s )/ 2 = 9.5 m/s.Taking this approximation as the average rate, the time required to travel 100 meters will be (100 meters) / (9.5 m/s) = 10.5 sec,approx..Note that simply averaging the 10 m/s and the 9 m/s might not be the best way to approximate the average rate--for example we if we knew enough about the situation we might expect that this runner would maintain the 10 m/s for most of the remaining 100 meters, and simply tire during the last few seconds.However we were not given this information, and we don't add extraneous assumptions without good cause.So the approximation we used here is pretty close to the best we can do with the given information.

You need to make note of anything in the given solution that you didn't understand when you solved the problem.If new ideas have been introduced in the solution, you need to note them.If you notice an error in your own thinking then you need to note that.In your own words, explain anything you didn't already understand and save your response as Notes.

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Self-critique (if necessary):ok

Self-critiqueRating:ok

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Question:`q012.We just averaged two quantities, adding them and dividing by 2, to find an average rate.We didn't do that before.Why we do it now?

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Your solution:

The reason we did this in this certain problem is because we were presented with more than one value of time. Before we had a set distance and speed, so to find the average speed traveled within that distance, we need only to take the distance traveled and divide it by the time taken to travel that distance. However in the problem above we were presented with a whole value split in to 3 values which we are to find the third. Taking the average speed from either the first or second increment of the race only would not give us an accurate answer for the third, because the time it took to run the 100 meter intervals were not the same. And depending on which interval of the race we took our average from the average speed could be higher or lower, giving us sporadic rates of speed. Therefore it is appropriate to take an average of the time values given to find the speed. For example if a car travels a distance of 300 miles and the first 100 miles was traveled 2 hours and the second 100 miles was traveled in 3 hours and the third 100 mile was traveled in1 hour then we would need to take an average of the time values to discover how fast the car was traveling. In this example the average time is 2 hours, where as if we take the average from just the second or third increment of the distance then we will get higher or lower traveling speed.

confidence rating #$&*:ok

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Given Solution:

In previous examples the quantities weren't rates.We were given the amount of change of some accumulating quantity, and the change in time or in some other quantity on which the first was dependent (e.g., dollars and months, miles and gallons).Here we are given 2 rates, 10 m/s and 9 m/s, in a situation where we need an average rate in order to answer a question.Within this context, averaging the 2 rates was an appropriate tactic.

You need to make note of anything in the given solution that you didn't understand when you solved the problem.If new ideas have been introduced in the solution, you need to note them.If you notice an error in your own thinking then you need to note that.In your own words, explain anything you didn't already understand and save your response as Notes.

STUDENT QUESTION:

I thought the change of an accumulating quantity was the rate?

INSTRUCTOR RESPONSE:

Quick response: The rate is not just the change in the accumulating quantity; if we're talking about a 'time rate' it's the change in the accumulating quantity divided by the time interval (or in calculus the limiting value of this ratio as the time interval approaches zero).

More detailed response: If quantity A changes with respect to quantity B, then the average rate of change of A with respect to B (i.e., change in A / change in B) is 'the rate'. If the B quantity is clock time, then 'the rate' tells you 'how fast' the A quantity accumulates. However the rate is not just the change in the quantity A (i.e., the change in the accumulating quantity), but change in A / change in B.

For students having had at least a semester of calculus at some level: Of course the above generalizes into the definition of the derivative. y ' (x) is the instantaneous rate at which the y quantity changes with respect to x. y ' (x) is the rate at which y accumulates with respect to x.

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Self-critique (if necessary):ok

Self-critiqueRating:ok

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&#Good work. Let me know if you have questions. &#