rates

#$&*

course Mth 279

6/8 9:06 pm

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:

 

 $50 in 5 hours is a ratio. 50/5 = 10 dollars per hour

 

 

confidence rating #$&*:3

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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 because almosteveryone is 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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Self-critique Rating: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:

 There are 12 months in a year. In order to find how much money you make per month, the amount per year must be divided by the number of months in a year

(Money/year) * (year/month) = (money/month)

60000/12 = $5000 per month

 

 

confidence rating #$&*:3

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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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Self-critique Rating: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:

 It would be more accurate to say that the business makes $5000 per month. By solving for $5000 per month we are assuming that the business will make the same amount each month.

If a company makes $60,000 in the first month, then absolutely nothing in the next 11 months, they still made $60,000 for the year.

 

 

confidence rating #$&*:3

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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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Self-critique Rating: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:

 total distance/total time = average rate

300/6 = 50 mph

This is the average rate because we do not know how fast the car was going during the 6 hours. It could have driven 200 miles in the first hour, then slowed down and dove the remaining 100 miles in the last 5 hours.

 

 

confidence rating #$&*:3

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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, obtaining ave rate = 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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Self-critique Rating: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:

 gas used/miles traveled = rate

60/1200 = 0.05 gallons per mile

 

 

confidence rating #$&*:3

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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 / mile tells you 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

 

 Almost fell for it too

 

 

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Self-critique Rating: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:

 

 In the above examples we were not given two rates to average. We were given two totals (in two different units) to use to find the average (which is a combination of the two units).

 

 

confidence rating #$&*:2

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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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Self-critique Rating: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:

 Interesting question.

If I read that right, you have to find the average rate of lifting strength per daily pushup for each group, then average those two numbers.

First group

147 pounds/10 pushups = 14.7 pounds/pushup

Second group

162 pounds/50 pushups = 3.24 pounds/pushup

...that doesn't seem right.

I don't think I read the question right but I'm not sure how else to solve the problem

(14.7+3.24)/2 = 8.97 pounds/pushup

 

confidence rating #$&*:0

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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):

 

 Not too sure why you did what you did. It seemed like you graphed lifting strength with respect to pushups, then found the slope of those two points? ???

Why did you do this???

Actually, I think I just don't understand the question. Could you explain a little more of what you're asking maybe???

 

 

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Self-critique Rating:0

@&

The average rate of change of y with respect to x is (change in y) / (change in x). This is the entire basis of the derivative.

The rate of change of one quantity with respect to another is a well-defined mathematical term with a universal meaning.

*@

 

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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:

 Doing the same thing you did in the previous problem (as I understood it at least):

10 pound weight, 171 pounds

30 pound weight, 188 pounds

Difference of 20 pounds of weight, difference of 17 pounds lifted.

20/17 = 1.1765

 

confidence rating #$&*:1

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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):

 This problem is similar to the previous one and I think I understand it less.

Could you perhaps explain this question a little further too???

 

 I do see my division error though. You asked for the rate in terms of shoulder weight and I gave it in terms of weight lifted.  

 

 

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Self-critique Rating:

 

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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:

 total distance/total time = rate

Keeping in mind that the time difference is 10 s not 22 s

100 meters/ 10 seconds = 10 meters per second

 

 

confidence rating #$&*:2

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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

 Tricky problem. I'm glad I'm reading them more closely now.

 

 

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Self-critique Rating: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:

 We have the rate she is going at the beginning of the interval and the rate she is going at the end. We can average those to find how long it takes.

(10+9)/2 = 9.5 meters per second

Meters/(meters/second) = seconds

100/9.5 = 10.5263 seconds

 

 

confidence rating #$&*:2

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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 speed is, 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

 

 

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Self-critique Rating:OK

 

@&

@&

I wouldn't penalize it, but you asked awhile ago about significant figures, and that's too many for the given situation.

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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:

 We were already given the rates. To find the average rate when given two rates, you add them and divide by two. If we were given two values with different units, we would find the rate by dividing them. But this was not the case in the last question.

 

 

confidence rating #$&*:3

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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

 I understood the question and solution better than the instructor response, but I understand why you included it.

 

@&

That response goes a little deeper than most, in response to a student who had demonstrated the potential to understand it.

*@

 

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Self-critique Rating:OK

Your  

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Question: `q013. The volume of water in a container increases from 1400 cm^3 to 1600 cm^3 as the depth of the water in the container changes from 10 cm to 14 cm. At what average rate was the volume changing with respect to depth?

 

Optional question:  What does this rate tell us about the container?

 

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Your solution:

(1400-1600)/(10-14) = 50 cm^3

Not too sure about this one. I think it's similar to the other two above that I didn't quite get.

 

The container is bigger/wider at the top than the bottom? Like an upside down cone?

@&

Your solution is correct.

The container clearly isn't a cylinder. Its shape could indeed be a section of a cone, and the cone would have to be upside down. Good thinking.

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confidence rating #$&*:0

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Question: `q014. An athlete's rate of doing work increases more or less steadily from 340 Joules / second to 420 Joules / second during a 6-minute event. How many Joules of work did she do during this time?

 

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Your solution:

Since I was already given to rates, I average them? Then solve for total joules?

(340+420)/2 = 380 joules per second

Joules per second/60 seconds per minute = total joules

380/60 = 6.3333 joules in a minute* 6 minutes = 38 joules

@&

If you were making 380 dollars / minute, you would make a lot more than 6 1/3 dollars in an hour.

At 380 Joules / second there will be a lot more than 6 1/3 Joules in a minute.

Related to your previous question about fractions vs. decimals, 6 1/3 is preferable here. However when typed out that mixed number doesn't look great, so we could express that number as 19/3 = 6.33 (approx.)

Other that the error I pointed out, your reasoning was correct.

*@

confidence rating #$&*:0

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Self-critique Rating:none of that seemed right. I'm not sure what to do with this problem ???"

&#Your work looks good. See my notes. Let me know if you have any questions. &#