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`xKX֡ Student Name: assignment #001
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18:41:02 `q001. It will be very important in this course for your instructor to see and understand the process of visualization and reasoning you use when you solve problems. This exercise is designed to give you a first experience with these ideas, and your instructor a first look at your work.
Answer the following questions and explain in commonsense terms why your answer makes sense.......!!!!!!!!...................................
RESPONSE --> ok
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18:44:49 For each question draw a picture to make sense out of the situation, and include a description of the picture.
Samples Sample question and response Question: If a bundle of shingles covers 30 square feet, how many bundles are required to cover a 600 square foot roof? Response: We might draw a picture of a rectangle representing the area, dividing the rectangle into a number of smaller rectangles each representing the area covered by a single bundle. This makes it clear that we are dividing the roof area into 1-bundle areas, and makes it clear why we are going to have to divide. Reasoning this problem out in words, we can say that a single bundle would cover 30 square feet. Two bundles would cover 60 square feet. Three bundles would cover 90 square feet. We could continue in this manner until we reach 600 square feet. However, this would be cumbersome. It is more efficient to use the ideas of multiplication and division. We imagine grouping the 600 square feet into 30 square foot patches. There will be 600 / 30 patches and each will require exactly one bundle. We therefore require 600 / 30 bundles = 20 bundles. {}Your responses might not be as clear as the above, though they might be even more clear. I won't be looking for perfection, though I wouldn't object to it, but for a first effort at visualizing a situation and communicating a reasoning process. This is not something you are used to doing and it might take a few attempts before you can achieve good results, but you will get better every time you try. {}You might be unsure of what to do on a specific question. In such a case specific questions and expressions of confusion are also acceptable responses. Such a response must include your attempts to come up with a picture and reason out an explanation. For example your response might be Sample expression of confusion: I've drawn a picture of a pile of bundles and a roof but I'm not sure how to connect the two. I tried multiplying the number of bundles by the square feet of the roof but I got 18,000, and I know it won't take 18,000 bundles to cover the roof. How do you put the area covered by a bundle together with the roof area to get the number of bundles required? A poor response would be something like 'I don't know how to do #17'. This response reveals nothing of your attempt to understand the question and the situation. Nor does it ask a specific question. Incidentally, you might be tempted to quote rules or formulas about rates and velocities in answering these questions. Don't. This exercise isn't about being able to memorize rules and quote them. It is about expanding your ability to visualize, reason and communicate.......!!!!!!!!...................................
RESPONSE --> 600/30=20 bundles of shingles are require to cover the 600 square foot roof, since each bundle covers 30 square feet
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18:45:22 In your own words briefly summarize the instructions and the intent of this exercise.
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RESPONSE --> Allow the instructor to know that the student is able to visualize and answer the problem
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18:46:45 `q001. If you earn 50 dollars in 5 hours, at what average rate are you earning money, in dollars per hour?
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RESPONSE --> 50 dollars/5 hours=10 dollars per hour in rate divided by time
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18:50:33 If you travel 300 miles in 6 hours, at what average rate are you traveling, in miles per hour?
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RESPONSE --> 300 miles/6 hours= an average of 50 miles per hour, distance/time=rate
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18:51:28 `q002. If a ball rolling down a grooved track travels 40 centimeters in 5 seconds, at what average rate is the ball moving, in centimeters per second?
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RESPONSE --> 40cm/5sec=8cm per second, distance/time=rate
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18:52:11 The preceding three questions illustrate the concept of a rate. In each case, to find the rate we divided the change in some quantity (the number of dollars or the distance, in these examples) by the time required for the change (the number of hours or seconds, in these examples). Explain in your own words what is meant by the idea of a rate.
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RESPONSE --> It is a unit used to establish a time or distance
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18:53:38 `q003. If you are earning money at the average rate of 15 dollars per hour, how much do you earn in 6 hours?
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RESPONSE --> 15 dollarsx6 hours=90 dollars in hourly wage times hours worked=income
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18:54:36 If you are traveling at an average rate of 60 miles per hour, how far do you travel in 9 hours?
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RESPONSE --> 60 milesx9 hours=540 miles ratextime=distance
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18:55:46 `q004. If a ball travels at and average rate of 13 centimeters per second, how far does it travel in 3 seconds?
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RESPONSE --> 13cmx3sec=39cm, ratextime=distance
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18:56:20 In the preceding three exercises you turned the concept of a rate around. You were given the rate and the change in the clock time, and you calculated the change in the quantity. Explain in your own words how this increases your understanding of the concept of a rate.
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RESPONSE --> rate is a function that can be used to find both time and distance
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18:57:19 `q005. How long does it take to earn 100 dollars at an average rate of 4 dollars per hour?
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RESPONSE --> 100dollars/4dollars per hour=25 hours to earn 100dollars at an hourly wage of 4 dollars per hour
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18:58:35 How long does it take to travel 500 miles at an average rate of 25 miles per hour?
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RESPONSE --> 500miles/25miles per hour=20hours is the time required given the distance and rate
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18:59:44 `q006. How long does it take a rolling ball to travel 80 centimeters at an average rate of 16 centimeters per second?
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RESPONSE --> 80cm/16cm=5 seconds to travel 80cm with a rate of 16cm a second
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19:00:36 In the preceding three exercises you again expanded your concept of the idea of a rate. Explain how these problems illustrate the concept of a rate.
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RESPONSE --> gave distance and time in order to find the rate(Inversely)
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Ոxmƍ| Student Name: assignment #002
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19:09:21 `q001. Solve the following problem by describing a picture, and by constructing a graph. Describe your picture and your graph in detail. Do the same on subsequent problems:
1. If your total earnings up to week 6 are 56.2 dollars, and your total earnings up to week 11 are 66.2 dollars, then what are your average earnings per week? Why do we say average earnings per week rather than just earnings per week?......!!!!!!!!...................................
RESPONSE --> Earnings in the first six weeks are acheived at a higher rate than later 5 weeks. The rate of the first six weeks =9.36 dollars per week and the overall 11 weeks are accumulated at a lower rate of 6.01 dollars per week. The average is suggested because the worker may have earned more than the average in one week but less than the average in another
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19:11:05 The most common errors on this problem involve things like dividing 56.2 dollars by 6 weeks and/or dividing 66.2 dollars by 11 weeks. Having made these errors there are various ways is to put the results together to get an incorrect answer.
The reason such calculations are meaningless is that you have no information about the total earnings at week 0. Total earnings might be measured from some time other than week 0 (e.g., fiscal years and calendar years don't usually coincide). So you don't know that it took 6 weeks to earn the 56.2 dollars, or 11 weeks to earn the 66.2 dollars. What you do know is that between week 6 and week 11 earnings increased from 56.2 dollars to 66.2 dollars, which means that you earned 10 dollars in the 5 intervening weeks. That's the only information given; any other assumption is extraneous and could well be wrong. So the conclusion is that 10 dollars were make in 5 weeks, so the average rate is 2 dollars per week.......!!!!!!!!...................................
RESPONSE --> problem uses the interval between the two earnings to determine the average rate since thier is not set basis on which the earnings began
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19:15:04 `q002. Solve the following by describing a picture, and by constructing a graph:
If an automobile is at milepost 421 after having traveled for 8 hours and at milepost 475 after having traveled for 15 hours, then what is its average speed? Why do we say average speed rather than just speed?......!!!!!!!!...................................
RESPONSE --> 15 hr-8hr=7hr and 475-421=54miles So 54miles/7hrs=7.71average rate of miles per hour, because we dont know where the vehicle began before the 8 hours of travel to milepost 421. It is an average speed because the vehcile probably traveled faster than rate, and also slower than the rate to average out
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19:15:55 The common error here is to divide 421 miles by 8 hours, or 475 miles by 15 hours. Both calculations would make the clearly questionable assumption that whoever is traveling started from milepost 0. Most people who drive up and down an Interstate highway don't drive to the beginning before they start.
What is known is that someone traveled 54 miles in 7 hours. That makes the average rate 54 miles / (7 hrs) = 7.7 miles / hr, approximately. Bad for a car; impressive if the traveler is on foot.......!!!!!!!!...................................
RESPONSE --> Enter, as appropriate, an answer to the question, a critique of your answer in response to a given answer, your insights regarding the situation at this point, notes to yourself, or just an OK.ok
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19:18:58 `q003. If the water in a uniform cylinder has depth 199 cm at clock time 7 seconds and depth 228 cm at clock time 16 seconds, then at what average rate is the depth changing? Why do we we say average rate rather than just rate?
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RESPONSE --> 228cm-199cm=29cm and 16sec-7sec=9sec So 29cm/9sec=3.22cm per second Average is because their is not a set rate the drop in depth varies both higher and lower than the average rate
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19:19:17 A graph of depth vs. clock time will have the points (7, 199) and (16, 228). The rise of this graph is from 199 cm to 228 cm, a rise of 29 cm. The run of this graph is from 7 sec to 16 sec, a run of 9 sec. The slope, which is rise / run, is thus 29 cm / (9 sec) = 3.22 cm / sec.
The rise represents the change in depth, the run represents change in clock time. So the slope represents change in depth / change in clock time, or the rate at which depth changes.......!!!!!!!!...................................
RESPONSE --> ok
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19:22:25 `q004. If you are earning money at an average rate of 15 dollars per week, and if at the end of week 9 your total earnings are 133.16 dollars, then what will be your total earnings that the end of week 17?
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RESPONSE --> dont know the earnings for the first 9 weeks should be 135 dollars given the rate of 15 dollars per hour then after 17 weeks the earnings would be $225
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19:24:19 Between week 9 and week 17 there are 17 - 9 = 8 weeks. At 15 dollars per week the earnings would be 15 dollars / week * 8 weeks = 120 dollars. Adding this to the initial 133.16 dollars you end up with total earnings of 253.6 dollars.
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RESPONSE --> The problem uses the intervals betwee the two periods too establish the rate and then adds the amount to the 133.16 that was earned at week 9
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19:29:30 `q005. If an automobile is traveling at the rate of 48 mph, and is at milepost 285 after having traveled for 6 hours, then at what milepost would it be after having traveled for 11 hours?
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RESPONSE --> 11hrs-6hrs=5hrs and 48mphx5hrs=240miles So Milepost 285+240=525milepost
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19:29:40 Between hour 6 and hour 11 is 5 hours. The distance traveled during that time will be 48 miles / hr * 5 hr = 240 miles. This will put the car at 285 miles + 240 miles = 525 miles.
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RESPONSE --> ok
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19:32:32 `q006. If the depth of water in a uniform cylinder is changing at an average rate of 21 cm/second between clock times 9 s and 19 s, and if its depth at clock time 9 seconds is 178 cm, then what is its depth at clock time 19 seconds?
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RESPONSE --> 19sec-9sec=10sec and 10secx21cm/sec=210cm so 210cm+178cm=388cm
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19:34:14 A graph of depth vs. clock time would contain the point (9, 178), representing the 178 cm depth at clock time 9 sec, and would have slope 21, representing the rate at which depth changes with respect to clock time.
The graph point correspoinding to clock time 19 seconds will lie 10 units to the right of the first point. The run between the points will therefore be 10, representing 10 seconds. If the run is 10 and the slope is 21, then since rise / run = slope we have rise / 10 = 21, so the rise is 210, representing the change in the depth. Thus the depth changes by 210 cm.......!!!!!!!!...................................
RESPONSE --> I see, I found the depth change but made an extra step and added it to 178cm that already existed, I realize why the answer is incorrect, I should have known it when the depth did not say whether or not it was rising or falling
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21:05:11 `q007. If your total earnings up to week 4 are 20.91 dollars, and if you earn money at a constant rate of 6 dollars per week, then at what week will your total earnings be 36.91 dollars?
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RESPONSE --> 36.91-20.91=16$ $16/$6=2.66 2.66+4wk=6.66weeks to earn the 36.91 dollars
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21:06:50 Earnings change by 16 dollars, changing at a rate of 6 dollars per week. It therefore takes 16 dollars / (6 dollars / week) = 2.67 weeks for earnings to change from 20.91 dollars to 36.91 dollars.
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RESPONSE --> the answer is 2.67, I added this to the prior 4 weeks of income, that was already established
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21:09:00 8. If an automobile is traveling at the rate of 51 mph, and is at milepost 202 after having traveled for 4 hours, then after how many hours of travel will it have reached milepost 15?
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RESPONSE --> 202-15=187milepost /51mph=3.66hr
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21:09:11 From milepost 202 to milepost 15 is 187 miles. At 51 mph it will take(187 miles / (51 mph) = 3.7 hours to travel this distance.
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RESPONSE --> ok
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21:12:17 9. If the depth of water in a uniform cylinder is changing at an average rate of 12 cm/second as it depth changes from 61.7 cm at clock time 6 seconds to 93.7 cm at an unknown clock time, then what is the unknown clock time?
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RESPONSE --> 93.7cm-61.7cm=32cm depth change divided by the rate is 32cm/12cm/sec=2.66 6sec+2.66sec=8.7sec
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21:13:19 A graph of depth vs. clock time will include the point (6, 61.7), representing the depth of 61.7 cm at clock time 6 sec, and will have a slope of 12, representing the 12 cm/s rate at which depth is changing.
The point at which depth is 93.7 cm will lie 32 units above the first point, representing the 32 cm depth change. Since slope = rise / run we have 12 = 32 / run so run = 32 / 12 = 2.75. Since run is change in clock time, this run represents the 2.75 second required for the depth change.......!!!!!!!!...................................
RESPONSE --> I added the additional 2.75 to the previous 6 seconds for an overall time
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