course Mth 173 dÛžð©óã½Çc§E£Xô±ë¤…ø|å½Ñassignment #001
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15:02:47 `qNote that there are four questions in this assignment. `q001. If your stocks are worth $5000 in mid-March, $5300 in mid-July and $5500 in mid-December, during which period, March-July or July-December was your money growing faster?
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RESPONSE --> sm = 5000 sj = 5300 sd = 5500 From March-July, my money grew $300 dollars in 4 months. An average of $75/month. From March-December, my money grew $200 in 5 months. An average of $40/month. My money grew faster from March-July. confidence assessment: 3
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15:03:36 `q002. What were the precise average rates of change during these two periods?
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RESPONSE --> March-July = $75/month July-December = $40/month confidence assessment: 3
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15:03:44 From mid-March thru mid-July is 4 months. A change of $300 in four months gives an average rate of change of $300 / (4 months) = $75/month. From mid-July through mid-December is 5 months, during which the value changes by $200, giving an average rate of $200 / (5 months) = $40 / month. Thus the rate was greater during the first period than during the second.
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RESPONSE --> ok self critique assessment: 3
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15:08:57 `q003. If the water in a uniform cylinder is leaking from a hole in the bottom, and if the water depths at clock times t = 10, 40 and 90 seconds are respectively 80 cm, 40 cm and 20 cm, is the depth of the water changing more quickly or less quickly, on the average, between t = 10 sec and t = 40 sec, or between t = 40 sec and t = 90 sec?
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RESPONSE --> The water is changing more quickly between t = 10 and t = 40. 4cm/3s (aprox. 1.333cm/sec) The water is changing less quickly between t = 40 and t = 90. 2cm/5s (.4cm/sec) confidence assessment: 3
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15:10:22 Between clock times t = 10 sec and t = 40 sec the depth changes by -40 cm, from 80 cm to 40 cm, in 30 seconds. The average rate is therefore -40 cm / (30 sec) = -1.33 cm/s. Between t = 40 sec and t = 90 sec the change is -20 cm and the time interval is 50 sec so the average rate of change is -20 cm/ 50 sec = -.4 cm/s, approx. The depth is changing more quickly during the first time interval.
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RESPONSE --> My mistake was, instead, of subtracting 80 from 40, I subtracted 40 from 80. And the same for the other set of numbers. self critique assessment: 2
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15:12:11 `q004. How are the two preceding questions actually different versions of the same question? How it is the mathematical reasoning the same in both cases?
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RESPONSE --> In both cases, I had to find the difference in averages of two situations. The first was the average of money versus time, the second was depth versus time. confidence assessment: 2
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15:12:22 In each case we were given a quantity that changed with time. We were given the quantities at three different clock times and we were asked to compare the rates of change over the corresponding intervals. We did this by determining the changes in the quantities and the changes in the clock times, and for each interval dividing change in quantity by change in clock time. We could symbolize this process by representing the change in clock time by `dt and the change in the quantity by `dQ. The rate is thus rate = `dQ /`dt.
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RESPONSE --> ok self critique assessment:
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22:41:24 `qNote that there are four questions in this assignment. `q001. Recall the stock value problem, where March, July and December values were $5000, $5300 and $5500. Construct a graph of stock value vs. number of month (e.g., 1 for Jan, 2 for Feb, etc.). You will have three points on your graph, one corresponding to the March value, one to the July value, and one to the December value. Stock value will be on the y axis and month number on the x axis. Your first point, for example, will be (3, 5000), corresponding to $5000 in March. Connect your three points with straight lines--i.e., connect the first point to the second and the second to the third. What is the slope of your line between the first and second point, and what is the slope of your line between the second in the third point? Recall that slope is rise / run.
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RESPONSE --> Points being (3, 5000), (7, 5300), and (12, 5500) The slope between the first and second point can be found with the equation (7-3)/(5300-5000) = 4/300 = 1/75 The slope between the second and the third point can be found with the equation (12-7)/(5500-5300) = 5/200 = 1/40 confidence assessment: 3
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22:42:09 The three points on the graph are (3, 5000), (7, 5300) and (12, 5500). The rise between the first point and the second is from 5000 to 5300, or 300, and the run is from 3 to 7, or 4, so the slope is 300 / 4 = 75. Note that the 300 represents $300 and the 4 represents 4 months, so the slope represents $300 / (4 months) = $75 / month, which is the average rate of change during the first time interval. The rise between the second point and the third is from 5300 to 5500, or 200, and the run from 7 to 12 is 5, so the slope is 200 / 5 = 40. This slope represents the $40/month average rate of change during the second time interval. Click on 'Next Picture' to see the graph.
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RESPONSE --> Mental mistake: slope = (delta y)/(delta x) self critique assessment: 2
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22:47:32 `q002. Look at your results for the slopes, and look the results for the average rates of change. What do you notice? In what way then does the graph represent the average rate of change?
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RESPONSE --> The average rates of change and the slopes are the same. The graph represents the average rate of change because it shows how the amount of money changes over a period of time; be it 4 or 5 months. confidence assessment: 3
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22:47:40 We see from this example that the slope of a graph of value vs. clock time represents the rate at which value is changing with respect to clock time.
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RESPONSE --> ok self critique assessment:
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22:51:22 `q003. To what extent do you think your graph, consisting of 3 points with straight line segments between them, accurately depicts the detailed behavior of the stocks over the 9-month period?
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RESPONSE --> I have never really relied 100% on graphs that depict points with y or x values that are spread over a graph. In this case however, I feel that since the price of stocks can change in a matter of days that there may be an increase or decrease in the price of stock during the months not graphed (not graphed being april, may, june, august, september, october, november). If the amount of money not graphed in this function follows the same curve, then of course it is an accurate depiction. confidence assessment: 2
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22:51:33 Stocks can do just about anything from day to day-they can go up or down more in a single day than their net change in a month or even a year. So based on the values several months apart we can't say anything about what happens from day to day or even from month to month. We can only say that on the average, from one time to another, the stocks changed at a certain rate.
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RESPONSE --> self critique assessment:
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22:52:13 `q004. From the given information, do you think you can accurately infer the detailed behavior of the stock values over the nine-month period?
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RESPONSE --> No, stock values change too much. confidence assessment: 3
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22:52:18 Not on a day-to-day basis, and not even on a month-to-month basis. All we can see from the given information is what might be an average trend.
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RESPONSE --> self critique assessment:
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