course MTH 271
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12:51:08 Note 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 --> X Y 3 5000 7 5300 12 5500 Slope of first and second point: (5300 - 5000) / (7 - 3) = (300/4) = 75 Slope of second and third point: (5500 - 5300) / (12 - 7) = (200/5) = 40
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13:27:46 `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 graph shows the change in value with respect to the continuation of time.
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13:27:52 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
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13:28:57 `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 --> Not very accurately for a 9-month period, because there could be fluctuations that are unknown between those two points.
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13:29:09 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 --> OK
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13:29:19 `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
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13:29:27 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 --> OK
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ʤ^jKN~㬧q֒} Student Name: assignment #003 003.
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13:32:15 Note that there are four questions in this assignment. `q001. Sketch a graph similar to that you constructed for the stock values, this time for the depth of the water vs. clock time (depths 80, 40, 20 at clock times 10, 40, 90). Your first point, for example, will be (10, 80). Connect these points with straight lines and determine the slopes of the lines.
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RESPONSE --> X Y 10 80 40 40 90 20 Slope 1: (40 - 80) / (40 - 10) = -4/3 Slope 2: (20 - 40) / (90 - 40) = -2/5
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13:35:42 `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 graph descends, and represents the depth change over the course of time.
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13:36:17 The slopes and the rates of change are numerically equal. For example between the second and third points the rise of -20 represents the -20 cm change in depth and the run of 50 represents the 50 seconds required to make this change, so the slope represents the -20 cm / (50 sec) average rate of change over the second time interval. We therefore see that slope represents average rate of change.
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RESPONSE --> Acknowledged.
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13:38:41 `q003. To what extent do you think your graph with three points and straight line segments between them accurately depicts the detailed behavior of the water over the 80-second period of observation? How do you think the actual behavior of the system differs from that of the graph? How do you think the graph of the actual behavior of the system would differ from that of the graph you made?
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RESPONSE --> Somewhat accurately, but will not show the gradual slowing of depth change. The actual behavior would be more of a quadratic form. It would begin with a steady drop, then slow down.
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13:38:52 The straight line segments would indicate a constant rate of change of depth. It is fairly clear that as depth decreases, the rate of change of depth will decrease, so that the rate of change of depth will not be constant. The graph will therefore never be straight, but will be a curve which is decreasing at a decreasing rate.
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RESPONSE --> OK.
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13:39:51 `q004. From the given information, do you think you can accurately infer the detailed behavior of the water depth over the 80-second period? Do you think you can infer the detailed behavior better than you could the values of the stocks? Why or why not?
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RESPONSE --> Not accurately, but it would certainly be much more accurate than the changing stocks. Because the stocks were over a much broader time, as opposed to the 80 seconds of the depth rate experiment.
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13:40:10 It will turn out that three data points will be sufficient to infer the detailed behavior, provided the data are accurate. However you might or might not be aware of that at this point, so you could draw either conclusion. However it should be clear that the behavior of the water depth is much more predictable than the behavior of the stock market. We don't know on a given day whether the market will go up or down, but we do know that if we shoot a hole in the bottom of a full bucket the water level will decrease, and we expect that identical holes in identical buckets should result in the same depth vs. clock time behavior.
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RESPONSE --> OK.
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QUERY PROGRAM Lz觞Vw}xxB}nܻ assignment #002 oϯ⬡_ǧq| Applied Calculus I 01-29-2006
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14:04:38 What were temperature and time for the first, third and fifth data points (express as temp vs clock time ordered pairs)?
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RESPONSE --> (0, 95) (20, 60) (40, 41)
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14:04:41 ** Continue to the next question **
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RESPONSE -->
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14:09:26 According to your graph what would be the temperatures at clock times 7, 19 and 31?
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RESPONSE --> (7, 82.0625) (19, 62.9375) (31, 47.89107)
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14:09:29 ** Continue to the next question **
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RESPONSE -->
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14:14:29 What three points did you use as a basis for your quadratic model (express as ordered pairs)?
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RESPONSE --> (0, 95) (30, 49) (70, 26)
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14:15:24 ** A good choice of points `spreads' the points out rather than using three adjacent points. For example choosing the t = 10, 20, 30 points would not be a good idea here since the resulting model will fit those points perfectly but by the time we get to t = 60 the fit will probably not be good. Using for example t = 10, 30 and 60 would spread the three points out more and the solution would be more likely to fit the data. The solution to this problem by a former student will be outlined in the remaining `answers'. STUDENT SOLUTION (this student probably used a version different from the one you used; this solution is given here for comparison of the steps) For my quadratic model, I used the three points (10, 75) (20, 60) (60, 30). **
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RESPONSE --> OK
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14:16:33 What is the first equation you got when you substituted into the form of a quadratic?
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RESPONSE --> a(0)^2+b(0)+c=95
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14:16:45 ** STUDENT SOLUTION CONTINUED: The equation that I got from the first data point (10,75) was 100a + 10b +c = 75.**
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RESPONSE --> OK
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14:17:21 What is the second equation you got when you substituted into the form of a quadratic?
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RESPONSE --> 900a+30+c=49
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14:17:26 ** STUDENT SOLUTION CONTINUED: The equation that I got from my second data point was 400a + 20b + c = 60 **
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RESPONSE --> OK
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14:18:17 What is the third equation you got when you substituted into the form of a quadratic?
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RESPONSE --> 4900a+70b+c=26
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14:18:21 ** STUDENT SOLUTION CONTINUED: The equation that I got from my third data point was 3600a + 60b + c = 30. **
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RESPONSE --> OK
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14:19:56 What multiple of which equation did you first add to what multiple of which other equation to eliminate c, and what is the first equation you got when you eliminated c?
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RESPONSE --> The second equation multiplied by a -1 added to the first equation to eliminate C, leaving me with -900a-30b=46
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14:20:01 ** STUDENT SOLUTION CONTINUED: First, I subtracted the second equation from the third equation in order to eliminate c. By doing this, I obtained my first new equation 3200a + 40b = -30. **
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RESPONSE --> ok
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14:21:01 To get the second equation what multiple of which equation did you add to what multiple of which other quation, and what is the resulting equation?
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RESPONSE --> The third equation multiplied by a -1 added to the second equation to remove C, leaving me with -4000a-40b=23
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14:21:06 ** STUDENT SOLUTION CONTINUED: This time, I subtracted the first equation from the third equation in order to again eliminate c. I obtained my second new equation: 3500a + 50b = -45**
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RESPONSE --> OK
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14:25:56 Which variable did you eliminate from these two equations, and what was its value?
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RESPONSE --> Then combined the following equations to remove B -900a-30b=46 -4000a-40b=23 Multiplied the top by a negative 40 and the bottom by a positive 30. Giving 36000a + 1200b = -1840 -12000a - 1200b = 690 and combining those -84000a = -1150 or -1150/-8400 = 23/1680 or about 0.1369
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14:26:12 ** STUDENT SOLUTION CONTINUED: In order to solve for a and b, I decided to eliminate b because of its smaller value. In order to do this, I multiplied the first new equation by -5 -5 ( 3200a + 40b = -30) and multiplied the second new equation by 4 4 ( 3500a + 50b = -45) making the values of -200 b and 200 b cancel one another out. The resulting equation is -2000 a = -310. **
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RESPONSE --> OK
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14:33:00 What equation did you get when you substituted this value into one of the 2-variable equations, and what did you get for the other variable?
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RESPONSE --> -900(115/8400)-30b=46 -12.32-30b=46 add subtract -12.32 from both sides. -30b = 58.321 Divide by -30 b = ~ -1.944
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14:33:22 ** STUDENT SOLUTION CONTINUED: After eliminating b, I solved a to equal .015 a = .015 I then substituted this value into the equation 3200 (.015) + 40b = -30 and solved to find that b = -1.95. **
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RESPONSE --> OK
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14:34:43 What is the value of c obtained from substituting into one of the original equations?
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RESPONSE --> (115/8400)(0)^2 + (-1.944)(0) + C = 95 0 + 0 + C = 95 C = 95
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14:35:19 ** STUDENT SOLUTION CONTINUED: By substituting both a and b into the original equations, I found that c = 93 **
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RESPONSE --> OK
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14:35:53 What is the resulting quadratic model?
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RESPONSE --> 0.1369x^2-1.944x+95
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14:36:11 ** STUDENT SOLUTION CONTINUED: Therefore, the quadratic model that I obtained was y = (.015) x^2 - (1.95)x + 93. **
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RESPONSE --> OK
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14:37:16 What did your quadratic model give you for the first, second and third clock times on your table, and what were your deviations for these clock times?
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RESPONSE --> (0, 95) No deviation (10, 77) -2 (20, 61.6) -1.6
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14:37:39 ** STUDENT SOLUTION CONTINUED: This model y = (.015) x^2 - (1.95)x + 93 evaluated for clock times 0, 10 and 20 gave me these numbers: First prediction: 93 Deviation: 2 Then, since I used the next two ordered pairs to make the model, I got back }the exact numbers with no deviation. So. the next two were Fourth prediction: 48 Deviation: 1 Fifth prediction: 39 Deviation: 2. **
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RESPONSE --> OK
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14:37:49 What was your average deviation?
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RESPONSE --> My average deviation was 0.45
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14:37:55 ** STUDENT SOLUTION CONTINUED: My average deviation was .6 **
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RESPONSE --> OK.
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14:38:46 Is there a pattern to your deviations?
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RESPONSE --> They seem to decrease as they near the middle point (30, 49) I chose to use in my equations.
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14:39:00 ** STUDENT SOLUTION CONTINUED: There was no obvious pattern to my deviations. INSTRUCTOR NOTE: Common patterns include deviations that start positive, go negative in the middle then end up positive again at the end, and deviations that do the opposite, going from negative to positive to negative. **
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RESPONSE --> OK.
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14:39:10 Have you studied the steps in the modeling process as presented in Overview, the Flow Model, Summaries of the Modeling Process, and do you completely understand the process?
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RESPONSE --> Yes.
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14:39:14 ** STUDENT SOLUTION CONTINUED: Yes, I do completely understand the process after studying these outlines and explanations. **
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RESPONSE -->
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14:45:50 Have you memorized the steps of the modeling process, and are you gonna remember them forever? Convince me.
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RESPONSE --> 1. Obtain and graph data Orient the data Observe trends in the data Organize data Graph data 2. Obtain a model Analyze Select representative points Obtain equation for each point Solve the system of equations 3. Validate the model Graph the model Anazlye differences between original and model.
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14:46:04 ** STUDENT SOLUTION CONTINUED: Yes, sir, I have memorized the steps of the modeling process at this point. I also printed out an outline of the steps in order to refresh my memory often, so that I will remember them forever!!! INSTRUCTOR COMMENT: OK, I'm convinced. **
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RESPONSE --> OK
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15:27:46 Query Completion of Model first problem: Completion of model from your data.Give your data in the form of depth vs. clock time ordered pairs.
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RESPONSE --> (3.5, 70.6) (7, 67) (10.5, 63.9) (14, 60.5) (17.5, 58.6) (21, 56.9)
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15:27:53 ** STUDENT SOLUTION: Here are my data which are from the simulated data provided on the website under randomized problems. (5.3, 63.7) (10.6. 54.8) (15.9, 46) (21.2, 37.7) (26.5, 32) (31.8, 26.6). **
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RESPONSE --> ok
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15:28:15 What three points on your graph did you use as a basis for your model?
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RESPONSE --> (3.5, 70.6) (10.5, 63.9) (21, 56.9)
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15:28:18 ** STUDENT SOLUTION CONTINUED: As the basis for my graph, I used ( 5.3, 63.7) (15.9, 46) (26.5, 32)**
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RESPONSE -->
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15:28:30 Give the first of your three equations.
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RESPONSE --> 12.25a + 3.5b + c = 70.6
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15:28:35 ** STUDENT SOLUTION CONTINUED: The point (5.3, 63.7) gives me the equation 28.09a + 5.3b + c = 63.7 **
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RESPONSE -->
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15:28:50 Give the second of your three equations.
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RESPONSE --> 110.25a + 10.5b + c = 63.9
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15:28:54 ** STUDENT SOLUTION CONTINUED: The point (15.9, 46) gives me the equation 252.81a +15.9b + c = 46 **
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RESPONSE -->
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15:29:19 Give the third of your three equations.
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RESPONSE --> 441a + 21b + c = 56.9
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15:29:21 ** STUDENT SOLUTION CONTINUED: The point (26.5,32) gives me the equation 702.25a + 26.5b + c = 32. **
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RESPONSE -->
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15:29:57 Give the first of the equations you got when you eliminated c.
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RESPONSE --> Eliminated using equation's 1 and 2 -98a - 7b = 6.7
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15:30:06 ** STUDENT SOLUTION CONTINUED: Subtracting the second equation from the third gave me 449.44a + 10.6b = -14. **
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RESPONSE -->
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15:30:58 Give the second of the equations you got when you eliminated c.
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RESPONSE --> Third equation subtracted by the second. 330.75a + 10.5b = -7
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15:31:00 ** ** STUDENT SOLUTION CONTINUED: Subtracting the first equation from the third gave me 674.16a + 21.2b = -31.7. **
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RESPONSE -->
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15:34:18 Explain how you solved for one of the variables.
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RESPONSE --> I cancelled out b in the following equations: 10.5 * (-98a-7b=6.7) 7 * (330.75+10.5b=-7) Which left me with: 2315.25a + 73.5b = -49 -1029 a - 73.5b = 70.35 Then combined those to get 1286.25a = 21.35 Then my answer for variable A became 61/3675 or c about .0165
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15:34:24 ** STUDENT SOLUTION CONTINUED: In order to solve for a, I eliminated b by multiplying the first equation by 21.2, which was the b value in the second equation. Then, I multiplied the seond equation by -10.6, which was the b value of the first equation, only I made it negative so they would cancel out. **
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RESPONSE --> OK
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15:35:20 What values did you get for a and b?
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RESPONSE --> A = (61 / 3675) or about 0.165 B = -(1249/1050) or about -1.189
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15:35:27 ** STUDENT SOLUTION CONTINUED: a = .0165, b = -2 **
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RESPONSE --> ok
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15:35:37 What did you then get for c?
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RESPONSE --> C = 74.56
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15:35:42 ** STUDENT SOLUTION CONTINUED: c = 73.4 **
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RESPONSE -->
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15:36:20 What is your function model?
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RESPONSE --> 0.165x^2 - 1.189x + 74.56
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15:36:31 ** STUDENT SOLUTION CONTINUED: y = (.0165)x^2 + (-2)x + 73.4. **
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RESPONSE -->
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15:37:32 What is your depth prediction for the given clock time (give clock time also)?
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RESPONSE --> (3.5, 70.6) (7, 67) (10.5, 63.9) (14, 61.1) (17.5, 58.8) (21, 56.9)
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15:51:23 ** STUDENT SOLUTION CONTINUED: The given clock time was 46 seconds, and my depth prediction was 16.314 cm.**
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RESPONSE --> Accidentally thought you wanted the models values for the clocktimes that were given, my answer follows: The clock time given was 46 seconds, but my model begins to curve back up, so therefore my model is probably invalid after approximately 36 or so seconds.
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15:58:52 What clock time corresponds to the given depth (give depth also)?
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RESPONSE --> Given depth is 14 centimeters, but again, my model is not able to calculate that depth, because solving using the quadratic equation will give an imaginary answer and visually inspecting my graph shows that the answer will not be rational.
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15:59:53 ** The specifics will depend on your model and the requested depth. For your model y = (.0165)x^2 + (-2)x + 73.4, if we wanted to find the clock time associated with depth 68 we would note that depth is y, so we would let y be 68 and solve the resulting equation: 68 = .01t^2 - 1.6t + 126 using the quadratic formula. There are two solutions, x = 55.5 and x = 104.5, approximately. **
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RESPONSE --> OK, I understand how to solve all of this, I just couldn't do so with my model, because of the lack of data.
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16:04:33 Completion of Model second problem: grade average Give your data in the form of grade vs. clock time ordered pairs.
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RESPONSE --> (0, 1) (10, 1.790569) (20, 2.118034) (30, 2.369306) (40, 2.581139) (50, 2.767767) (60, 2.936492) (70, 3.09165) (80, 3.236068) (90, 3.371708) (100, 3.5)
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16:04:35 ** STUDENT SOLUTION: Grade vs. percent of assignments reviewed (0, 1) (10, 1.790569) (20, 2.118034) (30, 2.369306) (40, 2.581139) (50, 2.767767) (60, 2.936492) (70, 3.09165) (80, 3.236068) (90, 3.371708) (100, 3.5). **
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RESPONSE -->
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16:06:10 What three points on your graph did you use as a basis for your model?
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RESPONSE --> (10, 1.790569) (60, 2.936492) (100, 3.5)
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16:06:14 ** STUDENT SOLUTION CONTINUED: (20, 2.118034) (50, 2.767767) (100, 3.5)**
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RESPONSE -->
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16:08:47 Give the first of your three equations.
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RESPONSE --> (10, 1.79) 100a + 10b + c = 1.79
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16:08:50 ** STUDENT SOLUTION CONTINUED: 400a + 20b + c = 2.118034**
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RESPONSE -->
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16:09:16 Give the second of your three equations.
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RESPONSE --> (60, 2.936) 3600a + 60b + c = 2.936
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16:09:31 ** STUDENT SOLUTION CONTINUED: 2500a + 50b + c = 2.767767 **
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RESPONSE -->
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16:10:06 Give the third of your three equations.
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RESPONSE --> (100, 3.5) 10000a + 100b + c = 3.5
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16:10:10 ** STUDENT SOLUTION CONTINUED: 10,000a + 100b + c = 3.5 **
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RESPONSE -->
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16:11:20 Give the first of the equations you got when you eliminated c.
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RESPONSE --> Multiplied equation 1 by -1 and added it to equation 2, giving me: 3500a + 50b = 1.1455
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16:11:23 ** STUDENT SOLUTION CONTINUED: 7500a + 50b = .732233. **
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RESPONSE -->
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16:12:46 Give the second of the equations you got when you eliminated c.
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RESPONSE --> Multiplied the second equation by a -1 and added it to the third equation, giving me: 6400a + 40b = .564
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16:12:48 ** STUDENT SOLUTION CONTINUED: Subracting the first equation from the third I go 9600a + 80b = 1.381966 **
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RESPONSE -->
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16:37:57 Explain how you solved for one of the variables.
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RESPONSE --> -40 * (3500a + 50b = 1.1455) 40 * (6400a + 40b = .564) Which gives: -140000a - 2000b = -45.82 320000a + 2000b = 28.2 And adding the two equations together: 180000a = -17.62 a = -0.000097888
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16:38:05 ** STUDENT SOLUTION CONTINUED: In order to solve for a, I eliminated the variable b. In order to do this, I multiplied the first new equation by 80 and the second new equation by -50. **
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RESPONSE --> OK
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16:38:25 What values did you get for a and b?
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RESPONSE --> a = -0.000097 b = 0.0297
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16:38:31 ** STUDENT SOLUTION CONTINUED: a = -.0000876638 b = .01727 **
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RESPONSE -->
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16:38:43 What did you then get for c?
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RESPONSE --> c = 1.5026
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16:38:48 ** STUDENT SOLUTION CONTINUED: c = 1.773. **
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RESPONSE -->
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16:40:51 What is your function model?
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RESPONSE --> -0.000097x^2 + 0.0297x + 1.502
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16:40:56 ** STUDENT ANSWER: y = (0) x^2 + (.01727)x + 1.773 **
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RESPONSE --> ok
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17:11:35 What is your percent-of-review prediction for the given range of grades (give grade range also)?
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RESPONSE --> The range is 3 to 4, however, 4 is not valid because it exceeds the maxima of the graph, and whenever the quadratic equation is applied, an imaginary number is the answer. The percent of review for 3 is 63.683 percent.
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17:11:44 ** The precise solution depends on the model desired average. For example if the model is y = -.00028 x^2 + .06 x + .5 (your model will probably be different from this) and the grade average desired is 3.3 we would find the percent of review x corresponding to grade average y = 3.3 then we have 3.3 = -.00028 x^2 + .06 x + .5. This equation is easily solved using the quadratic formula, remembering to put the equation into the required form a x^2 + b x + c = 0. We get two solutions, x = 69 and x = 146. Our solutions are therefore 69% grade review, which is realistically within the 0 - 100% range, and 146%, which we might reject as being outside the range of possibility. To get a range you would solve two equations, on each for the percent of review for the lower and higher ends of the range. In many models the attempt to solve for a 4.0 average results in an expression which includes the square root of a negative number; this indicates that there is no real solution and that a 4.0 is not possible. **
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RESPONSE --> OK
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17:13:10 What grade average corresponds to the given percent of review (give grade average also)?
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RESPONSE --> The given percentage is 80%, and plugging in values, we get 3.25 as the grade average.
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17:13:13 ** Here you plug in your percent of review for the variable. For example if your model is y = -.00028 x^2 + .06 x + .5 and the percent of review is 75, you plug in 75 for x and evaluate the result. The result gives you the grade average corresponding to the percent of review according to the model. **
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RESPONSE -->
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17:22:10 How well does your model fit the data (support your answer)?
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RESPONSE --> By taking and comparing the model data to the actual data, I figured that the data defers by an average of -.0331, which is rather inaccurate for the size of the graph. However, the data does a job of giving a general idea of what the graph should look like.
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17:22:21 ** You should have evaluated your function at each given percent of review-i.e., at 0, 10, 20, 30, . 100 to get the predicted grade average for each. Comparing your results with the given grade averages shows whether your model fits the data. **
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RESPONSE --> OK
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17:38:27 illumination vs. distance Give your data in the form of illumination vs. distance ordered pairs.
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RESPONSE --> (1, 935.13) (2, 264.44) (3, 105.12) (4, 61.01) (5, 43.06) (6, 25.91) (7, 19.92) (8, 16.27) (9, 11.28) (10, 9.48)
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17:38:30 ** STUDENT SOLUTION: (1, 935.1395) (2, 264..4411) (3, 105.1209) (4, 61.01488) (5, 43.06238) (6, 25.91537) (7, 19.92772) (8, 16.27232) (9, 11.28082) (10, 9.484465)**
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RESPONSE -->
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17:39:45 What three points on your graph did you use as a basis for your model?
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RESPONSE --> (1, 935.13) (6, 25.97) (10, 9.48)
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17:39:47 ** STUDENT SOLUTION CONTINUED: (2, 264.4411) (4, 61.01488) (8, 16.27232) **
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RESPONSE -->
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17:40:07 Give the first of your three equations.
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RESPONSE --> 1a + 1 b + c = 935.13
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17:40:09 ** STUDENT SOLUTION CONTINUED: 4a + 2b + c = 264.4411**
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RESPONSE -->
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17:40:24 Give the second of your three equations.
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RESPONSE --> 36 a + 6b + c = 25.91
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17:40:26 ** STUDENT SOLUTION CONTINUED: 16a + 4b + c = 61.01488**
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RESPONSE -->
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17:40:41 Give the third of your three equations.
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RESPONSE --> 100a + 10b + c = 9.48
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17:40:43 ** STUDENT SOLUTION CONTINUED: 64a + 8b + c = 16.27232**
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RESPONSE -->
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17:41:02 Give the first of the equations you got when you eliminated c.
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RESPONSE --> 35a + 5b = -909.22
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17:41:04 ** STUDENT SOLUTION CONTINUED: 48a + 4b = -44.74256**
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RESPONSE -->
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17:41:16 Give the second of the equations you got when you eliminated c.
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RESPONSE --> -99a + 9b = -925.65
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17:41:20 ** STUDENT SOLUTION CONTINUED: 60a + 6b = -248.16878**
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RESPONSE -->
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17:42:44 Explain how you solved for one of the variables.
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RESPONSE --> Take the two equations and eliminate B -9 * (35 a + 5b = -9009.22) 5 * ( 99a + 9b = -925.65) Gives -315a - 45b = 8182.98 495a + 45b = -4628.25 Adding together: 180a + 0 = 3554.73 Divide by 180: a = 19.7485
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17:42:49 ** STUDENT SOLUTION CONTINUED: I solved for a by eliminating the variable b. I multiplied the first new equation by 4 and the second new equation by -6 **
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RESPONSE -->
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17:43:03 What values did you get for a and b?
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RESPONSE --> a = 19.7485 b = -320.035
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17:43:05 ** STUDENT SOLUTION CONTINUED: a = 15.088, b = -192.24 **
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RESPONSE -->
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17:43:13 What did you then get for c?
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RESPONSE --> c = 1235.4165
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17:43:15 ** STUDENT SOLUTION CONTINUED: c = 588.5691**
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RESPONSE -->
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17:43:46 What is your function model?
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RESPONSE --> 19.7485x^2-320.035x+1235.4165
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17:43:49 ** STUDENT SOLUTION CONTINUED: y = (15.088) x^2 - (192.24)x + 588.5691 **
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RESPONSE -->
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17:53:13 What is your illumination prediction for the given distance (give distance also)?
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RESPONSE --> My illumination prediction is 773.1 for a distance of 1.6 amu
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17:53:37 ** STUDENT SOLUTION CONTINUED: The given distance was 1.6 Earth distances from the sun. My illumination prediction was 319.61 w/m^2, obtained by evaluating my function model for x = 1.6. **
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RESPONSE -->
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18:02:52 What distances correspond to the given illumination range (give illumination range also)?
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RESPONSE --> The illumination range is from 25-100 watts, using the quadratic equation, we find that the values for 25 are 6.0139 and 10.192, since the comet won't get brighter as it moves away, 10.192 is invalid and thrown out and the value for 100 watts is 5.246 and 10.96, and since 10.96 is further away, we throw it out as well. The range is 5.246 - 6.0139 AMU
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18:03:18 ** The precise solution depends on the model and the range of averages. For example if the model is y =9.4 r^2 - 139 r + 500 and the illumination range is 25 to 100 we would find the distance r corresponding to illumination y = 25, then the distance r corresponding to illumination y = 100, by solving the equations 25=9.4 r^2 - 139 r + 500 and 100 =9.4 r^2 - 139 r + 500 Both of these equations are easily solved using the quadratic formula, remembering to put both into the required form a r^2 + b r + c = 0. Both give two solutions, only one solution of each having and correspondence at all with the data. The solutions which correspond to the data are r = 3.9 when y = 100 and r = 5.4 when y = 25. So when the distance x has range 50% - 69% if the illumination range is 25 to 100. Note that a quadratic model does not fit this data well. Sometimes data is quadratic in nature, sometimes it is not. We will see as the course goes on how some situations are accurately modeled by quadratic functions, while others are more accurately modeled by exponential or power functions. **
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RESPONSE --> Acknowledged.
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ͽ۱NfߙoD assignment #002 oϯ⬡_ǧq| Applied Calculus I 01-29-2006
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18:48:29 ppCal1 Section 0.2 EXTRA QUESTION. What is the midpoint between two points What are your points and what is the midpoint? How did you find the midpoint?{}{}What is the midpoint between the points (3, 8) and (7, 12)?
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RESPONSE --> The midpoint is (5,10) (3 + 7) / 2 = 5 (12+8) / 2 = 10
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18:48:45 ** You are given two points. The points each have two coordinates. You have to average the x coordinates to get the x coordinate of the midpoint, then average the y coordinates to get the y coordinate of the midpoint. For example if the points are (3, 8) and (7, 12), the average of the x coordinates is (3 + 7) / 2 = 5 and the average of the y coordinates is (7 + 12) / 2 = 9.5 so the coordinates of the midpoint are (5, 9.5). **
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RESPONSE --> OK
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18:54:36 0.2.22 (was 0.2.14 solve abs(3x+1) >=4
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RESPONSE --> abs(3x+1) >=4 -4 <= 3x+1 3x+1 >= 4 -5 <= 3x 3x >= 3 -5/3 <= x x >= 1 The solution set is [-5/3, 1]
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18:55:04 ** abs(a) >= b translates to a >= b OR a <= -b. In this case abs(3x+1) > 4 gives you 3x + 1 >= 4 OR 3x + 1 <= -4, which on solution for x gives x >= 1 OR x < = -5/3. **
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RESPONSE --> Acknowledged.
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18:55:42 ** the given inequality is equivalent to the two inequalities 3x+1 >= 4 and 3x+1 =< -4. The solution to the first is x >= 1. The solution to the second is x <= -5/3. Thus the solution is x >= 1 OR x <= -5/3. COMMON ERROR: -5/3 > x > 1 INSTRUCTOR COMMENT: It isn't possible for -5/3 to be greater than a quantity and to have that same quantity > 1. Had the inequality read |3x+1|<4 you could have translated it to -4 < 3x+1 <4, but you can't reverse these inequalities without getting the contradiction pointed out here. **
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RESPONSE --> OK
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18:59:10 0.2.24 (was 0.2.16 solve abs(2x+1)<5. What inequality or inequalities did you get from the given inequality, and are these 'and' or 'or' inequalities? Give your solution.
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RESPONSE --> The inequalities I got were: -5 < 2x+1 <5 They are AND inequalities. -5 < 2x+1 2x+1 < 5 -6 < 2x 2x < 4 -3 < x x < 2
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18:59:21 ** abs(a) < b means a < b AND -b < -a so from the given inequality abs(2x+1) < 5 you get -5 < 2x+1 AND 2x+1 < 5. These can be combined into the form -5 < 2x+1 < 5 and solved to get your subsequent result. Subtracting 1 from all expressions gives us -6 < 2x < 4, then dividing through by 2 we get -3 < x < 2. **
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RESPONSE --> Acknowledged.
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19:00:07 0.2.5 (was 0.2.23 describe [-2,2 ]
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RESPONSE --> abs(x)<= 2
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19:00:15 ** The interval [-2, 2] is centered at the midpoint between x=-2 and x=2. You can calculate this midpoint as (-2 + 2) / 2 = 0. It is also clear from a graph of the interval that it is centered at x = 0 The center is at 0. The distance to each endpoint is 2. The interval is | x - center | < distance to endpoints. So the interval here is | x - 0 | < 2, or just | x | < 2. **
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RESPONSE --> Acknowledged.
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19:00:42 0.2.10 (was 0.2.28 describe [-7,-1]
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RESPONSE --> abs(x-4) < 3
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19:00:45 ** the interval is centered at -4 (midpt between -7 and -1). The distance from the center of the interval to -7 is 3, and the distance from the center of the interval to -1 is 3. This translates to the inequality | x - (-4) | < 3, which simplifies to give us | x + 4 | < 3. **
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RESPONSE -->
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19:01:16 0.2.12 (was 0.2.30) describe (-infinity, 20) U (24, infinity)
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RESPONSE --> abs(x - 22) > 2
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19:04:40 0.2.42 (was 0.2 #36 collies, interval abs( (w-57.5)/7.5 ) < 1
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RESPONSE --> First, multiply by 7.5 w - 57.5 <= 7.5 w-57.5 >= -7.5 add 57.5 to both sides w <= 65 w >= 50 50 <= w <= 65
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19:04:47 ** The inequality is translated as -1<=(w-57.5)/7.5<=1. Multiplying through by 7.5 we get -7.5<=w-57.5<=7.5 Now add 57.5 to all expressions to get -7.5 + 57.5 <= x <= 7.5 + 57.5 or 50 < x < 65, which tells you that the dogs weigh between 50 and 65 pounds. **
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RESPONSE --> Acknowledged.
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19:05:57 0.2.40 (was 0.2.38 stocks vary from 33 1/8 by no more than 2. What absolute value inequality or inequalities correspond(s) to this prediction?
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RESPONSE --> The solution is abs(p - 33.125) <= 2
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19:06:06 ** this statement says that the 'distance' between a stock price and 33 1/8 must not be more than 2, so this distance is <= 2 The distance between a price p and 33 1/8 is | p - 33 1/8 |. The desired inequality is therefore | p = 33 1/8 | < = 2. **
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RESPONSE --> OK
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yD⯘ assignment #003 oϯ⬡_ǧq| Applied Calculus I 01-29-2006
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19:08:48 0.3.24 (was 0.3.24 simplify z^-3 (3z^4)
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RESPONSE --> 3z^-3z^4 = 3z
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19:08:58 ** z^-3 ( 3 z^4) = 3 * z^-3 * z^4 = 3 * z^(4-3) = 3 z. **
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RESPONSE --> Acknowledged.
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19:13:48 0.3.30 (was 0.3.30 simplify(12 s^2 / (9s) ) ^ 3
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RESPONSE --> (12s^2/(9s)) ^ 3 (12^3*s^5)/(9^3*s^3) (1728 s ^(6 - 3 ))/729 (1728 s^3) / 729 ((64s^3)/27) ^ 1/3 = 4s/3
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19:14:53 ** Starting with (12 s^2 / (9s) ) ^ 3 we simplify inside parentheses to get ( 4 s / 3) ^ 3, which is equal to 4^3 * s^3 / 3^3 = 64 s^3 / 27 It is possible to expand the cube without first simplifying inside, but the subsequent simplification is a little more messy and error-prone; however done correctly it gives the same result. It's best to simplify inside the parentheses first. **
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RESPONSE --> Acknowledged.
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19:19:23 0.3.38 (was 0.3.38 simplify ( (3x^2 y^3)^4) ^ (1/3) and (54 x^7) ^ (1/3)
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RESPONSE --> ((3x^2 y^3)^4) ^ (1/4) = 3x^2abs(y^3) (54x^7)^ (1/3) B. = 3x (2x^4)^ (1/3)
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19:19:32 ** To simplify (54 x^7)^(1/3) you have to find the maximum factor inside the parentheses which a perfect 3d power. First factor 54 into its prime factors: 54 = 2 * 27 = 2 * 3 * 3 * 3 = 2 * 3^3. Now we have (2 * 3^3 * x^7)^(1/3). 3^3 and x^6 are both perfect 3d Powers. So we factor 3^3 * x^6 out of the expression in parentheses to get ( (3^3 * x^6) * 2x ) ^(1/3). This is equal to (3^3 * x^6)^(1/3) * (2x)^(1/3). Simplifying the perfect cube we end up with 3 x^2 ( 2x ) ^ (1/3) For the second expression: The largest cube contained in 54 is 3^3 = 27 and the largest cube contained in x^7 is x^6. Thus you factor out what's left, which is 2x. Factoring 2x out of (54 x^7)^(1/3) gives you 2x ( 27 x^6) so your expression becomes [ 2x ( 27 x^6) ] ^(1/3) = (2x)^(1/3) * [ 27 x^6 ] ^(1/3) = (2x)^(1/3) * [ (27)^(1/3) (x^6)^(1/3) ] = (2x)^(1/3) * 3 x^2, which in more traditional order is 3 x^2 ( 2x)^(1/3). **
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RESPONSE --> Acknowledged.
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19:24:42 0.3.58 (was 0.3.54 factor P(1+r) from expression P(1+r) + P(1+r)^2 + P(1+r)^3 + ...
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RESPONSE --> P(1+r) * (1 + 1^2 + 1^3 .... + 1^n)
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19:25:47 ** Few students get this one. If you didn't you've got a lot of company; if you did congratulations. It's important to understand how this problem illustrates the essence of factoring. It's important also because expressions of this form occur throughout calculus. Factor out P * (1 + r). Divide each term by P ( 1 + r), and your result is P (1 + r) * your quotient. Your quotient would be 1 + (1+r) + (1+r)^2 + (1+r)^3 + ... . The factored form would therefore be P(1+r) [ 1 + (1+r) + (1+r)^2 + (1+r)^3 + ... ]. You can verify that this is identical to the original expression if you multiply it back out. Analogy with different exponents and only three terms: A x^3 + A x^4 + A x^5 can be divided by A x^2 to give quotient x + x^2 + x^3, so the factored expression is A ( x + x^2 + x^3). **
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RESPONSE --> Acknowledged, I forgot to leave the remaing r variables in.
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