course Mth 271 Jän÷Š±oÖ¼„wźÖIĪÉŅ“assignment #002
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12:52:46 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 --> (118, 0), (96.23313, 32), (79.50665, 64) confidence assessment: 3
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12:55:14 According to your graph what would be the temperatures at clock times 7, 19 and 31?
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RESPONSE --> At 7, it would be 118. At 19, it would be 105. At 31, it would be 94. confidence assessment: 1
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12:58:04 What three points did you use as a basis for your quadratic model (express as ordered pairs)?
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RESPONSE --> (118, 16), (79.50665, 64), and (66.6534, 96) confidence assessment: 3
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12:58:21 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. self critique assessment: 1
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12:59:48 What is the first equation you got when you substituted into the form of a quadratic?
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RESPONSE --> 256 a + 16b + c = 118. confidence assessment: 2
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13:00:01 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. self critique assessment: 1
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13:01:33 What is the second equation you got when you substituted into the form of a quadratic?
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RESPONSE --> 4096a + 64b + c = 79.50665 confidence assessment: 2
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13:01:43 STUDENT SOLUTION CONTINUED: The equation that I got from my second data point was 400a + 20b + c = 60 **
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RESPONSE --> ok. self critique assessment: 1
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13:03:12 What is the third equation you got when you substituted into the form of a quadratic?
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RESPONSE --> 9216a + 96b + c = 66.6534 confidence assessment: 2
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13:03:22 STUDENT SOLUTION CONTINUED: The equation that I got from my third data point was 3600a + 60b + c = 30. **
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RESPONSE --> ok. self critique assessment: 1
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13:05:52 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 --> I subtracted equation 2 from equation 1 to get 3840a + 48b = 38.49335 confidence assessment: 2
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13:06:05 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 self critique assessment: 1
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13:07:59 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 --> I subtracted the third from the second. 5120a + 32b = 12.85325 confidence assessment: 2
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13:08:09 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 self critique assessment: 1
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13:09:25 Which variable did you eliminate from these two equations, and what was its value?
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RESPONSE --> I eliminated variable a. It was equal to 0.00500350091 confidence assessment: 2
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13:09:34 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 self critique assessment: 1
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13:12:45 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 --> to find b I substituted into the equation 3840(0.0050035091) + 48b = -38.49335 confidence assessment: 3
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13:12:54 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 self critique assessment: 1
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13:14:26 What is the value of c obtained from substituting into one of the original equations?
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RESPONSE --> c = 135.95471 confidence assessment: 3
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13:14:35 STUDENT SOLUTION CONTINUED: By substituting both a and b into the original equations, I found that c = 93 **
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RESPONSE --> ok self critique assessment:
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13:17:38 What is the resulting quadratic model?
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RESPONSE --> it was y = 0.00500350091t^2 + -1.202225524t + 135.95471 confidence assessment: 3
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13:17:50 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 self critique assessment: 1
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13:20:25 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 --> clock times: 135.95, 118, 102.61 deviations: 17.95, 11.5991, 6.37687 confidence assessment: 3
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13:20:35 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 self critique assessment: 1
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13:21:37 What was your average deviation?
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RESPONSE --> 5.25871125 confidence assessment: 3
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13:21:46 STUDENT SOLUTION CONTINUED: My average deviation was .6 **
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RESPONSE --> ok self critique assessment: 1
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13:22:24 Is there a pattern to your deviations?
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RESPONSE --> It appears that the numbers keep decreasing. confidence assessment: 3
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13:22:39 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 self critique assessment: 1
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13:24:42 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 --> I do understand for the most part but I think that I should study them a little more so I will know exactly what I am doing. Since I have so much trouble when it comes to tests, is there a certain way that you Would suggest for me to accurately study this material? confidence assessment: 2
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13:24:51 STUDENT SOLUTION CONTINUED: Yes, I do completely understand the process after studying these outlines and explanations. **
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RESPONSE --> ok self critique assessment: 1
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13:25:21 Have you memorized the steps of the modeling process, and are you gonna remember them forever? Convince me.
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RESPONSE --> I asked about this in the previous question. confidence assessment: 2
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13:25:48 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 self critique assessment: 1
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13:37:06 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 --> (40, 3.5) (34.4, 7) (28.5, 10.5) (23.9, 14) (19.1, 17.5) (15.8, 21) confidence assessment: 3
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13:37:27 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 self critique assessment: 3
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13:38:48 What three points on your graph did you use as a basis for your model?
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RESPONSE --> (34.4, 7) (23.9, 14) (15.8, 21) confidence assessment: 3
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13:39:03 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 --> ok self critique assessment: 3
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13:40:20 Give the first of your three equations.
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RESPONSE --> 7 = 1183.36a + 34.4b + c confidence assessment: 3
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13:40:29 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 --> ok self critique assessment: 3
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13:41:05 Give the second of your three equations.
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RESPONSE --> 14 = 571.21a + 23.9b + c confidence assessment: 3
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13:41:14 STUDENT SOLUTION CONTINUED: The point (15.9, 46) gives me the equation 252.81a +15.9b + c = 46 **
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RESPONSE --> ok self critique assessment: 3
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13:41:55 Give the third of your three equations.
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RESPONSE --> 21 = 249.64a + 15.8b + c confidence assessment: 3
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13:42:02 STUDENT SOLUTION CONTINUED: The point (26.5,32) gives me the equation 702.25a + 26.5b + c = 32. **
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RESPONSE --> ok self critique assessment: 3
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13:43:07 Give the first of the equations you got when you eliminated c.
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RESPONSE --> 7 = -612.15a + -10.5b confidence assessment: 3
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13:43:23 STUDENT SOLUTION CONTINUED: Subtracting the second equation from the third gave me 449.44a + 10.6b = -14. **
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RESPONSE --> 3 self critique assessment: 3
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13:44:15 Give the second of the equations you got when you eliminated c.
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RESPONSE --> 7 = -321.57a + -8.1b confidence assessment: 3
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13:44:22 ** STUDENT SOLUTION CONTINUED: Subtracting the first equation from the third gave me 674.16a + 21.2b = -31.7. **
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RESPONSE --> ok self critique assessment: 3
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13:49:10 Explain how you solved for one of the variables.
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RESPONSE --> 7 = -612.15(0.0106199389) + -10.5b 7 = -6.500995598 + -10.5b 13.5009956 = -10.5b -1.285809105 = b confidence assessment: 3
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13:49:19 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 self critique assessment: 3
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13:51:05 What values did you get for a and b?
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RESPONSE --> a = 0.0106199389 b = -1.285809105 confidence assessment: 3
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13:51:15 STUDENT SOLUTION CONTINUED: a = .0165, b = -2 **
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RESPONSE --> ok self critique assessment: 3
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13:51:37 What did you then get for c?
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RESPONSE --> 38.66462231 confidence assessment: 3
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13:51:47 STUDENT SOLUTION CONTINUED: c = 73.4 **
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RESPONSE --> ok self critique assessment: 3
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13:53:32 What is your function model?
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RESPONSE --> y = 0.010619938t^2 + -1.285809105t + 38.66462231 confidence assessment: 3
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13:53:41 STUDENT SOLUTION CONTINUED: y = (.0165)x^2 + (-2)x + 73.4. **
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RESPONSE --> ok self critique assessment: 3
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13:56:48 What is your depth prediction for the given clock time (give clock time also)?
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RESPONSE --> The given clock time is 46 sec. and predictated depth is 1.9892. confidence assessment: 3
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13:56:56 STUDENT SOLUTION CONTINUED: The given clock time was 46 seconds, and my depth prediction was 16.314 cm.**
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RESPONSE --> ok self critique assessment: 3
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13:57:27 What clock time corresponds to the given depth (give depth also)?
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RESPONSE --> 23.709 seconds confidence assessment: 3
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13:57:37 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 self critique assessment: 3
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