course Mth 163 xԏ徙̉{ɪassignment #002
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13:14:42 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 --> (90,0) (55,20) (40,40) confidence assessment: 2
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13:14:47 ** Continue to the next question **
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RESPONSE --> ok self critique assessment: 3
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13:17:43 According to your graph what would be the temperatures at clock times 7, 19 and 31?
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RESPONSE --> 86.49; 65.61; 47.61. confidence assessment: 2
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13:17:48 ** Continue to the next question **
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RESPONSE --> ok self critique assessment: 3
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13:18:21 What three points did you use as a basis for your quadratic model (express as ordered pairs)?
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RESPONSE --> (10,75) (30,49) (50,35) confidence assessment: 2
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13:19:18 ** 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 --> I see how it is done and I got the same answer. self critique assessment: 3
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13:19:57 What is the first equation you got when you substituted into the form of a quadratic?
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RESPONSE --> 100a+10b+c=75 confidence assessment: 2
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13:20:12 ** 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 --> I see how that is done. I got the same answer. self critique assessment: 3
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13:20:45 What is the second equation you got when you substituted into the form of a quadratic?
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RESPONSE --> 400a + 20b + c = 60 confidence assessment: 2
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13:21:03 ** STUDENT SOLUTION CONTINUED: The equation that I got from my second data point was 400a + 20b + c = 60 **
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RESPONSE --> I see how it is done and I was able to get the sam answer. self critique assessment: 3
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13:21:38 What is the third equation you got when you substituted into the form of a quadratic?
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RESPONSE --> 3600a+60b+c=30 confidence assessment: 2
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13:21:52 ** STUDENT SOLUTION CONTINUED: The equation that I got from my third data point was 3600a + 60b + c = 30. **
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RESPONSE --> I see how it is done. I was able to get the same answer. self critique assessment: 2
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13:22:37 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 --> 3200a+40b=-30 confidence assessment: 2
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13:22:54 ** 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 --> I see how it is done and I was able to get the same answer. self critique assessment: 3
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13:23:27 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 used 3600a+60b+c=30 and subtracted 100a+10b+c=75 from it. I got 3500a+50b=-45 for the new equation. confidence assessment: 2
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13:23:44 ** 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 --> I used 3600a+60b+c=30 and subtracted 100a+10b+c=75 from it. I got 3500a+50b=-45 for the new equation. self critique assessment: 3
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13:24:05 Which variable did you eliminate from these two equations, and what was the value of the variable for which you solved these equations?
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RESPONSE --> I eliminated b from the equation to a=.015. confidence assessment: 2
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13:24:21 ** 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 --> I reworked the problems and I was getting the same thing until I got to the answer. I am not getting the answer -310. I am getting -30/-2,000 which equals 0.015. self critique assessment: 3
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13:24:58 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 --> I used the equation 3200a+40b=-30 and I solved for b and I got b=24,799.25. confidence assessment: 2
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13:25:13 ** 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 --> After realizing that I had the right answer for the last problem but I went too far for what the answer was. I reworked the problem using the answer for a=.015 and then substiuted it in and got b=-1.95. self critique assessment: 3
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13:26:13 What is the value of c obtained from substituting into one of the original equations?
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RESPONSE --> I used the equation 100a+10b+c=75. When I substituted the answers in for a and b I worked it out and got c=93. confidence assessment: 2
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13:26:42 ** STUDENT SOLUTION CONTINUED: By substituting both a and b into the original equations, I found that c = 93 **
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RESPONSE --> I see what I did and got the same answer. self critique assessment: 3
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13:26:55 What is the resulting quadratic model?
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RESPONSE --> The model would be .015t^2-1.95+93. confidence assessment: 2
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13:27:30 ** STUDENT SOLUTION CONTINUED: Therefore, the quadratic model that I obtained was y = (.015) x^2 - (1.95)x + 93. **
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RESPONSE --> I see what is done and I got the same answer. The only thing I didn't do was put in y=. self critique assessment: 3
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13:28:33 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 --> 93 and the diviation is 2 48 and the diviation is 1 39 and the diviation is 2 confidence assessment: 2
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13:28:53 ** 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 --> I see how it is done and I was able to get the same answer. self critique assessment: 3
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13:29:28 What was your average deviation?
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RESPONSE --> 0.6 confidence assessment: 2
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13:29:49 ** STUDENT SOLUTION CONTINUED: My average deviation was .6 **
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RESPONSE --> I see how it is done and I was able to get the same answer. self critique assessment: 3
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13:30:23 Is there a pattern to your deviations?
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RESPONSE --> There is no pattern in the deviations. confidence assessment: 2
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13:30:43 ** 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 --> There is no pattern in the deviations. Some of them are positive and then there are some that are negative. self critique assessment: 3
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13:31:08 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 and I understand all of it. confidence assessment: 3
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13:31:25 ** STUDENT SOLUTION CONTINUED: Yes, I do completely understand the process after studying these outlines and explanations. **
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RESPONSE --> I see how it is. self critique assessment: 3
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13:31:54 Have you memorized the steps of the modeling process, and are you gonna remember them forever? Convince me.
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RESPONSE --> Yes ans I weill remember it for future reference and never forget it. confidence assessment: 3
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13:32:13 ** 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 --> Yes. I will print them out and keep them with me for as long as I can. I will remember them forever! self critique assessment: 3
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13:32:38 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 --> (5.3, 63.7) (10.6. 54.8) (15.9, 46) (21.2, 37.7) (26.5, 32) (31.8, 26.6) confidence assessment: 2
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13:32:54 ** 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 --> I see how it is done and I was able to get the same answer. self critique assessment: 3
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13:33:22 What three points on your graph did you use as a basis for your model?
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RESPONSE --> ( 5.3, 63.7) (15.9, 46) (26.5, 32) confidence assessment: 2
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13:33:37 ** 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 --> I see how it is done and I was able to get the same answer. self critique assessment: 3
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13:34:10 Give the first of your three equations.
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RESPONSE --> 28.09a+5.3b+c=63.7 confidence assessment: 2
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13:34:22 ** 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 --> I see how it is done and I was able to get the same answer. self critique assessment: 3
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13:34:38 Give the second of your three equations.
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RESPONSE --> 252.81a+15.9b+c=46. confidence assessment: 2
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13:34:51 ** STUDENT SOLUTION CONTINUED: The point (15.9, 46) gives me the equation 252.81a +15.9b + c = 46 **
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RESPONSE --> I see how it was done and I got the same answer. self critique assessment: 3
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13:35:31 Give the third of your three equations.
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RESPONSE --> 702.25a+26.5b+c=32. confidence assessment: 2
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13:35:46 ** STUDENT SOLUTION CONTINUED: The point (26.5,32) gives me the equation 702.25a + 26.5b + c = 32. **
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RESPONSE --> I see how it was done and I got the same answer. self critique assessment: 3
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13:36:01 Give the first of the equations you got when you eliminated c.
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RESPONSE --> I took 702.25a+26.5b+c=32 and subtracted 252.81a+15.96b+c=46 from it and I got 449.44a+10.54b=-14. confidence assessment: 2
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13:36:21 ** STUDENT SOLUTION CONTINUED: Subtracting the second equation from the third gave me 449.44a + 10.6b = -14. **
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RESPONSE --> I see how it is done and I got the same answer self critique assessment: 3
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13:36:36 Give the second of the equations you got when you eliminated c.
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RESPONSE --> I got 674.16a+21.2b=-31.7 confidence assessment: 2
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13:36:51 ** STUDENT SOLUTION CONTINUED: Subtracting the first equation from the third gave me 674.16a + 21.2b = -31.7. **
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RESPONSE --> I see how it is done and I got the same answer. self critique assessment: 3
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13:37:15 Explain how you solved for one of the variables.
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RESPONSE --> I multiplied 21.2 by 449.44a+10.54b=-14 and multiplied -10.54 by 674.16a+21.2b=-31.7. I then took he two and subtracted it from each other and arrived at a=.015. confidence assessment: 2
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13:37:36 ** 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 --> I see how it was done and did it the same except I did not round the second number off. self critique assessment: 3
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13:38:01 What values did you get for a and b?
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RESPONSE --> a = .0165, b = -2 confidence assessment: 2
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13:38:18 ** STUDENT SOLUTION CONTINUED: a = .0165, b = -2 **
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RESPONSE --> I see how it is done and I was able to get the same answer. self critique assessment: 3
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13:38:34 What did you then get for c?
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RESPONSE --> 73.4 confidence assessment: 2
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13:38:52 ** STUDENT SOLUTION CONTINUED: c = 73.4 **
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RESPONSE --> I see how it is done adn I was able to get the same answer. self critique assessment: 3
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13:39:27 What is your function model?
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RESPONSE --> 0.0165t^2-2t+73.4 confidence assessment: 2
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13:39:41 ** STUDENT SOLUTION CONTINUED: y = (.0165)x^2 + (-2)x + 73.4. **
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RESPONSE --> I got the same answer and I see how it was done. I just forgot to put the y= in. self critique assessment: 3
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13:41:18 What is your depth prediction for the given clock time (give clock time also)?
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RESPONSE --> The clock time is46 seconds and the depth is 16.314cm. confidence assessment: 2
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13:41:40 ** STUDENT SOLUTION CONTINUED: The given clock time was 46 seconds, and my depth prediction was 16.314 cm.**
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RESPONSE --> I see how it is done and I was able to get the sam answer. self critique assessment: 3
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13:43:41 What clock time corresponds to the given depth (give depth also)?
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RESPONSE --> Use the equation 68 = .01t^2 - 1.6t + 126 to get x=55.5 and x=104. confidence assessment: 2
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13:43:57 ** INSTRUCTOR COMMENT: The exercise should have specified a depth. 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 --> I see how it is done and I was able to get the same answer. self critique assessment: 3
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13:49:44 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 --> (1,0) (1.8,10) (2.12,20) (2.37,30) (2.58,40) (2.77,50) (2.94,60) (3.09,70) (3.24,80) (3.37,90) (3.5,100) confidence assessment: 2
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13:50:33 ** 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 --> I see how it is done. I realize that I rounded the numbers and I shouldn't have. self critique assessment: 3
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13:53:35 What three points on your graph did you use as a basis for your model?
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RESPONSE --> (10, 1.790569) (20, 2.118034) (60, 2.936492) confidence assessment: 2
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13:54:32 ** STUDENT SOLUTION CONTINUED: (20, 2.118034) (50, 2.767767) (100, 3.5)**
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RESPONSE --> I see where I made my mistake and should have used more widely spaced numbers. self critique assessment: 3
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13:56:59 Give the first of your three equations.
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RESPONSE --> 400a+20b+c=2.118034 confidence assessment: 2
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13:57:16 ** STUDENT SOLUTION CONTINUED: 400a + 20b + c = 2.118034**
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RESPONSE --> I see how it is done adn I got the same answer. self critique assessment: 3
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13:58:38 Give the second of your three equations.
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RESPONSE --> 2500a+50b+c=2.767767 confidence assessment: 2
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13:58:50 ** STUDENT SOLUTION CONTINUED: 2500a + 50b + c = 2.767767 **
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RESPONSE --> I see how it is done and I got the same answer. self critique assessment: 3
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13:59:43 Give the third of your three equations.
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RESPONSE --> 10,000a+100b+c=3.5 confidence assessment: 2
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14:00:00 ** STUDENT SOLUTION CONTINUED: 10,000a + 100b + c = 3.5 **
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RESPONSE --> I see how it is done and I got the same answer. self critique assessment: 3
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14:02:00 Give the first of the equations you got when you eliminated c.
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RESPONSE --> 9,600a+80b=1.381966 confidence assessment: 2
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14:04:40 ** STUDENT SOLUTION CONTINUED: 7500a + 50b = .732233. **
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RESPONSE --> I see how it is done. I used the wrong equation for the second one and when I went back and reworked the problems I was able to get the same answer. self critique assessment: 3
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14:05:03 Give the second of the equations you got when you eliminated c.
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RESPONSE --> 9,600a+80b=1.381966 confidence assessment: 2
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14:05:18 ** STUDENT SOLUTION CONTINUED: Subracting the first equation from the third I go 9600a + 80b = 1.381966 **
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RESPONSE --> I see how it is done and I got the same answer. self critique assessment: 3
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14:11:09 Explain how you solved for one of the variables.
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RESPONSE --> I took the equation 9600a+80b=1.381966 and multiplied it by 50 and I took the equation 7500a+50b=.732233 and multiplied it by -80. I then solved for a and got -8.77. confidence assessment: 2
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14:11:29 ** 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 --> I see how this is done and I did it the same way. self critique assessment: 3
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14:14:11 What values did you get for a and b?
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RESPONSE --> a=-8.77 b=1,052.42 confidence assessment: 2
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14:15:56 ** STUDENT SOLUTION CONTINUED: a = -.0000876638 b = .01727 **
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RESPONSE --> I see what I did and I went back and reworked the problem and I was able to come up with the same answers. self critique assessment: 3
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14:19:00 What did you then get for c?
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RESPONSE --> c=1.73756848 confidence assessment: 2
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14:19:35 ** STUDENT SOLUTION CONTINUED: c = 1.773. **
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RESPONSE --> I see that I rounded the number wrong. I rounded it again and then I was able to get the same answer. self critique assessment: 2
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14:20:59 What is your function model?
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RESPONSE --> .0000876638t^2+.01727t+1.773 confidence assessment: 2
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14:21:46 ** y = -.0000876638 x^2 + (.01727)x + 1.773 **
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RESPONSE --> I see how it is done and I didn't put in y= and I used the wrong variable. I should have used x instead of t. self critique assessment: 3
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14:28:51 What grade average corresponds to the given percent of review (give grade average also)?
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RESPONSE --> 69% confidence assessment: 2
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14:31:30 ** 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 --> I went back and reworked the problem and I was able to get the correct answer. self critique assessment: 3
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14:35:11 illumination vs. distance Give your data in the form of illumination vs. distance ordered pairs.
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RESPONSE --> (935.1395,1) (264,4411,2) (105.1209,3) (61.01488,4) (43,06238,5) (25.91537,6) (19.92772,7) (16.27232,8) (11.28082,9) (9.484465,10) confidence assessment: 2
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14:35:31 ** 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 --> I see how it id done and I was able to get the same answers. self critique assessment: 3
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14:37:08 What three points on your graph did you use as a basis for your model?
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RESPONSE --> (2,264.4411) (5,43.06238) (10,9.484465) confidence assessment: 2
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14:38:19 ** STUDENT SOLUTION CONTINUED: (2, 264.4411) (4, 61.01488) (8, 16.27232) **
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RESPONSE --> I see wat I have done and I should have choosen different numbers that what I did. self critique assessment: 3
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14:41:12 Give the first of your three equations.
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RESPONSE --> 4a+2b+c=264.4411 confidence assessment: 2
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14:41:24 ** STUDENT SOLUTION CONTINUED: 4a + 2b + c = 264.4411**
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RESPONSE --> I see how it id done adn I got the same answer. self critique assessment: 3
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14:41:48 Give the second of your three equations.
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RESPONSE --> 16a+4b+c=61.01488 confidence assessment: 2
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14:42:02 ** STUDENT SOLUTION CONTINUED: 16a + 4b + c = 61.01488**
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RESPONSE --> I see how it is done and I got the same answer. self critique assessment: 3
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14:42:19 Give the third of your three equations.
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RESPONSE --> 64a+8b+c=16.27232 confidence assessment: 2
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14:42:30 ** STUDENT SOLUTION CONTINUED: 64a + 8b + c = 16.27232**
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RESPONSE --> I see how it is done and I got the same answer. self critique assessment: 3
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14:44:30 Give the first of the equations you got when you eliminated c.
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RESPONSE --> 48a+4b=-44.74256 confidence assessment: 2
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14:44:50 ** STUDENT SOLUTION CONTINUED: 48a + 4b = -44.74256**
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RESPONSE --> I see how it is done and I was able to get the same answer. self critique assessment: 3
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14:46:33 Give the second of the equations you got when you eliminated c.
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RESPONSE --> 60a+6b=-248.16878 confidence assessment: 2
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14:46:49 ** STUDENT SOLUTION CONTINUED: 60a + 6b = -248.16878**
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RESPONSE --> I see how it is done and I got the same answer. self critique assessment: 3
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14:50:07 Explain how you solved for one of the variables.
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RESPONSE --> I multiplied the first equation by 4 and the second equation by -6 and solve for a. confidence assessment: 2
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14:50:22 ** 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 --> I see how it is done and I was able toget the same answer. self critique assessment: 3
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14:52:55 What values did you get for a and b?
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RESPONSE --> a=.3429010833 b=-75.65217167 confidence assessment: 2
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14:54:03 ** STUDENT SOLUTION CONTINUED: a = 15.088, b = -192.24 **
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RESPONSE --> I did the problem wrong. I went back and reworked the problems and found my mistake and fixed it. I was able to get the same answers. self critique assessment: 3
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14:56:04 What did you then get for c?
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RESPONSE --> c=588.5691 confidence assessment: 2
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14:56:24 ** STUDENT SOLUTION CONTINUED: c = 588.5691**
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RESPONSE --> I see how it was done and I was able to get the same answer. self critique assessment: 3
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14:57:33 What is your function model?
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RESPONSE --> y=15.088x^2-192.24x+588.5691 confidence assessment: 2
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14:58:00 ** STUDENT SOLUTION CONTINUED: y = (15.088) x^2 - (192.24)x + 588.5691 **
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RESPONSE --> I see how it is done and I was able to get the same answer. self critique assessment: 3
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14:58:54 What is your illumination prediction for the given distance (give distance also)?
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RESPONSE --> 1.6 confidence assessment: 2
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14:59:45 ** 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 --> I see how it is done. I also used the distance of the Earth from the sun which is 1.6. self critique assessment: 3
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15:01:12 What distances correspond to the given illumination range (give illumination range also)?
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RESPONSE --> 61 confidence assessment: 2
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15:02:11 ** 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 3.9 - 5.4 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 --> I see how it is done. I realize that I didn't read the questions right. I had between 25 and 100 and for some reason I did not put that as my answer. self critique assessment: 3
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