course MTH 173 I didn't understand how the quary was associated with this experiment, but I answered the first questions with the data I had. Then when I got to the second set of questions it pertained to my data, but I didn't reanswer them since I had already did it in the earlier responses. ȋqUDzwsassignment #001
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11:33:36 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 --> I used time vs. depth! ((5.3,63.7),(15.9,46),(26.5,32)
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11:33:38 ** Continue to the next question **
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11:33:47 According to your graph what would be the temperatures at clock times 7, 19 and 31?
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RESPONSE --> 60,40,27.5
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11:33:49 ** Continue to the next question **
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11:34:13 What three points did you use as a basis for your quadratic model (express as ordered pairs)?
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RESPONSE --> (5.3,63.7),(10.6,54.8),(15.9,46)
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11:34:14 ** 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 -->
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11:34:44 What is the first equation you got when you substituted into the form of a quadratic?
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RESPONSE --> 63.7=28.09a+5.3b+c
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11:34:45 ** 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 -->
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11:35:00 What is the second equation you got when you substituted into the form of a quadratic?
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RESPONSE --> 54.8=112.36a+10.6b+c
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11:35:02 ** STUDENT SOLUTION CONTINUED: The equation that I got from my second data point was 400a + 20b + c = 60 **
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RESPONSE -->
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11:35:18 What is the third equation you got when you substituted into the form of a quadratic?
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RESPONSE --> 46=252.81a+15.9b+c
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11:35:20 ** STUDENT SOLUTION CONTINUED: The equation that I got from my third data point was 3600a + 60b + c = 30. **
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RESPONSE -->
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11:36:17 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 --> c was one so I didn't have to add a multiple. -8.9=84.26a+5.3b
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11:36:26 ** 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 -->
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11:36:59 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 --> No multiples -.8.=140.45a+5.3b
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11:37:03 ** 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 -->
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11:37:31 Which variable did you eliminate from these two equations, and what was its value?
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RESPONSE --> c In the end it equaled 72.76
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11:37:42 ** 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 -->
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11:38:31 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 --> -8.8=0.25+5.3b b=-1.71
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11:38:33 ** 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 -->
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11:38:45 What is the value of c obtained from substituting into one of the original equations?
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RESPONSE --> c=72.76
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11:38:49 ** STUDENT SOLUTION CONTINUED: By substituting both a and b into the original equations, I found that c = 93 **
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RESPONSE -->
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11:39:30 What is the resulting quadratic model?
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RESPONSE --> y=1.78E-3t^2+-1.71t+72.76
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11:39:32 ** STUDENT SOLUTION CONTINUED: Therefore, the quadratic model that I obtained was y = (.015) x^2 - (1.95)x + 93. **
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RESPONSE -->
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11:42:39 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 --> 63.75 (+.05),54.83 (+.03),46.02 (+.02)
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11:42:41 ** 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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11:42:51 What was your average deviation?
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RESPONSE --> 0.03
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11:42:54 ** STUDENT SOLUTION CONTINUED: My average deviation was .6 **
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RESPONSE -->
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11:43:03 Is there a pattern to your deviations?
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RESPONSE --> no
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11:43:06 ** 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 -->
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11:43:16 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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11:43:18 ** 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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11:46:37 Have you memorized the steps of the modeling process, and are you gonna remember them forever? Convince me.
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RESPONSE --> I start with the data, then I make 3 qudratic equations with 3 of the data sets, then I subtract the 2-1 and then the 3-2 to get two equations with 2 variables, then I multiply those two equations by a multipler to then subtract them and have only one equation with one variable. Then you solve the one variable and pulg it back into the last equation and solve for the other variable, then you plug both of the variables back into the first equation to get the last variable.
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11:46:48 ** 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 -->
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11:47:06 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 -->
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11:47:31 ** 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 --> (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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11:47:58 What three points on your graph did you use as a basis for your model?
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RESPONSE --> the first three
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11:48:01 ** 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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11:48:25 Give the first of your three equations.
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RESPONSE --> I already did this in the previous questions.
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11:48:26 ** 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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11:48:29 Give the second of your three equations.
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RESPONSE --> I already did this in the previous questions.
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11:48:31 ** 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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11:48:34 Give the third of your three equations.
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RESPONSE --> I already did this in the previous questions.
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11:48:35 ** 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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11:48:39 Give the first of the equations you got when you eliminated c.
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RESPONSE --> I already did this in the previous questions.
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11:48:40 ** 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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11:48:43 Give the second of the equations you got when you eliminated c.
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RESPONSE --> I already did this in the previous questions.
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11:48:44 ** ** 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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11:48:48 Explain how you solved for one of the variables.
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RESPONSE --> I already did this in the previous questions.
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11:48:50 ** 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 -->
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11:48:54 What values did you get for a and b?
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RESPONSE --> I already did this in the previous questions.
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11:48:56 ** STUDENT SOLUTION CONTINUED: a = .0165, b = -2 **
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RESPONSE -->
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11:49:00 What did you then get for c?
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RESPONSE --> I already did this in the previous questions.
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11:49:02 ** STUDENT SOLUTION CONTINUED: c = 73.4 **
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RESPONSE -->
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11:49:07 What is your function model?
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RESPONSE --> I already did this in the previous questions.
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11:49:09 ** STUDENT SOLUTION CONTINUED: y = (.0165)x^2 + (-2)x + 73.4. **
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RESPONSE -->
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11:49:19 What is your depth prediction for the given clock time (give clock time also)?
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RESPONSE --> I already did this in the previous questions.
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11:49:20 ** STUDENT SOLUTION CONTINUED: The given clock time was 46 seconds, and my depth prediction was 16.314 cm.**
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RESPONSE -->
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11:49:26 What clock time corresponds to the given depth (give depth also)?
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RESPONSE --> I already did this in the previous questions.
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11:49:27 ** 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 -->
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