course Mth 271 I completed this assignment until the questions of using grades as comparison to assignments. I couldn't find the data for these problems. Also I couldn't find anything on the illumination problems. Am I missing something? ?????~??{????assignment #002002. `Query 2
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11:28:59 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 --> 1st: (95, 0) 3rd: (60, 20) 5th: (41, 40) confidence assessment: 3
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11:48:26 According to your graph what would be the temperatures at clock times 7, 19 and 31?
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RESPONSE --> Clock Time 7 d=.14t^2 - 5.7t + 118 d= 14 (7)^2 - 5.7 (7) + 118 d= 14 (49) -39.9 +118 d= 686-39.9 + 118 d=764.1 Clock Time 19 d= 14t^2 - 5.7 t + 118 d= 14 (19)^2 -5.7 (19) + 118 D= 14 (361) -108.3 + 118 d= 5054-108.3+118 d= 5063.7 Clock Time 31 d=14t^2 - 5.7t +118 d= 14 (31)^2 - 5.7 (31) + 118 d= 14 (961) -176.7 + 118 d= 13454 - 176.7 +118 d= 13395.3 These clock times doesn't seem to fit the model very closely at all. I arrived at this equation from eliminating c from the original equations, then eliminating b because it was the easiest coefficient. Then plugging a back into the set of two equations to arrive at b, then substituting a and b in the original set to acquire c. confidence assessment: 2
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11:48:31 Continue to the next question **
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RESPONSE --> self critique assessment: 3
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11:49:51 What three points did you use as a basis for your quadratic model (express as ordered pairs)?
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RESPONSE --> I used (10, 75) , (20, 60) , and (30, 49) confidence assessment: 3
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11:51:04 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 now that the answers probably don't make sense because I used 3 points that were close together. I need to choose 3 points that are spread out to have a better chance at fitting the data. self critique assessment: 2
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11:52:14 What is the first equation you got when you substituted into the form of a quadratic?
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RESPONSE --> (10, 75) 100a + 10b + c (20, 60) 400a + 20b + c (30, 49) 700a + 30b+ c confidence assessment: 3
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11:52:37 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 --> Sorry. Ignore the second 2 data points. self critique assessment: 3
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11:53:09 What is the second equation you got when you substituted into the form of a quadratic?
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RESPONSE --> Data point (20,60) 400a + 20b + c confidence assessment: 3
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11:53:16 STUDENT SOLUTION CONTINUED: The equation that I got from my second data point was 400a + 20b + c = 60 **
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RESPONSE --> self critique assessment: 3
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11:53:41 What is the third equation you got when you substituted into the form of a quadratic?
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RESPONSE --> The third data point (30,49) 900a + 30b + c confidence assessment: 3
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11:53:51 STUDENT SOLUTION CONTINUED: The equation that I got from my third data point was 3600a + 60b + c = 30. **
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RESPONSE --> self critique assessment: 3
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11:55:15 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 didn't multiply the equations by anything other than -1 to cross eliminate c from all equations. The first equation I got without c was: 300a + 10b = -15 confidence assessment: 3
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11:55:25 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 --> self critique assessment: 3
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11:56:04 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 --> To get the second equation I simply subtracted the third and second equation giving me the equation: 500a + 10b = -11 confidence assessment: 3
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11:56:12 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 --> self critique assessment: 3
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11:56:50 Which variable did you eliminate from these two equations, and what was its value?
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RESPONSE --> I eliminated b to obtain a. Which the value of a was .14 confidence assessment: 3
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12:02:06 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 --> Not enough detail. The first step I took in eliminating b was multiplying the first equation by 2, and multiplying the second equation by -2. Thus bringing the first equation to: 2 (300a +10b= -15) 600a +20b = 30 The second equation : -2 (500a + 10b =-11) -1000 - 20 b = 22 The b's eliminate and results in -400a= 55 * As a side note, I just realized another mathmatical mistake I made in the previous equations. I had .14t^2, when in fact, this step shows it is actually -.14t^2 self critique assessment: 2
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12:04:33 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 first equation with 2 variable to substitute back in: 300a +10b =-15 300 (-.14) +10b =-15 -420 + 10b = -15 10b = 405 b = 40.5 confidence assessment: 3
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12:04:44 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 --> self critique assessment: 3
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12:08:58 What is the value of c obtained from substituting into one of the original equations?
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RESPONSE --> I substituted a and b back into the first original equation to obtain c. 100a + 10b + c = 75 100(-.14) + 10(40.5) + c = 75 -14 + 405 + c = 75 391 + c = 75 c = -316 confidence assessment: 2
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12:11:10 STUDENT SOLUTION CONTINUED: By substituting both a and b into the original equations, I found that c = 93 **
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RESPONSE --> self critique assessment: 3
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12:11:49 What is the resulting quadratic model?
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RESPONSE --> The resulting quadratic model is Temp = -.14t^2 = 40.5t -316 confidence assessment: 3
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12:12:32 STUDENT SOLUTION CONTINUED: Therefore, the quadratic model that I obtained was y = (.015) x^2 - (1.95)x + 93. **
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RESPONSE --> I kept temp and t, instead of y= - .14x^2 + 40.5x - 316 self critique assessment: 2
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12:16:54 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 --> My 1st, 2nd, and 3rd points are as follows. ( 0, 95) which results in 316, a 221 degree difference, which shows me that I most likely have done a calculation wrong in creating the model. ( 10, 75) results in 75, which does not deviate at all. ( 20, 60) results in 438, a 378 degree difference, which also lends to the thought process that I have made a wrong calculation somewhere. confidence assessment: 3
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12:17:27 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 --> self critique assessment: 3
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12:18:00 What was your average deviation?
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RESPONSE --> My average deviation was 112.125. confidence assessment: 2
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12:18:06 STUDENT SOLUTION CONTINUED: My average deviation was .6 **
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RESPONSE --> self critique assessment: 3
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12:18:33 Is there a pattern to your deviations?
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RESPONSE --> No, I believe this is because of a mistake in calculations. confidence assessment: 1
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12:18:43 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 --> self critique assessment: 2
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12:19:07 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 I have, and yes I understand the process. confidence assessment: 3
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12:19:14 STUDENT SOLUTION CONTINUED: Yes, I do completely understand the process after studying these outlines and explanations. **
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RESPONSE --> self critique assessment: 3
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12:20:16 Have you memorized the steps of the modeling process, and are you gonna remember them forever? Convince me.
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RESPONSE --> I have memorized the steps of the modeling process and will remember them forever. I will remember them forever because they are going to be a basis of this course, and used repeatedly. confidence assessment:
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12:20:48 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 --> I too printed the outline. self critique assessment: 2
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12:21:52 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 --> (7, 34.4) (14, 23.9) (21, 15.8) confidence assessment: 3
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12:24:07 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 --> Sorry, didn't quite understand the question My complete data were taken from the simulated data on the website. (3.5, 40) (7, 34.4) (10.5, 28.5) (14, 23.9) (17.5, 19.1) (21, 15.8) self critique assessment: 2
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12:24:48 What three points on your graph did you use as a basis for your model?
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RESPONSE --> The three points I unsed were (7,34.4) (14, 23.9) (21, 15.8) confidence assessment: 3
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12:24:56 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 --> self critique assessment: 3
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12:26:05 Give the first of your three equations.
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RESPONSE --> The first equation is from point ( 7, 34.4) a (7^2) + b(7) +c = 34.4 49a +7b + c confidence assessment: 3
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12:26:13 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 --> self critique assessment: 3
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12:26:52 Give the second of your three equations.
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RESPONSE --> The second equation comes from point (14, 23.9) 196a + 14b + c = 23.9 confidence assessment: 3
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12:26:55 STUDENT SOLUTION CONTINUED: The point (15.9, 46) gives me the equation 252.81a +15.9b + c = 46 **
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RESPONSE --> self critique assessment: 3
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12:27:34 Give the third of your three equations.
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RESPONSE --> The third equation comes from point (21, 15.8) 441a + 21b + c = 15.8 confidence assessment: 3
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12:27:37 STUDENT SOLUTION CONTINUED: The point (26.5,32) gives me the equation 702.25a + 26.5b + c = 32. **
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RESPONSE --> self critique assessment: 3
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12:29:59 Give the first of the equations you got when you eliminated c.
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RESPONSE --> I eliminated c by first subtracting the second equation from the first equation, which gave me: 147a +7b = -10.5 confidence assessment: 3
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12:30:06 STUDENT SOLUTION CONTINUED: Subtracting the second equation from the third gave me 449.44a + 10.6b = -14. **
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RESPONSE --> self critique assessment:
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12:36:07 Give the second of the equations you got when you eliminated c.
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RESPONSE --> I arrived at the second equation by subtracting the third equation from the second equation giving the equation: -152a +7b = -8.1 confidence assessment: 3
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12:36:13 ** STUDENT SOLUTION CONTINUED: Subtracting the first equation from the third gave me 674.16a + 21.2b = -31.7. **
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RESPONSE --> self critique assessment: 3
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12:44:13 Explain how you solved for one of the variables.
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RESPONSE --> Using the previous 2 equations, I solved for a by multiplying the first equation by 2, and the second equation by -2. 2(147a +7b = -10.5) = 294a + 14b = -21 -2(-152a +7b = -8.1)= 384a - 14b = 16.2 By subtracting these 2 equations, we get -90a = -37.2 a= .41 confidence assessment: 3
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12:44:30 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 --> self critique assessment: 3
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12:46:05 What values did you get for a and b?
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RESPONSE --> The value for a was .41. The value for b was -7.18 confidence assessment: 3
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12:46:11 STUDENT SOLUTION CONTINUED: a = .0165, b = -2 **
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RESPONSE --> self critique assessment: 3
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12:48:13 STUDENT SOLUTION CONTINUED: c = 73.4 **
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RESPONSE --> I got c=64.57. self critique assessment: 2
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12:48:57 What is your function model?
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RESPONSE --> My function model is y= .41x^2 - 7.8x +64.57 confidence assessment: 3
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12:49:04 STUDENT SOLUTION CONTINUED: y = (.0165)x^2 + (-2)x + 73.4. **
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RESPONSE --> self critique assessment: 3
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12:54:23 What is your depth prediction for the given clock time (give clock time also)?
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RESPONSE --> My prediction for the first clock time of 7 was 34.4, the second time of 14 was 44.41 and the third time of 21 was 94.6 confidence assessment: 3
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12:54:34 STUDENT SOLUTION CONTINUED: The given clock time was 46 seconds, and my depth prediction was 16.314 cm.**
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RESPONSE --> self critique assessment: 2
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12:54:56 What clock time corresponds to the given depth (give depth also)?
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RESPONSE --> The clock time that corresponds to the given depth was 7. confidence assessment: 2
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12:55:15 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 --> self critique assessment: 3
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12:55:33 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 --> confidence assessment:
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12:57:22 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 cannot find this assigned problem. If you want me to complete it I will be more than happy to, if you can let me know where it is, I will complete and submit it. self critique assessment: 3
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12:57:41 What three points on your graph did you use as a basis for your model?
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RESPONSE --> See previous note. confidence assessment: 0
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12:57:49 STUDENT SOLUTION CONTINUED: (20, 2.118034) (50, 2.767767) (100, 3.5)**
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RESPONSE --> self critique assessment: 3
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12:58:00 Give the first of your three equations.
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RESPONSE --> See previous note. confidence assessment:
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12:58:15 Give the first of the equations you got when you eliminated c.
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RESPONSE --> confidence assessment:
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12:58:17 STUDENT SOLUTION CONTINUED: 7500a + 50b = .732233. **
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RESPONSE --> self critique assessment:
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12:58:20 Give the second of the equations you got when you eliminated c.
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RESPONSE --> confidence assessment:
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