course MTH 163 Wط Fq^ة؍ʵassignment #002
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01:08:20 What were temperature and time for the first, third and fifth data points (express as temp vs clock time ordered pairs)?
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RESPONSE --> (0 , 95) (20 , 60) (40 , 41) confidence assessment: 3
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01:11:09 According to your graph what would be the temperatures at clock times 7, 19 and 31?
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RESPONSE --> The temperature at clock time 7 would be about 79. The temperature at clock time 19 would be about 63. The temperature at clock time 31 would be about 49. confidence assessment: 3
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01:11:56 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) (40 , 41) (70 , 26) confidence assessment: 3
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01:12:14 ** 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 --> I understand. self critique assessment: 2
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01:12:58 What is the first equation you got when you substituted into the form of a quadratic?
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RESPONSE --> My first quadratic equation was 100a + 10b + c = 75 confidence assessment: 3
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01:13:04 ** 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 understand. self critique assessment: 2
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01:13:31 What is the second equation you got when you substituted into the form of a quadratic?
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RESPONSE --> My second quadratic equation was 1600a + 40b + c = 41 confidence assessment: 3
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01:13:37 ** STUDENT SOLUTION CONTINUED: The equation that I got from my second data point was 400a + 20b + c = 60 **
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RESPONSE --> I understand. self critique assessment: 2
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01:14:03 What is the third equation you got when you substituted into the form of a quadratic?
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RESPONSE --> My third quadratic equation was 4900a + 70b + c = 26 confidence assessment: 3
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01:14:11 ** STUDENT SOLUTION CONTINUED: The equation that I got from my third data point was 3600a + 60b + c = 30. **
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RESPONSE --> I understand. self critique assessment: 2
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01:14:56 What multiple of which equation did you first add to what multiple of which other equation to eliminate c, and what is the first equation you got when you eliminated c?
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RESPONSE --> I first subtracted the first quadratic from the third quadratic, giving me a result of: 4800a + 60b = -49 confidence assessment: 3
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01:15:10 ** 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 understand. self critique assessment: 2
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01:15:45 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 --> Next I subtracted the second quadratic from the third to get: 3300a + 30b = -15 confidence assessment: 3
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01:15:50 ** 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 understand. self critique assessment: 2
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01:18:30 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 chose to eliminate b. I did so by multiplying my second equation by -2: -6600a - 60b = 30 Then subtracting it from the first equation: (4800a + 60b = -49) + (-6600a - 60b = 30) For a result of: -1800a = -19 I then divided a out to reach: a = 0.0106 confidence assessment: 3
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01:18:40 ** 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 understand. self critique assessment: 2
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01:20:07 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 plugged the a value into a previous equation: 3300(0.0106) + 30b = -15 34.833 + 30b = -15 30b = -49.833 b = -1.661 confidence assessment: 3
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01:20:16 ** 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 --> I understand. self critique assessment: 2
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01:21:36 What is the value of c obtained from substituting into one of the original equations?
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RESPONSE --> I then plugged both the a and b values into an original quadratic: 100(0.0106) + 10(-1.661) + c = 75 1.06 - 16.61 + c = 75 -15.55 + c = 75 c = 90.55 confidence assessment: 3
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01:21:45 ** STUDENT SOLUTION CONTINUED: By substituting both a and b into the original equations, I found that c = 93 **
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RESPONSE --> I understand. self critique assessment: 2
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01:22:19 What is the resulting quadratic model?
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RESPONSE --> y = 0.0106(x^2) - 1.661(x) + 90.55 confidence assessment: 3
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01:23:14 ** 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 understand. self critique assessment: 2
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01:26:09 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 --> For the first clock time (0) I recieved 90.55 For the second clock time (10) I recieved 75 For the third clock time (20) I recieved 61.57 The deviation for the first clock time was -4.45 The deviation for the second clock time was 0 The deviation for the third clock time was +1.57 confidence assessment: 3
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01:26:23 ** 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 understand. self critique assessment: 2
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01:27:38 What was your average deviation?
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RESPONSE --> My average deviation was -0.96 confidence assessment: 3
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01:27:46 ** STUDENT SOLUTION CONTINUED: My average deviation was .6 **
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RESPONSE --> I understand. self critique assessment: 2
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01:27:57 Is there a pattern to your deviations?
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RESPONSE --> There doesn't appear to be. confidence assessment: 3
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01:28:06 ** 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 --> I understand. self critique assessment: 2
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01:28:19 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 do. confidence assessment: 3
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01:28:27 ** STUDENT SOLUTION CONTINUED: Yes, I do completely understand the process after studying these outlines and explanations. **
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RESPONSE --> I understand. self critique assessment: 2
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01:28:54 Have you memorized the steps of the modeling process, and are you gonna remember them forever? Convince me.
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RESPONSE --> Yes I have, and yes I'll do my best. First we familiarize ourselves with the problem. Then we establish and set up our data. We make a chart of the data. We sketch a graph, pick three reference points on it, then begin plugging those points into a quadratic formula. We solve for a, b, and c. We take those values and plug them into the original quadratic formula. We plug our reference points into the model, take the results, and compare them to the original data, looking for deviations, and compare them to the graph. We check to make sure the results make good sense. I know that's not how the procedure goes word for word, but that was the gist I could gather off the top of my head at this point. confidence assessment: 3
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01:29:13 ** 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 --> I understand. self critique assessment: 2
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01:31:07 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 --> (5.2 , 80.1) (10.4 , 70.6) (15.6 , 62.2) (20.8 , 55) (26 , 48.9) (31.2 , 44.2) confidence assessment: 3
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01:31:58 ** 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 understand. I too received my data from the randomized problems. self critique assessment: 2
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01:32:27 What three points on your graph did you use as a basis for your model?
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RESPONSE --> (10.4 , 70.6) (20.8 , 55) (31.2 , 44.2) confidence assessment: 3
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01:32:38 ** 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 understand. self critique assessment: 2
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01:33:05 Give the first of your three equations.
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RESPONSE --> 108.16a + 10.4b + c = 70.6 confidence assessment: 3
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01:33:32 ** 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 understand. That equation was for my first point (10.4 , 70.6) self critique assessment: 2
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01:34:08 Give the second of your three equations.
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RESPONSE --> The point (20.8 , 55) gives me: 432.64a + 20.8b + c = 55 confidence assessment: 3
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01:34:18 ** STUDENT SOLUTION CONTINUED: The point (15.9, 46) gives me the equation 252.81a +15.9b + c = 46 **
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RESPONSE --> I understand. self critique assessment: 2
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01:34:49 Give the third of your three equations.
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RESPONSE --> The point (31.2 , 44.2) gives me: 973.44a + 31.2b + c = 44.2 confidence assessment: 3
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01:34:54 ** STUDENT SOLUTION CONTINUED: The point (26.5,32) gives me the equation 702.25a + 26.5b + c = 32. **
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RESPONSE --> I understand. self critique assessment: 2
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01:36:43 Give the first of the equations you got when you eliminated c.
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RESPONSE --> I subtracted the first from the third equation: (973.44a + 31.2b + c = 44.2) + (-108.16a - 10.4b - c = -70.6) To get: 865.28a + 20.8b = -26.4 confidence assessment: 3
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01:36:51 ** STUDENT SOLUTION CONTINUED: Subtracting the second equation from the third gave me 449.44a + 10.6b = -14. **
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RESPONSE --> I understand. self critique assessment: 2
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01:38:08 Give the second of the equations you got when you eliminated c.
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RESPONSE --> I then subtracted the second from the third equation: (973.44a + 31.2b + c = 44.2) + (-432.64a - 20.8b - c = -55) To get: 540.8a + 10.4b = -10.8 confidence assessment: 3
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01:38:13 ** STUDENT SOLUTION CONTINUED: Subtracting the first equation from the third gave me 674.16a + 21.2b = -31.7. **
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RESPONSE --> I understand. self critique assessment: 2
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01:41:21 Explain how you solved for one of the variables.
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RESPONSE --> I chose to eliminate a by multiplying the first equation by 540.8 and the second equation by -865.28 to get: (467943.424a + 11248.64b = -14277.12) + (-467943.424a - 8998.912b = 9345.024) To get: 2249.728b = -4932.096 I then divided out: b = -2.192 confidence assessment: 3
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01:41:32 ** 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 understand. self critique assessment: 2
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01:43:10 What values did you get for a and b?
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RESPONSE --> I take the b value and plug it in to a previous equation: 540.8a + 10.4(-2.192) = -10.8 540.8a - 22.797 = -10.8 540.8a = 11.997 a = .0222 confidence assessment: 3
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01:43:31 ** STUDENT SOLUTION CONTINUED: a = .0165, b = -2 **
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RESPONSE --> I understand. a = .0222 b = -2.192 self critique assessment: 2
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01:44:52 What did you then get for c?
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RESPONSE --> I then plugged both a and b into an original quadratic: 108.16(.0222) + 10.4(-2.192) + c = 70.6 2.399 - 22.797 + c = 70.6 -20.398 + c = 70.6 c = 90.998 confidence assessment: 3
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01:44:59 ** STUDENT SOLUTION CONTINUED: c = 73.4 **
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RESPONSE --> I understand. self critique assessment: 2
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01:45:29 What is your function model?
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RESPONSE --> y = .0222(x^2) - 2.192(x) + 90.998 confidence assessment: 3
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01:45:36 ** STUDENT SOLUTION CONTINUED: y = (.0165)x^2 + (-2)x + 73.4. **
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RESPONSE --> I understand. self critique assessment: 2
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01:59:30 What is your depth prediction for the given clock time (give clock time also)?
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RESPONSE --> The given clock time being 46 seconds. y = .0222(46^2) - 2.192(46) + 90.998 y = 46.975 - 100.832 + 90.998 y = 37.141 confidence assessment: 3
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01:59:50 ** STUDENT SOLUTION CONTINUED: The given clock time was 46 seconds, and my depth prediction was 16.314 cm.**
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RESPONSE --> I understand. self critique assessment: 2
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02:07:24 What clock time corresponds to the given depth (give depth also)?
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RESPONSE --> The given depth being 14 centimeters: 14 = .0222(x^2) - 2.192(x) + 90.998 0 = .0222(x^2) - 2.192(x) + 76.998 t = [2.192 +- `sqrt(-2.192^2-4*.0222*90.998)] / (2*.0222) The + case gives us 130.225 The - case gives us -31.486 Obviously clock time cannot be negative, so the corresponding clock time would be 130.225 confidence assessment: 3
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02:07:38 ** 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 understand. self critique assessment: 2
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02:10:12 Completion of Model second problem: grade average Give your data in the form of grade vs. clock time ordered pairs.
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RESPONSE --> (0 , 1) (10 , 1.790569) (20 , 2.118034) (30 , 2.369306) (40 , 2.581139) (50 , 2.767767) (60 , 2.936492) (70 , 3.09165) (80 , 3.236068) (90 , 3.371708) (100 , 3.5) confidence assessment: 3
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02:10:23 ** 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 understand. self critique assessment: 2
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02:11:01 What three points on your graph did you use as a basis for your model?
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RESPONSE --> (30 , 2.369) (60 , 2.936) (90 , 3.372) confidence assessment: 3
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02:11:09 ** STUDENT SOLUTION CONTINUED: (20, 2.118034) (50, 2.767767) (100, 3.5)**
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RESPONSE --> I understand. self critique assessment: 2
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02:11:28 Give the first of your three equations.
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RESPONSE --> 900a + 30b + c = 2.369 confidence assessment: 3
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02:11:40 ** STUDENT SOLUTION CONTINUED: 400a + 20b + c = 2.118034**
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RESPONSE --> I understand. self critique assessment: 2
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02:11:49 Give the second of your three equations.
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RESPONSE --> 3600a + 60b + c = 2.936 confidence assessment: 3
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02:11:57 ** STUDENT SOLUTION CONTINUED: 2500a + 50b + c = 2.767767 **
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RESPONSE --> I understand. self critique assessment: 2
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02:12:07 Give the third of your three equations.
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RESPONSE --> 8100a + 90b + c = 3.372 confidence assessment: 3
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02:12:15 ** STUDENT SOLUTION CONTINUED: 10,000a + 100b + c = 3.5 **
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RESPONSE --> I understand. self critique assessment: 2
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02:12:44 Give the first of the equations you got when you eliminated c.
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RESPONSE --> I subtracted the first from the third equation: (8100a + 90b + c = 3.372) - (900a + 30b + c = 2.369) For a result of: 7200a + 60b = 1.003 confidence assessment: 3
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02:12:52 ** STUDENT SOLUTION CONTINUED: 7500a + 50b = .732233. **
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RESPONSE --> I understand. self critique assessment: 2
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02:13:08 Give the second of the equations you got when you eliminated c.
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RESPONSE --> I then subtracted the second from the third equation: (8100a + 90b + c = 3.372) - (3600a + 60b + c = 2.936) For a result of: 4500a + 30b = 0.436 confidence assessment: 3
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02:13:18 ** STUDENT SOLUTION CONTINUED: Subracting the first equation from the third I go 9600a + 80b = 1.381966 **
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RESPONSE --> I understand. self critique assessment: 2
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02:13:35 Explain how you solved for one of the variables.
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RESPONSE --> 30 fits nicely into 60, so I multiplied the second equation by -2 to get: -9000a - 60b = -0.872 I then subtracted that equation from the first: (7200a + 60b = 1.003) (-9000a - 60b = -0.872) For a result of: -1800a = .131 I then divided to get: a = -7.278 confidence assessment: 3
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02:13:47 ** 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 understand. self critique assessment: 2
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02:14:02 What values did you get for a and b?
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RESPONSE --> a = -7.278 For b I plugged my a vaule into a prefious equation: 4500 (-7.278) + 30b = 0.436 -.328 + 30b = 0.436 30b = .7635 b = .0255 confidence assessment: 3
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02:14:10 ** STUDENT SOLUTION CONTINUED: a = -.0000876638 b = .01727 **
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RESPONSE --> I understand. self critique assessment: 2
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02:14:25 What did you then get for c?
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RESPONSE --> c = 1.671 confidence assessment: 3
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02:14:45 ** STUDENT SOLUTION CONTINUED: c = 1.773. **
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RESPONSE --> I understand. self critique assessment: 2
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02:14:57 What is your function model?
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RESPONSE --> y = -7.278(x^2) + .02545(x) + 1.671 confidence assessment: 3
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02:15:07 ** y = -.0000876638 x^2 + (.01727)x + 1.773 **
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RESPONSE --> I understand. self critique assessment: 2
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02:44:57 What is your percent-of-review prediction for the given range of grades (give grade range also)?
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RESPONSE --> The given grade ranges are 3.0 and 4.0 3 = -7.278(x^2) + .02545(x) + 1.671 0 = -7.278(x^2) + .02545(x) - 1.329 y = [-.02545 +- `sqrt(.02545^2-4*-7.278*-1.329)] / (2*-7.278) The + case gives us 64.237 The - case gives us 285.458 64.237 seems to make the most good sense of the two possible answers. (64.237 , 3.0) 4 = -7.278(x^2) + .02545(x) + 1.671 0 = -7.278(x^2) + .02545(x) - 2.329 y = [-.02545 +- `sqrt(.02545^2-4*-7.278*-2.329)] / (2*-7.278) The + case gives us 137.061 The - case gives us 212.634 Neither of these solutions seem to make sense. In referring to the graph it appears that 100 could be the highest possible percent. If no one received a grade average higher than 3.5, then 0% received a grade average of 4.0 (0 , 4.0) (64.237 , 3.0) confidence assessment: 3
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02:45:13 ** The precise solution depends on the model desired average. For example if the model is y = -.00028 x^2 + .06 x + .5 (your model will probably be different from this) and the grade average desired is 3.3 we would find the percent of review x corresponding to grade average y = 3.3 then we have 3.3 = -.00028 x^2 + .06 x + .5. This equation is easily solved using the quadratic formula, remembering to put the equation into the required form a x^2 + b x + c = 0. We get two solutions, x = 69 and x = 146. Our solutions are therefore 69% grade review, which is realistically within the 0 - 100% range, and 146%, which we might reject as being outside the range of possibility. To get a range you would solve two equations, on each for the percent of review for the lower and higher ends of the range. In many models the attempt to solve for a 4.0 average results in an expression which includes the square root of a negative number; this indicates that there is no real solution and that a 4.0 is not possible. **
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RESPONSE --> I understand. self critique assessment: 2
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02:47:27 What grade average corresponds to the given percent of review (give grade average also)?
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RESPONSE --> The given percent being 80, a look at the graph tells us that 3.236068 is the corresponding grade average. A check with the model confirms this. confidence assessment: 3
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02:47:42 ** 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 understand. self critique assessment: 2
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02:53:22 How well does your model fit the data (support your answer)?
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RESPONSE --> With an average deviation of +.01 I'd say the model fits pretty well. confidence assessment: 3
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02:53:52 ** You should have evaluated your function at each given percent of review-i.e., at 0, 10, 20, 30, . 100 to get the predicted grade average for each. Comparing your results with the given grade averages shows whether your model fits the data. **
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RESPONSE --> I understand. self critique assessment: 2
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02:54:24 illumination vs. distance Give your data in the form of illumination vs. distance ordered pairs.
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RESPONSE --> (1 , 935.1395) (2 , 264.4411) (3 , 150.1209) (4 , 61.01488) (5 , 43.06238) (6 , 25.91537) (7 , 19.92772) (8 , 16.27232) (9 , 11.28082) (10 , 9.484465) confidence assessment: 3
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02:54:43 ** 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 understand. self critique assessment: 2
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02:55:17 What three points on your graph did you use as a basis for your model?
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RESPONSE --> (3 , 150.1209) (6 , 25.91537) (9 , 11.28082) confidence assessment: 3
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02:55:29 ** STUDENT SOLUTION CONTINUED: (2, 264.4411) (4, 61.01488) (8, 16.27232) **
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RESPONSE --> I understand. self critique assessment: 2
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02:55:49 Give the first of your three equations.
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RESPONSE --> 9a + 3b + c = 105.121 confidence assessment: 3
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02:55:54 ** STUDENT SOLUTION CONTINUED: 4a + 2b + c = 264.4411**
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RESPONSE --> I understand. self critique assessment: 2
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02:56:10 Give the second of your three equations.
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RESPONSE --> 36a + 6b + c = 25.915 confidence assessment: 3
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02:56:16 ** STUDENT SOLUTION CONTINUED: 16a + 4b + c = 61.01488**
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RESPONSE --> I understand. self critique assessment: 2
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02:56:30 Give the third of your three equations.
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RESPONSE --> 81a + 9b + c = 11.281 confidence assessment: 3
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02:56:36 ** STUDENT SOLUTION CONTINUED: 64a + 8b + c = 16.27232**
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RESPONSE --> I understand. self critique assessment: 2
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02:57:55 Give the first of the equations you got when you eliminated c.
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RESPONSE --> I subtracted the first equation from the third: (81a + 9b + c = 11.281) + (-9a - 3b - c = -105.121) To get: -72a + 6b = -93.84 confidence assessment: 3
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02:58:01 ** STUDENT SOLUTION CONTINUED: 48a + 4b = -44.74256**
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RESPONSE --> I understand. self critique assessment: 2
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02:59:07 Give the second of the equations you got when you eliminated c.
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RESPONSE --> I then subtracted the second equation from the third: (81a + 9b + c = 11.281) + (-36a - 6b - c = -25.915) To get: 45a + 3b = -14.634 confidence assessment: 3
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02:59:12 ** STUDENT SOLUTION CONTINUED: 60a + 6b = -248.16878**
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RESPONSE --> I understand. self critique assessment: 2
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03:01:29 Explain how you solved for one of the variables.
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RESPONSE --> I multiplied the second equation by -2 so the b variable would cancel out clean: (72a + 6b = -93.84) + (-90a - 6b = 29.268) To get: -18a = -64.572 From there I divided: a = 3.587 confidence assessment: 3
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03:01:36 ** 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 understand. self critique assessment: 2
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03:02:59 What values did you get for a and b?
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RESPONSE --> Next I plugged the a value (3.587) in to a previous equation: 45(3.587) + 3b = -14.634 161.43 + 3b = -14.634 3b = -176.064 b = -58.688 confidence assessment: 3
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03:03:05 ** STUDENT SOLUTION CONTINUED: a = 15.088, b = -192.24 **
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RESPONSE --> I understand. self critique assessment: 2
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03:04:22 What did you then get for c?
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RESPONSE --> I then plugged both variables into an origianl quadratic: 9(3.587) + 3(-58.688) + c = 105.121 32.283 - 176.064 + c = 105.121 -143.781 + c = 105.121 c = 248.902 confidence assessment: 3
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03:04:28 ** STUDENT SOLUTION CONTINUED: c = 588.5691**
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RESPONSE --> I understand. self critique assessment: 2
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03:05:01 What is your function model?
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RESPONSE --> y = 3.587(x^2) - 58.688(x) + 248.902 confidence assessment: 3
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03:05:07 ** STUDENT SOLUTION CONTINUED: y = (15.088) x^2 - (192.24)x + 588.5691 **
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RESPONSE --> I understand. self critique assessment: 2
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03:09:11 What is your illumination prediction for the given distance (give distance also)?
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RESPONSE --> The given distance is 1.6 y = 3.587(1.6^2) - 58.688(1.6) + 248.902 y = 9.183 - 93.901 + 248.902 y = 164.184 confidence assessment: 2
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03:09:28 ** 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 understand. self critique assessment: 2
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03:19:27 What distances correspond to the given illumination range (give illumination range also)?
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RESPONSE --> The illumination ranges given are 25 and 100 25 = 3.587(x^2) - 58.688(x) + 248.902 0 = 3.587(x^2) - 58.688(x) + 223.902 0 = [58.688 +- `sqrt(-58.688^2-4*3.587*248.902)] / (2*3.587) The + casegives us 19.856 The - case gives us -3.495 The distance could not possibly be negative. 19.856 makes the most sense. 100 = 3.587(x^2) - 58.688(x) + 248.902 0 = 3.587(x^2) - 58.688(x) + 148.902 0 = [58.688 +- `sqrt(-58.688^2-4*3.587*148.902)] / (2*3.587) the + casegives us 18.594 The - case gives us -2.233 Distance cannot possibly be negative. 18.594 makes sense. Between 18.594 and 19.856 AU would be comfortable for reading. confidence assessment: 3
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03:19:52 ** 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 understand. self critique assessment: 2
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