course Mth 163 Due to our agreed delay on my work and then major technical difficulties this was completed in multiple work sessions. I am pretty sure that I got everything copied and pasted together correctly. I have been having multiple problems getting to the exercises over the last week or so, I hope that it was just vhcc server problems due to weather. I will keep you up to date if these problems continue. I did send you an email because I could not even get this form to pull up to communicate with you.
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22:04:42 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 --> data point 1 (where x= time and y = temp) (0,118) Data point three (40, 91.96194) Data point five (80, 73.29324) confidence assessment: 2
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22:58:17 According to your graph what would be the temperatures at clock times 7, 19 and 31?
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RESPONSE --> Enter, as appropriate, an answer to the question, a critique of your answer in response to a given answer, your insights regarding the situation at this point, notes to yourself, or just an OK. Always critique your solutions by describing any insights you had or errors you makde, and by explaining how you can make use of the insight or how you now know how to avoid certain errors. Also pose for the instructor any question or questions that you have related to the problem or series of problems. At seven minutes the temperature would still be 118 degrees due to the fact that at twenty minutes the temp was still 118 degrees. At 19 minutes the temperature would be still be 118 degrees because the temp remained at 118 degrees until after the 20 minute mark At 31 minutes the temperature would be 97.01642 degrees confidence assessment: 2
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23:03:00 What three points did you use as a basis for your quadratic model (express as ordered pairs)?
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RESPONSE --> Enter, as appropriate, an answer to the question, a critique of your answer in response to a given answer, your insights regarding the situation at this point, notes to yourself, or just an OK. Always critique your solutions by describing any insights you had or errors you makde, and by explaining how you can make use of the insight or how you now know how to avoid certain errors. Also pose for the instructor any question or questions that you have related to the problem or series of problems. (40, 91.96194) (60, 81.85297) (80, 73.29324) I am not sure what you want as a critique. confidence assessment: 2
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23:04:39 ** 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 chose points close together. I now understand why that was not a good choice. It may have made the calculations easier, but it could mess up the reliability. self critique assessment: 2
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23:08:41 What is the first equation you got when you substituted into the form of a quadratic?
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RESPONSE --> equation 1 for point (40, 91.96194) 1600a + 40b + c = 91.96194 confidence assessment: 3
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23:09:15 ** STUDENT SOLUTION CONTINUED: The equation that I got from the first data point (10,75) was 100a + 10b +c = 75.**
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RESPONSE --> OK self critique assessment: 2
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23:09:38 What is the second equation you got when you substituted into the form of a quadratic?
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RESPONSE --> equation 2 for point (60, 81.85297) 3600a + 60b + c = 81.85297 confidence assessment: 3
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23:09:55 ** STUDENT SOLUTION CONTINUED: The equation that I got from my second data point was 400a + 20b + c = 60 **
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RESPONSE --> ok self critique assessment: 2
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23:10:11 What is the third equation you got when you substituted into the form of a quadratic?
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RESPONSE --> equation 3 for point (80, 73.29324) 6400a + 80b + c =73.29324 confidence assessment: 3
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23:10:30 ** STUDENT SOLUTION CONTINUED: The equation that I got from my third data point was 3600a + 60b + c = 30. **
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RESPONSE --> ok self critique assessment: 3
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23:14:50 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 substracted equation 1 from equation 2 3600a + 60b + c = 81.85297 1600a + 40b + c = 91.96194 this gave me the following equation after eliminating c 2000a + 20b = -10.10897 confidence assessment: 3
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23:15:16 ** STUDENT SOLUTION CONTINUED: First, I subtracted the second equation from the third equation in order to eliminate c. By doing this, I obtained my first new equation 3200a + 40b = -30. **
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RESPONSE --> ok self critique assessment: 3
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23:17:39 To get the second equation what multiple of which equation did you add to what multiple of which other quation, and what is the resulting equation?
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RESPONSE --> I subtracted equation 2 from equation 3 in order to eliminate c 6400a + 80b +c = 73.29324 3600a + 60b +c = 81.85297 this gives me a new equation of 2800a +20b = -8.55973 confidence assessment: 3
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23:17:52 ** 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:
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23:21:21 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 then eliminated b from both equations I first multiplied equation 2 by -1 then added the equations together 2800a + 20b = -8.55973 -2000a -20b = 10.10897 800a = 1.54924 a = 0.00193655 confidence assessment: 3
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23:21:53 ** STUDENT SOLUTION CONTINUED: In order to solve for a and b, I decided to eliminate b because of its smaller value. In order to do this, I multiplied the first new equation by -5 -5 ( 3200a + 40b = -30) and multiplied the second new equation by 4 4 ( 3500a + 50b = -45) making the values of -200 b and 200 b cancel one another out. The resulting equation is -2000 a = -310. **
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RESPONSE --> ok self critique assessment: 3
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23:24:38 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 then substituted a in equation 1 and solved for b 2000 (0.00193655) + 20b = -10.10897 3.8731 + 20b = -10.10897 20b = -13.98207 b = -0.6991035 confidence assessment: 3
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23:24:45 ** 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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23:27:50 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 in the original equation 1 1600(0.00193655) + 40(-0.6991035) + c= 91.96194 3.09848 + -27.96414 +c = 91.96194 -24.86566 + c = 91.96194 c = 116.8276 confidence assessment: 3
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23:28:05 ** STUDENT SOLUTION CONTINUED: By substituting both a and b into the original equations, I found that c = 93 **
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RESPONSE --> ok self critique assessment: 3
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23:32:31 What is the resulting quadratic model?
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RESPONSE --> y = 0.00193655 (t^2) + -0.6991035(t) + 116.8276 confidence assessment: 3
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23:32:54 ** STUDENT SOLUTION CONTINUED: Therefore, the quadratic model that I obtained was y = (.015) x^2 - (1.95)x + 93. **
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RESPONSE --> ok self critique assessment: 3
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23:49:12 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 --> First clock time is 0 and the predicted temp using my model was 116.8276 with a deviation of +1.1724 Second clock time is 20 and the predicted temp using my model was 103.62015 with a deviation of +0.28045 Third clock time is 40 and the predicted temp using my model was 91.96194 with a deviation of 0 The temperature on the first set of data points for the 20 minute mark does not match the temperature in the prediction graph. confidence assessment: 3
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23:49:46 ** STUDENT SOLUTION CONTINUED: This model y = (.015) x^2 - (1.95)x + 93 evaluated for clock times 0, 10 and 20 gave me these numbers: First prediction: 93 Deviation: 2 Then, since I used the next two ordered pairs to make the model, I got back }the exact numbers with no deviation. So. the next two were Fourth prediction: 48 Deviation: 1 Fifth prediction: 39 Deviation: 2. **
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RESPONSE --> ok self critique assessment: 2
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00:07:47 What was your average deviation?
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RESPONSE --> -0.2376991375 confidence assessment: 2
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00:08:08 ** STUDENT SOLUTION CONTINUED: My average deviation was .6 **
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RESPONSE --> ok my was not even close to that. self critique assessment: 1
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00:09:51 Is there a pattern to your deviations?
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RESPONSE --> To some degree yes +1.1724 +0.28045 0 0 0 -0.23742 -0.91331 -2.19789 confidence assessment: 2
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00:10:29 ** 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 --> mine was positive, zero, and then negative self critique assessment: 2
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00:12:02 Have you studied the steps in the modeling process as presented in Overview, the Flow Model, Summaries of the Modeling Process, and do you completely understand the process?
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RESPONSE --> I think so. I still need my notes though, I cannot do it from rote memory yet. As for understanding it, every time I think I understand it I get confused on the next step. I am a mathematical work in progress. confidence assessment: 3
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00:12:11 ** STUDENT SOLUTION CONTINUED: Yes, I do completely understand the process after studying these outlines and explanations. **
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RESPONSE --> self critique assessment:
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00:22:28 Have you memorized the steps of the modeling process, and are you gonna remember them forever? Convince me.
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RESPONSE --> Here we go, I can't seem to find my notes so we will see if I have it memorized or not. 1. Pick three data points to create three equations using the mathematical model of y = at^2 + bt + c 2. Eliminate c from the equations by manipulating the three equations. This gives you two new equations with only a and b 3. Eliminate a (or b) from the two new equations by multiplying by the lowest common denominator allowing the a (or b) to cancel out add the equations together. Solve for b (or a) 4. Substitute appropriate value in original a/b equation to solve for the other 5. Substitute a and b in original equationd and solve for c 4. confidence assessment: 1
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00:23:29 ** 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 --> Okay so I wasn't so convincing. I will have to be more convincing next time you ask. self critique assessment: 2
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00:24:44 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 --> confidence assessment: 0
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00:33:45 ** 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 did not understand the question. Now I do and here is my answer. I received my data using the random problems (5.2, 80.1) (10.4, 70.6) (15.6, 62.2) (20.8, 55) (26, 48.9) (31.2, 44.2) self critique assessment: 3
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01:12:24 What three points on your graph did you use as a basis for your model?
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RESPONSE --> (5.2, 80.1) (15.6, 62.2) (26, 48.9) confidence assessment: 3
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01:12:44 ** STUDENT SOLUTION CONTINUED: As the basis for my graph, I used ( 5.3, 63.7) (15.9, 46) (26.5, 32)**
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RESPONSE --> ok self critique assessment: 3
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01:13:44 Give the first of your three equations.
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RESPONSE --> eq 1 27.04a + 5.2b + c = 80.1 confidence assessment: 3
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01:15:00 ** STUDENT SOLUTION CONTINUED: The point (5.3, 63.7) gives me the equation 28.09a + 5.3b + c = 63.7 **
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RESPONSE --> ok self critique assessment: 3
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01:16:07 Give the second of your three equations.
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RESPONSE --> point (15.6, 62.2) gives me the following equation 243.36a + 15.6b + c = 62.2 confidence assessment: 3
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01:16:15 ** 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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01:17:08 Give the third of your three equations.
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RESPONSE --> point (26, 48.9) gives me the equation 676a + 26b + c = 48.9 confidence assessment: 3
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01:17:17 ** STUDENT SOLUTION CONTINUED: The point (26.5,32) gives me the equation 702.25a + 26.5b + c = 32. **
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RESPONSE --> ok self critique assessment: 3
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01:18:21 Give the first of the equations you got when you eliminated c.
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RESPONSE --> after elimimating c my first new equation was 216.32a + 10.4b = -17.9 confidence assessment:
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01:19:10 ** STUDENT SOLUTION CONTINUED: Subtracting the second equation from the third gave me 449.44a + 10.6b = -14. **
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RESPONSE --> I hit enter before I explained what I did. I subtracted eq. 1 from eq 2 self critique assessment: 3
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01:20:09 Give the second of the equations you got when you eliminated c.
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RESPONSE --> After subtracting eq 1 from eq 3 I got the new equation of 648.96a + 20.8b = -31.2 confidence assessment: 3
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01:20:19 ** STUDENT SOLUTION CONTINUED: Subtracting the first equation from the third gave me 674.16a + 21.2b = -31.7. **
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RESPONSE --> ok self critique assessment: 3
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02:08:38 Explain how you solved for one of the variables.
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RESPONSE --> I multiplied the new eq1 by -20.8 to get -4499.456a + -216.32b = 324.48 I then multiplied the new eq2 by 10.4 to get 6749.184a + 216.32b = -372.32 then I added the equations together and solved for a a= -0.3097263 confidence assessment: 1
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02:09:04 ** STUDENT SOLUTION CONTINUED: In order to solve for a, I eliminated b by multiplying the first equation by 21.2, which was the b value in the second equation. Then, I multiplied the seond equation by -10.6, which was the b value of the first equation, only I made it negative so they would cancel out. **
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RESPONSE --> ok self critique assessment: 2
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03:01:41 What values did you get for a and b?
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RESPONSE --> I happened to look at my figures again when getting the numbers to answer this question and found a mathematical error. The a and b values I gave previously are incorrect. Here are my correct values: a = 0.02126479 b = -2.163461478 confidence assessment: 3
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03:01:55 ** STUDENT SOLUTION CONTINUED: a = .0165, b = -2 **
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RESPONSE --> ok self critique assessment: 2
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03:02:47 What did you then get for c?
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RESPONSE --> After substituting a and b in one of the original equations I got the followig value for c c = 90.774999764 confidence assessment: 3
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03:03:00 ** STUDENT SOLUTION CONTINUED: c = 73.4 **
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RESPONSE --> ok self critique assessment: 2
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03:05:27 What is your function model?
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RESPONSE --> y = 0.02126479 (x^2) + -2.163461478 (x) + 90.774999764 confidence assessment: 3
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03:06:17 ** STUDENT SOLUTION CONTINUED: y = (.0165)x^2 + (-2)x + 73.4. **
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RESPONSE --> I have been keeping too many decimal places I think self critique assessment: 3
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03:19:15 What is your depth prediction for the given clock time (give clock time also)?
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RESPONSE --> clock time predicted depth 5.2 80.1 (used this point for eq) 10.4 70.6 15.6 62.2 (used this point for eq) 20.8 55.0 26 48.9 (used this poit for eq) 31.2 44.0 confidence assessment: 2
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03:20:36 ** STUDENT SOLUTION CONTINUED: The given clock time was 46 seconds, and my depth prediction was 16.314 cm.**
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RESPONSE --> I think I missed a question I didn't see a specific clock time I completed the prediciton chart. self critique assessment: 1
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03:24:00 What clock time corresponds to the given depth (give depth also)?
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RESPONSE --> Where am I missing the given informaiton? confidence assessment: 0
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04:07:57 ** 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 have figured the depth for the given clock time ( I had forgotten about the exercises page) the depth for 46 seconds is 36.3 cm Now, as for the clock time for the depth of 14cm I am having problems. to find x I use the same model right? 14=0.02(x^2)+ (-2.16x)+90.77 14-90.77=0.02(x^2) +(-2.16x) -76.77= 0.02(x^2)+(-2.16x) Here is where I forget what to do next.
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04:35:37 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: 0
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04:42:17 ** 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 was confused by the information given in the previouse question, it asked for grades vs clock time. the data points I have are (0, .9258546) (10, 1.36495) (20, 1.750933) (30, 2.090227) (40, 2.388479) (50, 2.650655) (60, 2.881118) (70, 3.083704) (80, 3.261786) (90, 3.418326) (100, 3.555931) self critique assessment: 1
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??V???f????-H???assignment #002 002. `query2 Precalculus I 07-22-2008
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00:13:35 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 --> Grade average data points (0, .9258546) (10, 1.36495) (20, 1.750933) (30, 2.090227) (40, 2.388479) (50, 2.650655) (60, 2.881118) (70, 3.083704) (80, 3.261786) (90, 3.418326) (100, 3.555931) confidence assessment: 2
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00:13:56 ** 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 --> self critique assessment: 2
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00:55:08 What three points on your graph did you use as a basis for your model?
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RESPONSE --> (10, 1.36495) (40, 2.388479) (80, 3.261786) confidence assessment: 3
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00:55:24 ** STUDENT SOLUTION CONTINUED: (20, 2.118034) (50, 2.767767) (100, 3.5)**
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RESPONSE --> self critique assessment: 2
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00:59:16 Give the first of your three equations.
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RESPONSE --> 1.36495= a(10^2)+b(10)+c 100a+10b+c=1.36495 confidence assessment: 3
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00:59:30 ** STUDENT SOLUTION CONTINUED: 400a + 20b + c = 2.118034**
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RESPONSE --> self critique assessment: 3
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01:00:35 Give the second of your three equations.
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RESPONSE --> 1600a+40b+c=2.388479 confidence assessment: 3
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01:00:42 ** STUDENT SOLUTION CONTINUED: 2500a + 50b + c = 2.767767 **
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RESPONSE --> self critique assessment: 3
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01:01:19 Give the third of your three equations.
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RESPONSE --> 6400a+80b+c=3.261786 confidence assessment: 3
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01:01:31 ** STUDENT SOLUTION CONTINUED: 10,000a + 100b + c = 3.5 **
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RESPONSE --> self critique assessment: 3
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01:02:26 Give the first of the equations you got when you eliminated c.
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RESPONSE --> 6300a+70b=1.896836 confidence assessment: 3
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01:02:33 ** STUDENT SOLUTION CONTINUED: 7500a + 50b = .732233. **
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RESPONSE --> self critique assessment: 3
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01:03:10 Give the second of the equations you got when you eliminated c.
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RESPONSE --> 4800a+40b=.873307 confidence assessment: 3
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01:03:19 ** STUDENT SOLUTION CONTINUED: Subracting the first equation from the third I go 9600a + 80b = 1.381966 **
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RESPONSE --> self critique assessment: 3
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01:06:02 Explain how you solved for one of the variables.
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RESPONSE --> I solved for a by eliminating b. I multiplied one equation by 70 and the other by 40 in order to get a b value of 2800 in both equations and then subtracted to eliminate b I then solved for a confidence assessment: 2
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01:06:20 ** 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 --> self critique assessment: 3
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01:07:15 What values did you get for a and b?
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RESPONSE --> a= -.0001754994 b= .4289260314 confidence assessment: 3
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01:07:25 ** STUDENT SOLUTION CONTINUED: a = -.0000876638 b = .01727 **
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RESPONSE --> self critique assessment: 3
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01:07:51 What did you then get for c?
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RESPONSE --> c= .9535739086 confidence assessment: 3
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01:07:58 ** STUDENT SOLUTION CONTINUED: c = 1.773. **
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RESPONSE --> self critique assessment: 3
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01:09:57 What is your function model?
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RESPONSE --> y = - .0001754994(x^2) + .04289260314X + .9535739086 confidence assessment: 2
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01:10:16 ** y = -.0000876638 x^2 + (.01727)x + 1.773 **
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RESPONSE --> self critique assessment: 3
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01:19:42 What is your percent-of-review prediction for the given range of grades (give grade range also)?
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RESPONSE --> Given range of grades 3.0 and 4.0 For a grade of 3.0 the function model gave me a percentage of 64.857 For a grade of 4.0 the function model gave me a percentage of 74 these answers do not fit the data given. confidence assessment: 1
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01:21:57 ** 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 --> Other than forgeting that you cannot take the square root of a negative number, the process used in the student solution is the same process I followed. self critique assessment: 2
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01:23:29 What grade average corresponds to the given percent of review (give grade average also)?
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RESPONSE --> The given percentage of review was 80. The grade average for that percentage that I obtained was 3.266 confidence assessment: 2
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01:23:46 ** 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 --> self critique assessment: 2
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02:15:37 How well does your model fit the data (support your answer)?
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RESPONSE --> My function model fits quite well with the data provided. My predicted values were very similar with the data points given with low diviations from the given data. Deviations range from .007 to .0609 with Residuals from -.0609 to .0281. confidence assessment: 2
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02:15:53 ** 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 --> self critique assessment: 2
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03:57:34 illumination vs. distance Give your data in the form of illumination vs. distance ordered pairs.
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RESPONSE --> Distance, Illumination (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) confidence assessment: 3
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03:57:52 ** 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 --> self critique assessment: 3
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03:59:22 What three points on your graph did you use as a basis for your model?
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RESPONSE --> I used the following three data points as the basis for my model. (2, 264.4411) (5, 43.06238) (8, 16.27232) confidence assessment: 3
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03:59:32 ** STUDENT SOLUTION CONTINUED: (2, 264.4411) (4, 61.01488) (8, 16.27232) **
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RESPONSE --> self critique assessment: 3
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04:00:22 Give the first of your three equations.
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RESPONSE --> eq 1 4a + 2b + c = 264.4411 confidence assessment: 3
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04:00:33 ** STUDENT SOLUTION CONTINUED: 4a + 2b + c = 264.4411**
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RESPONSE --> self critique assessment: 3
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04:01:11 Give the second of your three equations.
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RESPONSE --> eq2 25a + 5b + c = 43.06238 confidence assessment: 3
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04:01:18 ** STUDENT SOLUTION CONTINUED: 16a + 4b + c = 61.01488**
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RESPONSE --> self critique assessment: 3
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04:02:02 Give the third of your three equations.
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RESPONSE --> eq3 64a + 8b + c = 16.27232 confidence assessment: 3
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04:02:09 ** STUDENT SOLUTION CONTINUED: 64a + 8b + c = 16.27232**
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RESPONSE --> self critique assessment: 3
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04:03:04 Give the first of the equations you got when you eliminated c.
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RESPONSE --> eq1a 60a + 6b = -248.16878 confidence assessment: 3
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04:03:46 ** STUDENT SOLUTION CONTINUED: 48a + 4b = -44.74256**
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RESPONSE --> self critique assessment: 2
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04:04:34 Give the second of the equations you got when you eliminated c.
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RESPONSE --> eq2a 39a + 3b = -26.79006 confidence assessment: 3
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04:04:59 ** STUDENT SOLUTION CONTINUED: 60a + 6b = -248.16878**
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RESPONSE --> self critique assessment: 3
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04:05:58 Explain how you solved for one of the variables.
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RESPONSE --> I subtracted eq1 from eq3 to eliminate c confidence assessment: 3
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04:07:50 ** 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 --> wrong equation To solve for a I multiplied eq2a by -2 to give me a -6b value to eliminate b from the equation and then simpliflied the equation to solve for a. self critique assessment: 3
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04:08:23 What values did you get for a and b?
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RESPONSE --> a = 10.8105 b = -149.4665 confidence assessment: 3
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04:09:19 ** STUDENT SOLUTION CONTINUED: a = 15.088, b = -192.24 **
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RESPONSE --> self critique assessment: 2
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04:10:05 What did you then get for c?
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RESPONSE --> c = 520.1321 confidence assessment: 2
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04:10:14 ** STUDENT SOLUTION CONTINUED: c = 588.5691**
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RESPONSE --> self critique assessment: 2
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04:11:36 What is your function model?
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RESPONSE --> y = 10.8105(x^2) + -149.4665(x) + 520.1321 confidence assessment: 3
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04:11:50 ** STUDENT SOLUTION CONTINUED: y = (15.088) x^2 - (192.24)x + 588.5691 **
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RESPONSE --> self critique assessment: 2
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04:13:31 What is your illumination prediction for the given distance (give distance also)?
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RESPONSE --> For the given distance of 1.6 AU the predicted illumination was 308.66058 W/m^2 confidence assessment: 3
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04:14:11 ** 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 am worried none of my answers are matching the student solutions. self critique assessment: 2
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04:16:46 What distances correspond to the given illumination range (give illumination range also)?
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RESPONSE --> For the illumination of 25W/M^2 my prediction was 8.3233 or 5.5028 AU and for the illumination of 100 W/m^2 my prediction was 9.9008 or 3.9253 AU confidence assessment: 2
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04:17:15 ** 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 --> self critique assessment: 2
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