course Mth 163 qLܐkT|څassignment #002
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08:47:58 What were temperature and time for the first, third and fifth data points (express as temp vs clock time ordered pairs)?
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RESPONSE --> I used (10,75) (40,41) (70,26) confidence assessment: 2
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08:55:05 According to your graph what would be the temperatures at clock times 7, 19 and 31?
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RESPONSE --> (7, 72) (19,83) (31, 97) confidence assessment: 0
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08:55:47 What three points did you use as a basis for your quadratic model (express as ordered pairs)?
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RESPONSE --> (10,75) (40,41) (70,26) confidence assessment: 3
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08:56:19 ** 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 --> OK self critique assessment: 3
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08:57:04 What is the first equation you got when you substituted into the form of a quadratic?
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RESPONSE --> 100a+10b+c=75 confidence assessment: 2
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08:57:20 ** 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 was correct self critique assessment: 3
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08:57:49 What is the second equation you got when you substituted into the form of a quadratic?
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RESPONSE --> 1600a+40b+c=41 confidence assessment: 3
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08:58:07 ** 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: 3
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08:58:34 What is the third equation you got when you substituted into the form of a quadratic?
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RESPONSE --> 4900a+70b+c=26 confidence assessment: 3
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08:58:45 ** 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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08:59:16 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 --> 1500a+30b=-31 confidence assessment: 3
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08:59:39 ** 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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09:00:02 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 --> 3300a+30b=-15 confidence assessment: 3
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09:00:16 ** 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 --> ok self critique assessment: 3
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09:03:35 ** 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 --> In order to solve for a and b, I eliminated b first. I multiplied each new equation by it's opposite coefficient 30 (1500a+b30=-34) -30 (3300A+b30=-15) 30b-30b=0 Then I added the two together to get the new equation -54000a=-570 self critique assessment: 3
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09:06:01 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 --> The last equation was -54000a=-570 a=.0105 I substituted this into 1600 (.0105)+b40=41 16.8+b40=41 b40=24.20 b=.605 confidence assessment: 2
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09:06:46 ** 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 --> ok self critique assessment: 3
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09:08:16 What is the value of c obtained from substituting into one of the original equations?
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RESPONSE --> The value of c was 67.9 75=1.05+6.05+c 75=7.1+c c=67.9 confidence assessment: 2
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09:08:29 ** 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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09:09:56 What is the resulting quadratic model?
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RESPONSE --> The resulting quadratic model is temperature=.0105t^2+.605t+67.9 confidence assessment: 2
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09:15:28 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 clock time 7 I found temperature to be 72 for clock time 19 I found temperature to be 83 and for clock time31 I found temperature to be 96 I'm not sure the deviation was. These times were not included in the chart to compare, but when using the parameters to solve for t=20 the temp was 84 with a deviation of 24 degrees confidence assessment: 1
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09:15:54 ** 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: 3
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09:20:44 What was your average deviation?
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RESPONSE --> I got an average deviation of 31.25 degrees. I worked this problem several times and could never get the data to fit the model well. confidence assessment: 1
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09:20:55 ** STUDENT SOLUTION CONTINUED: My average deviation was .6 **
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RESPONSE --> ok self critique assessment: 3
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09:21:29 Is there a pattern to your deviations?
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RESPONSE --> As the time increased the deviation increased. confidence assessment: 1
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09:21:51 ** 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 --> ok self critique assessment: 3
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09:22:11 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 confidence assessment: 3
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09:22:24 ** STUDENT SOLUTION CONTINUED: Yes, I do completely understand the process after studying these outlines and explanations. **
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RESPONSE --> ok self critique assessment: 3
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09:30:07 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 worked on memorizing the steps. There are the 3 Os Orient Observe Organize data Graph Postulate Select Point Obtain equations Solve equations substitute the parameters graph it again quantify pose and answer questions do the science or apply it to the real world. confidence assessment: 3
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09:30:23 ** 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 --> ok self critique assessment: 3
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09:31:51 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 --> depth vs clock time (0,95) 67.9 27.2 (10, 75) 0 0 (20, 60) 84 24 confidence assessment: 1
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09:32:09 ** 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 --> ok self critique assessment: 3
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09:32:50 What three points on your graph did you use as a basis for your model?
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RESPONSE --> (10,75) (40,41) (70,26) confidence assessment: 3
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09:33:02 ** 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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09:34:15 Give the first of your three equations.
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RESPONSE --> confidence assessment: 0
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09:36:05 Give the third of your three equations.
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RESPONSE --> ok, im sorry I didn't realize this was for a different problem confidence assessment: 0
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09:36:16 ** 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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09:38:54 Give the first of the equations you got when you eliminated c.
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RESPONSE --> 252.81a+5.3b=-5.7 confidence assessment: 2
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09:39:05 ** STUDENT SOLUTION CONTINUED: Subtracting the second equation from the third gave me 449.44a + 10.6b = -14. **
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RESPONSE --> ok self critique assessment: 3
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09:39:43 Give the second of the equations you got when you eliminated c.
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RESPONSE --> Subtracting the 2nd from the 3rd gave the equation 308.99a+5.3b=-5.4 confidence assessment: 3
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09:39:51 ** 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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09:41:31 Explain how you solved for one of the variables.
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RESPONSE --> Once I eliminated c I multiplied the two remain equations by their equal and opposite coefficient and then added them together to get the equation .297754a=-1.59 a=.00534 confidence assessment: 3
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09:41:44 ** 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: 3
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09:42:13 What values did you get for a and b?
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RESPONSE --> a=.00534 b=-1.33 confidence assessment: 3
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09:42:25 ** STUDENT SOLUTION CONTINUED: a = .0165, b = -2 **
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RESPONSE --> ok self critique assessment: 3
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09:42:46 What did you then get for c?
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RESPONSE --> c=63.496 confidence assessment: 3
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09:42:55 ** STUDENT SOLUTION CONTINUED: c = 73.4 **
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RESPONSE --> ok self critique assessment: 3
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09:43:30 What is your function model?
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RESPONSE --> y=.00534t^2-1.33t+63.496 confidence assessment: 3
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09:43:40 ** STUDENT SOLUTION CONTINUED: y = (.0165)x^2 + (-2)x + 73.4. **
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RESPONSE --> ok self critique assessment: 3
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09:44:49 What is your depth prediction for the given clock time (give clock time also)?
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RESPONSE --> My prediction for clock time t=5.3 was 56.597. confidence assessment: 2
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09:45:00 ** STUDENT SOLUTION CONTINUED: The given clock time was 46 seconds, and my depth prediction was 16.314 cm.**
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RESPONSE --> ok self critique assessment: 3
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09:45:37 What clock time corresponds to the given depth (give depth also)?
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RESPONSE --> For the depth of 54.8 cm the clock time was 10.6 secs confidence assessment: 2
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09:50:33 ** 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 --> Ok I just used a depth from the chart. I did not understand. So if using the model if I wanted to find clock time at a depth of 50cm 50=.00534t^2-1.33T+63.496 0=.00534T^2-133T+13.496 t=203.37 or t=45.93 45.93 more closely fits self critique assessment: 3
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09:51:58 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) confidence assessment: 3
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09:52:11 ** 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:
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09:53:17 What three points on your graph did you use as a basis for your model?
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RESPONSE --> (0,1) (10, 1.790569) (20, 2.118034) confidence assessment: 3
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09:53:27 ** STUDENT SOLUTION CONTINUED: (20, 2.118034) (50, 2.767767) (100, 3.5)**
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RESPONSE --> ok self critique assessment: 3
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09:53:51 Give the first of your three equations.
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RESPONSE --> a+b+c=1 confidence assessment: 3
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09:53:59 ** STUDENT SOLUTION CONTINUED: 400a + 20b + c = 2.118034**
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RESPONSE --> ok self critique assessment: 3
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09:54:26 Give the second of your three equations.
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RESPONSE --> a100+b10+c=1.790569 confidence assessment: 3
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09:54:32 ** STUDENT SOLUTION CONTINUED: 2500a + 50b + c = 2.767767 **
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RESPONSE --> self critique assessment: 3
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09:54:52 Give the third of your three equations.
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RESPONSE --> a400+b20+c=2.118034 confidence assessment: 3
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09:54:58 ** STUDENT SOLUTION CONTINUED: 10,000a + 100b + c = 3.5 **
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RESPONSE --> self critique assessment: 3
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09:55:31 Give the first of the equations you got when you eliminated c.
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RESPONSE --> a100+b10=1.790569 confidence assessment: 3
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09:55:38 ** STUDENT SOLUTION CONTINUED: 7500a + 50b = .732233. **
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RESPONSE --> self critique assessment: 3
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09:55:55 ** STUDENT SOLUTION CONTINUED: 7500a + 50b = .732233. **
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RESPONSE --> a0+b0=1 self critique assessment: 3
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09:57:48 Give the second of the equations you got when you eliminated c.
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RESPONSE --> a0+b0=1 confidence assessment: 3
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09:58:03 ** STUDENT SOLUTION CONTINUED: Subracting the first equation from the third I go 9600a + 80b = 1.381966 **
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RESPONSE --> self critique assessment: ok
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09:59:39 Explain how you solved for one of the variables.
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RESPONSE --> I multiplied each new equation by the opposite equal coefficient to eliminate b added them together and then solved for a confidence assessment: 3
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09:59:49 ** 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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10:00:12 What values did you get for a and b?
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RESPONSE --> a=-.00232 b=.20226 confidence assessment: 3
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10:00:17 ** STUDENT SOLUTION CONTINUED: a = -.0000876638 b = .01727 **
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RESPONSE --> self critique assessment: 3
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10:00:39 What did you then get for c?
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RESPONSE --> c=1 confidence assessment: 3
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10:01:34 What is your function model?
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RESPONSE --> The function model is y=-.00232t^2+.20226t+1 confidence assessment: 3
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10:01:40 ** y = -.0000876638 x^2 + (.01727)x + 1.773 **
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RESPONSE --> self critique assessment: 3
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10:03:40 What is your percent-of-review prediction for the given range of grades (give grade range also)?
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RESPONSE --> confidence assessment: 0
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10:05:21 What grade average corresponds to the given percent of review (give grade average also)?
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RESPONSE --> 3=-.00232t^2+.20226t+1 Solve the quation by turning it into standard form 0=-.00232t^2+.20226t-2 Solving for t t=[-.20226+-sqrt(.0409-4(-.00232)(-2)]/-.00464 t=(-.20226+.0476)/-.00464=33.332 t=(-.20226-.0476)/-.00464=53.849 t=53.849 will fit best with the model confidence assessment: 2
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10:06:30 ** Here you plug in your percent of review for the variable. For example if your model is y = -.00028 x^2 + .06 x + .5 and the percent of review is 75, you plug in 75 for x and evaluate the result. The result gives you the grade average corresponding to the percent of review according to the model. **
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RESPONSE --> ok self critique assessment: 3
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10:07:42 How well does your model fit the data (support your answer)?
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RESPONSE --> I did not think that this data set fit the model well. I determined that they could only received a 2.3328 GPA from reviewing 80% but that they could get a 4.0 from only 68.219% of the assignments. The results were not consistant. confidence assessment: 2
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10:07:56 ** 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: 3
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10:15:34 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, 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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10:15:45 ** 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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10:16:34 What three points on your graph did you use as a basis for your model?
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RESPONSE --> (1,935.1395) (2, 264.4411), (3, 105.1209) confidence assessment: 3
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10:16:40 ** 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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10:17:36 Give the first of your three equations.
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RESPONSE --> 4a+2b+c=264.4411 confidence assessment: 3
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10:17:42 ** STUDENT SOLUTION CONTINUED: 4a + 2b + c = 264.4411**
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RESPONSE --> self critique assessment: 3
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10:18:04 Give the second of your three equations.
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RESPONSE --> 1a+1b+c=935.1395 confidence assessment: 3
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10:18:10 ** STUDENT SOLUTION CONTINUED: 16a + 4b + c = 61.01488**
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RESPONSE --> self critique assessment: 3
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10:19:11 Give the third of your three equations.
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RESPONSE --> 3a+1b+c=-670.6984 confidence assessment: 3
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10:19:17 ** STUDENT SOLUTION CONTINUED: 64a + 8b + c = 16.27232**
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RESPONSE --> self critique assessment: 3
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10:20:24 Give the first of the equations you got when you eliminated c.
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RESPONSE --> 9a+3b=105.1209 confidence assessment: 3
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10:20:29 ** STUDENT SOLUTION CONTINUED: 48a + 4b = -44.74256**
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RESPONSE --> self critique assessment: 3
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10:20:46 Give the second of the equations you got when you eliminated c.
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RESPONSE --> 4a+2b=264.4411 confidence assessment: 3
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10:20:51 ** STUDENT SOLUTION CONTINUED: 60a + 6b = -248.16878**
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RESPONSE --> self critique assessment: 3
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10:22:57 Explain how you solved for one of the variables.
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RESPONSE --> acutally I entered the wrong equations on the previous screen they should have been 3a+1b=-670.6984 and 5a+1b=-159.3202 I multiplied the first equation by 1 and the second by -1 to eliminate b and then added to two equations togeter to get 3a=-670.6984 a=255.6891 confidence assessment: 3
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10:23:02 ** 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 --> self critique assessment: 3
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10:23:42 What values did you get for a and b?
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RESPONSE --> a=255.6891 b=379.15765 confidence assessment: 3
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10:23:47 ** STUDENT SOLUTION CONTINUED: a = 15.088, b = -192.24 **
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RESPONSE --> self critique assessment: 3
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10:24:05 What did you then get for c?
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RESPONSE --> c=300.29275 confidence assessment: 3
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10:24:13 ** STUDENT SOLUTION CONTINUED: c = 588.5691**
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RESPONSE --> self critique assessment: 3
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10:24:55 What is your function model?
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RESPONSE --> y=255.6891t^2+379.15765t+300.29275 confidence assessment: 3
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10:25:01 ** STUDENT SOLUTION CONTINUED: y = (15.088) x^2 - (192.24)x + 588.5691 **
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RESPONSE --> self critique assessment: 3
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10:26:48 What is your illumination prediction for the given distance (give distance also)?
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RESPONSE --> The illumination distance at 1.6 Earth distances would be 545.73225 w/m^2 confidence assessment: 3
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10:26:56 ** 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 --> self critique assessment: 3
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10:27:54 What distances correspond to the given illumination range (give illumination range also)?
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RESPONSE --> For the illumination range of 25 to 100 watts per square meter I can tell by the chart that the range would be 3-6 AU. confidence assessment: 2
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10:28:19 ** 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: 0
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