course Mth 173 _ꦬz뼸|f唡assignment #001
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03:46:13 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 --> First (95,0) Third (60,20) Fifth (41,40)
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03:46:19 ** Continue to the next question **
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RESPONSE --> ok
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03:52:14 According to your graph what would be the temperatures at clock times 7, 19 and 31?
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RESPONSE --> According to my graph I would estimate the temps when diplayed at temp vs. time to be (78,7) (62,19) (57,31)
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03:52:20 ** Continue to the next question **
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RESPONSE --> ok
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03:53:28 What three points did you use as a basis for your quadratic model (express as ordered pairs)?
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RESPONSE --> (60,20) (49,30) (41,40)
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03:54:46 ** 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 made the mistake of using 3 adjacent pairs which is probably why my model wasnt very accurate at the extreme ends.
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03:55:41 What is the first equation you got when you substituted into the form of a quadratic?
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RESPONSE --> 60=400a + 20b + c
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03:56:04 ** 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 --> ok
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03:56:28 What is the second equation you got when you substituted into the form of a quadratic?
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RESPONSE --> 49= 900a + 30b + c
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03:56:33 ** STUDENT SOLUTION CONTINUED: The equation that I got from my second data point was 400a + 20b + c = 60 **
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RESPONSE --> ok
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03:56:57 What is the third equation you got when you substituted into the form of a quadratic?
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RESPONSE --> 41= 1600a + 40b + c
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03:57:01 ** STUDENT SOLUTION CONTINUED: The equation that I got from my third data point was 3600a + 60b + c = 30. **
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RESPONSE --> ok
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03:58:49 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 used my first to equations to eliminate c and ended up with -11 = 500a + 10b
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03:58:53 ** 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
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04:01:24 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 next subtracted 49=900a+30b+c from 41=1600a+40b+c and ended up with -8= 700a+10b
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04:01:35 ** 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
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04:02:27 Which variable did you eliminate from these two equations, and what was its value?
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RESPONSE --> I eliminated b from these two equations to find a and found the value of a to be .015
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04:02:37 ** 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
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04:03:28 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 substituted a into the quation -11=500a+10b and got -1.85 for the value of b
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04:03:34 ** 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
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04:03:49 What is the value of c obtained from substituting into one of the original equations?
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RESPONSE --> c=91
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04:03:56 ** STUDENT SOLUTION CONTINUED: By substituting both a and b into the original equations, I found that c = 93 **
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RESPONSE --> ok
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04:04:35 What is the resulting quadratic model?
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RESPONSE --> temp= .015t^2-1.85t+91
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04:04:43 ** 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
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04:05:27 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 --> 91,74, and 60 which had deviations of +4,+1, and 0
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04:05:37 ** 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
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04:07:04 What was your average deviation?
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RESPONSE --> My average deviation was +2.125
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04:07:10 ** STUDENT SOLUTION CONTINUED: My average deviation was .6 **
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RESPONSE --> ok
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04:07:48 Is there a pattern to your deviations?
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RESPONSE --> The farther you got in either direction from the points I used for my model the higher the deviation seemed to get.
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04:08:01 ** 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
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04:08:25 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 have studied this and understand it.
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04:08:32 ** STUDENT SOLUTION CONTINUED: Yes, I do completely understand the process after studying these outlines and explanations. **
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RESPONSE --> ok
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04:11:46 Have you memorized the steps of the modeling process, and are you gonna remember them forever? Convince me.
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RESPONSE --> I have memorized the steps of the modeling process and I will remember them as long as I continue to use them. First, I take three points that are not beside each other and plug them into 3 separate quadratic equations. Then I use elimination to solve for a,b, and c. The result is my model.
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04:11:59 ** 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
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04:19:58 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 --> I used the simulated data version 1 which gave me (63.7,5.3) (54.8,10.6) (46,15.9) (37.7,21.2) (32,26.5) (26.6,31.8)
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04:31:23 ** 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
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14:18:41 What three points on your graph did you use as a basis for your model?
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RESPONSE --> (63.7,5.3) (46,15.9) (26.6,31.8)
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14:18:59 ** 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
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14:19:27 Give the first of your three equations.
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RESPONSE --> 63.7=28.09a+5.3b+c
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14:19:37 ** 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
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14:19:59 Give the second of your three equations.
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RESPONSE --> 46=252.8a+15.9b+c
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14:20:03 ** STUDENT SOLUTION CONTINUED: The point (15.9, 46) gives me the equation 252.81a +15.9b + c = 46 **
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RESPONSE --> ok
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14:20:26 Give the third of your three equations.
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RESPONSE --> 26.6=1011.24a+31.8b+c
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14:20:31 ** STUDENT SOLUTION CONTINUED: The point (26.5,32) gives me the equation 702.25a + 26.5b + c = 32. **
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RESPONSE --> ok
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14:20:57 Give the first of the equations you got when you eliminated c.
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RESPONSE --> -17.7=224.71a+10.6b
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14:21:04 ** STUDENT SOLUTION CONTINUED: Subtracting the second equation from the third gave me 449.44a + 10.6b = -14. **
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RESPONSE --> ok
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14:21:39 Give the second of the equations you got when you eliminated c.
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RESPONSE --> -19.4=758.44a+15.9b
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14:21:51 ** ** STUDENT SOLUTION CONTINUED: Subtracting the first equation from the third gave me 674.16a + 21.2b = -31.7. **
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RESPONSE --> ok
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14:23:07 Explain how you solved for one of the variables.
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RESPONSE --> I multiplyed each side by the amount of the opposite equations coefficient for b which gave both equations the same value for the b coefficient. Then I subtracted one equation from the other to eliminate b. Then I solved for a using algebra giving me a=.01
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14:23:19 ** 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
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14:23:37 What values did you get for a and b?
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RESPONSE --> a=.01 b=-1.9
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14:23:44 ** STUDENT SOLUTION CONTINUED: a = .0165, b = -2 **
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RESPONSE --> ok
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14:24:06 What did you then get for c?
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RESPONSE --> I then used a and b to get c which gave me c=73.5
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14:24:13 ** STUDENT SOLUTION CONTINUED: c = 73.4 **
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RESPONSE --> ok
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14:24:48 What is your function model?
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RESPONSE --> y=.01t^2-1.9t+73.5
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14:24:58 ** STUDENT SOLUTION CONTINUED: y = (.0165)x^2 + (-2)x + 73.4. **
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RESPONSE --> ok
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14:28:15 What is your depth prediction for the given clock time (give clock time also)?
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RESPONSE --> I plugged a time of 15.9 into the function and the result was a depth of 45.8
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14:28:25 ** STUDENT SOLUTION CONTINUED: The given clock time was 46 seconds, and my depth prediction was 16.314 cm.**
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RESPONSE --> ok
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14:28:57 What clock time corresponds to the given depth (give depth also)?
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RESPONSE --> This corresponded with the same clock time given which had a depth of 46cm
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14:29:03 ** 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
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14:37:57 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 --> (1,0) (1.79,10) (2.11,20) (2.36,30) (2.58,40) (2.76,50) (2.93,60) (3.09,70) (3.23,80) (3.37,90) (3.5,100)
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14:38:10 ** 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 --> ok
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14:39:23 What three points on your graph did you use as a basis for your model?
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RESPONSE --> (1,0) (2.118034,20) (2.581139,40)
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14:39:35 ** STUDENT SOLUTION CONTINUED: (20, 2.118034) (50, 2.767767) (100, 3.5)**
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RESPONSE --> ok
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14:40:25 Give the first of your three equations.
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RESPONSE --> c=1
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14:57:32 ** STUDENT SOLUTION CONTINUED: 400a + 20b + c = 2.118034**
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RESPONSE --> For the sake of making my work easier I am changing my first equation to use the point (2.118034,20) therefore 20=4.486068a+2.118034b+c
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14:57:58 Give the second of your three equations.
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RESPONSE --> 40=6.662278a+2.581139b+c
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15:12:41 ** STUDENT SOLUTION CONTINUED: 2500a + 50b + c = 2.767767 **
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RESPONSE --> I got my numbers back wards when entering my info into my equations therefore my first two equations would be 2.118034=400a+20b+c 2.581139=1600a+40b+c and my 3rd one would be 2.936492=3600a+60b+c
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15:13:03 Give the third of your three equations.
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RESPONSE --> 2.936492=3600a+60b+c
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15:13:09 ** STUDENT SOLUTION CONTINUED: 10,000a + 100b + c = 3.5 **
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RESPONSE --> ok
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15:13:31 Give the first of the equations you got when you eliminated c.
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RESPONSE --> .463105=1200a+20b
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15:13:35 ** STUDENT SOLUTION CONTINUED: 7500a + 50b = .732233. **
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RESPONSE --> ok
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15:13:50 Give the second of the equations you got when you eliminated c.
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RESPONSE --> .818458=3200a+40b
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15:13:58 ** STUDENT SOLUTION CONTINUED: Subracting the first equation from the third I go 9600a + 80b = 1.381966 **
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RESPONSE --> ok
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15:14:52 Explain how you solved for one of the variables.
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RESPONSE --> To solve for one of the variables i multiplyed each equation by the coefficient for b from the opposite equation. Therefore for a I got a=-.00013469
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15:15:00 ** 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 --> ok
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15:15:32 What values did you get for a and b?
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RESPONSE --> a=-.00013469 b=.01507385
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15:15:42 ** STUDENT SOLUTION CONTINUED: a = -.0000876638 b = .01727 **
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RESPONSE --> ok
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15:15:56 What did you then get for c?
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RESPONSE --> c=1.870433
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15:16:00 ** STUDENT SOLUTION CONTINUED: c = 1.773. **
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RESPONSE --> ok
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15:16:55 What is your function model?
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RESPONSE --> y=-.00013469t^2 + .01507385t + 1.870433
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15:17:36 y = (0) x^2 + (.01727)x + 1.773
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RESPONSE --> ok
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15:20:02 What is your percent-of-review prediction for the given range of grades (give grade range also)?
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RESPONSE --> I do not understand what this question is asking
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15:20:45 ** 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 --> ok
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15:20:59 What grade average corresponds to the given percent of review (give grade average also)?
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RESPONSE --> I understand what was done.
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15:21:04 ** 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
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15:25:43 How well does your model fit the data (support your answer)?
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RESPONSE --> I understand this. My model fits well.
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15:25:51 ** 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 --> ok
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15:40:12 illumination vs. distance Give your data in the form of illumination vs. distance ordered pairs.
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RESPONSE --> (935.1395,1) (246.4411,2) (105.1209,3) (61.01488.4) (43.06238,5) (25.91537,6) (19.92772,7) (16.27232,8) (11.28082,9) (9.484465,10)
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15:40:17 ** 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 --> ok
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15:40:54 What three points on your graph did you use as a basis for your model?
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RESPONSE --> (246.4411,2) (43.06238,5) (11.28082,9)
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15:40:58 ** STUDENT SOLUTION CONTINUED: (2, 264.4411) (4, 61.01488) (8, 16.27232) **
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RESPONSE --> ok
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15:41:21 Give the first of your three equations.
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RESPONSE --> 246.4411 = 4a + 2b + c
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15:41:24 ** STUDENT SOLUTION CONTINUED: 4a + 2b + c = 264.4411**
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RESPONSE --> ok
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15:41:44 Give the second of your three equations.
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RESPONSE --> 43.06238 = 25a + 5b + c
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15:41:48 ** STUDENT SOLUTION CONTINUED: 16a + 4b + c = 61.01488**
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RESPONSE --> ok
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15:42:06 Give the third of your three equations.
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RESPONSE --> 11.28082 = 81a + 9b + c
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15:42:11 ** STUDENT SOLUTION CONTINUED: 64a + 8b + c = 16.27232**
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RESPONSE --> ok
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15:42:31 Give the first of the equations you got when you eliminated c.
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RESPONSE --> 203.37872 = -24a - 3b
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15:42:35 ** STUDENT SOLUTION CONTINUED: 48a + 4b = -44.74256**
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RESPONSE --> ok
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15:42:56 Give the second of the equations you got when you eliminated c.
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RESPONSE --> 235.16028 = -77a - 7b
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15:43:01 ** STUDENT SOLUTION CONTINUED: 60a + 6b = -248.16878**
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RESPONSE --> ok
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15:43:32 Explain how you solved for one of the variables.
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RESPONSE --> I eliminated b by multiplying each equation by the coefficient for b from the opposite equation.
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15:43:37 ** 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 --> ok
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15:43:56 What values did you get for a and b?
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RESPONSE --> a= 11.39 b= -158.91
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15:44:06 ** STUDENT SOLUTION CONTINUED: a = 15.088, b = -192.24 **
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RESPONSE --> ok
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15:44:17 What did you then get for c?
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RESPONSE --> c= 475.64
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15:44:21 ** STUDENT SOLUTION CONTINUED: c = 588.5691**
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RESPONSE --> ok
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15:44:53 What is your function model?
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RESPONSE --> y = 11.39t^2 - 158.91t + 475.64
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15:44:58 ** STUDENT SOLUTION CONTINUED: y = (15.088) x^2 - (192.24)x + 588.5691 **
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RESPONSE --> ok
.................................................
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15:48:06 What is your illumination prediction for the given distance (give distance also)?
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RESPONSE --> When I figured a distance of 6 I got an output of an illumination = -67.78 which I would assume to be impossible
.................................................
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15:48:19 ** 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 --> ok
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15:49:10 What distances correspond to the given illumination range (give illumination range also)?
......!!!!!!!!...................................
RESPONSE --> They didnt correspond. I must have obtained a poor model using the points that I did, that is unless I messed up on my math which I have rechecked and did not find an error.
.................................................
......!!!!!!!!...................................
15:49:14 ** 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 50% - 69% if 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 --> ok
.................................................
"
course Mth 173 _ꦬz뼸|f唡assignment #001
......!!!!!!!!...................................
03:46:13 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 --> First (95,0) Third (60,20) Fifth (41,40)
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......!!!!!!!!...................................
03:46:19 ** Continue to the next question **
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RESPONSE --> ok
.................................................
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03:52:14 According to your graph what would be the temperatures at clock times 7, 19 and 31?
......!!!!!!!!...................................
RESPONSE --> According to my graph I would estimate the temps when diplayed at temp vs. time to be (78,7) (62,19) (57,31)
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03:52:20 ** Continue to the next question **
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RESPONSE --> ok
.................................................
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03:53:28 What three points did you use as a basis for your quadratic model (express as ordered pairs)?
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RESPONSE --> (60,20) (49,30) (41,40)
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......!!!!!!!!...................................
03:54:46 ** 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). **
......!!!!!!!!...................................
RESPONSE --> I made the mistake of using 3 adjacent pairs which is probably why my model wasnt very accurate at the extreme ends.
.................................................
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03:55:41 What is the first equation you got when you substituted into the form of a quadratic?
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RESPONSE --> 60=400a + 20b + c
.................................................
......!!!!!!!!...................................
03:56:04 ** STUDENT SOLUTION CONTINUED: The equation that I got from the first data point (10,75) was 100a + 10b +c = 75.**
......!!!!!!!!...................................
RESPONSE --> ok
.................................................
......!!!!!!!!...................................
03:56:28 What is the second equation you got when you substituted into the form of a quadratic?
......!!!!!!!!...................................
RESPONSE --> 49= 900a + 30b + c
.................................................
......!!!!!!!!...................................
03:56:33 ** STUDENT SOLUTION CONTINUED: The equation that I got from my second data point was 400a + 20b + c = 60 **
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RESPONSE --> ok
.................................................
......!!!!!!!!...................................
03:56:57 What is the third equation you got when you substituted into the form of a quadratic?
......!!!!!!!!...................................
RESPONSE --> 41= 1600a + 40b + c
.................................................
......!!!!!!!!...................................
03:57:01 ** STUDENT SOLUTION CONTINUED: The equation that I got from my third data point was 3600a + 60b + c = 30. **
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RESPONSE --> ok
.................................................
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03:58:49 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 used my first to equations to eliminate c and ended up with -11 = 500a + 10b
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......!!!!!!!!...................................
03:58:53 ** 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
.................................................
......!!!!!!!!...................................
04:01:24 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?
......!!!!!!!!...................................
RESPONSE --> I next subtracted 49=900a+30b+c from 41=1600a+40b+c and ended up with -8= 700a+10b
.................................................
......!!!!!!!!...................................
04:01:35 ** 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
.................................................
......!!!!!!!!...................................
04:02:27 Which variable did you eliminate from these two equations, and what was its value?
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RESPONSE --> I eliminated b from these two equations to find a and found the value of a to be .015
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04:02:37 ** 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
.................................................
......!!!!!!!!...................................
04:03:28 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 substituted a into the quation -11=500a+10b and got -1.85 for the value of b
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04:03:34 ** 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
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04:03:49 What is the value of c obtained from substituting into one of the original equations?
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RESPONSE --> c=91
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......!!!!!!!!...................................
04:03:56 ** STUDENT SOLUTION CONTINUED: By substituting both a and b into the original equations, I found that c = 93 **
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RESPONSE --> ok
.................................................
......!!!!!!!!...................................
04:04:35 What is the resulting quadratic model?
......!!!!!!!!...................................
RESPONSE --> temp= .015t^2-1.85t+91
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......!!!!!!!!...................................
04:04:43 ** 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
.................................................
......!!!!!!!!...................................
04:05:27 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 --> 91,74, and 60 which had deviations of +4,+1, and 0
.................................................
......!!!!!!!!...................................
04:05:37 ** 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
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04:07:04 What was your average deviation?
......!!!!!!!!...................................
RESPONSE --> My average deviation was +2.125
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04:07:10 ** STUDENT SOLUTION CONTINUED: My average deviation was .6 **
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RESPONSE --> ok
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04:07:48 Is there a pattern to your deviations?
......!!!!!!!!...................................
RESPONSE --> The farther you got in either direction from the points I used for my model the higher the deviation seemed to get.
.................................................
......!!!!!!!!...................................
04:08:01 ** 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
.................................................
......!!!!!!!!...................................
04:08:25 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 have studied this and understand it.
.................................................
......!!!!!!!!...................................
04:08:32 ** STUDENT SOLUTION CONTINUED: Yes, I do completely understand the process after studying these outlines and explanations. **
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RESPONSE --> ok
.................................................
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04:11:46 Have you memorized the steps of the modeling process, and are you gonna remember them forever? Convince me.
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RESPONSE --> I have memorized the steps of the modeling process and I will remember them as long as I continue to use them. First, I take three points that are not beside each other and plug them into 3 separate quadratic equations. Then I use elimination to solve for a,b, and c. The result is my model.
.................................................
......!!!!!!!!...................................
04:11:59 ** 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
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04:19:58 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 --> I used the simulated data version 1 which gave me (63.7,5.3) (54.8,10.6) (46,15.9) (37.7,21.2) (32,26.5) (26.6,31.8)
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......!!!!!!!!...................................
04:31:23 ** 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
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14:18:41 What three points on your graph did you use as a basis for your model?
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RESPONSE --> (63.7,5.3) (46,15.9) (26.6,31.8)
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14:18:59 ** 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
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14:19:27 Give the first of your three equations.
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RESPONSE --> 63.7=28.09a+5.3b+c
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14:19:37 ** 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
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14:19:59 Give the second of your three equations.
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RESPONSE --> 46=252.8a+15.9b+c
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14:20:03 ** STUDENT SOLUTION CONTINUED: The point (15.9, 46) gives me the equation 252.81a +15.9b + c = 46 **
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RESPONSE --> ok
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14:20:26 Give the third of your three equations.
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RESPONSE --> 26.6=1011.24a+31.8b+c
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14:20:31 ** STUDENT SOLUTION CONTINUED: The point (26.5,32) gives me the equation 702.25a + 26.5b + c = 32. **
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RESPONSE --> ok
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14:20:57 Give the first of the equations you got when you eliminated c.
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RESPONSE --> -17.7=224.71a+10.6b
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14:21:04 ** STUDENT SOLUTION CONTINUED: Subtracting the second equation from the third gave me 449.44a + 10.6b = -14. **
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RESPONSE --> ok
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14:21:39 Give the second of the equations you got when you eliminated c.
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RESPONSE --> -19.4=758.44a+15.9b
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14:21:51 ** ** STUDENT SOLUTION CONTINUED: Subtracting the first equation from the third gave me 674.16a + 21.2b = -31.7. **
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RESPONSE --> ok
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14:23:07 Explain how you solved for one of the variables.
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RESPONSE --> I multiplyed each side by the amount of the opposite equations coefficient for b which gave both equations the same value for the b coefficient. Then I subtracted one equation from the other to eliminate b. Then I solved for a using algebra giving me a=.01
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14:23:19 ** 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
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14:23:37 What values did you get for a and b?
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RESPONSE --> a=.01 b=-1.9
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14:23:44 ** STUDENT SOLUTION CONTINUED: a = .0165, b = -2 **
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RESPONSE --> ok
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14:24:06 What did you then get for c?
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RESPONSE --> I then used a and b to get c which gave me c=73.5
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14:24:13 ** STUDENT SOLUTION CONTINUED: c = 73.4 **
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RESPONSE --> ok
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14:24:48 What is your function model?
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RESPONSE --> y=.01t^2-1.9t+73.5
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14:24:58 ** STUDENT SOLUTION CONTINUED: y = (.0165)x^2 + (-2)x + 73.4. **
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RESPONSE --> ok
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14:28:15 What is your depth prediction for the given clock time (give clock time also)?
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RESPONSE --> I plugged a time of 15.9 into the function and the result was a depth of 45.8
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14:28:25 ** STUDENT SOLUTION CONTINUED: The given clock time was 46 seconds, and my depth prediction was 16.314 cm.**
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RESPONSE --> ok
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14:28:57 What clock time corresponds to the given depth (give depth also)?
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RESPONSE --> This corresponded with the same clock time given which had a depth of 46cm
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14:29:03 ** 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
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14:37:57 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 --> (1,0) (1.79,10) (2.11,20) (2.36,30) (2.58,40) (2.76,50) (2.93,60) (3.09,70) (3.23,80) (3.37,90) (3.5,100)
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14:38:10 ** 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 --> ok
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14:39:23 What three points on your graph did you use as a basis for your model?
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RESPONSE --> (1,0) (2.118034,20) (2.581139,40)
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14:39:35 ** STUDENT SOLUTION CONTINUED: (20, 2.118034) (50, 2.767767) (100, 3.5)**
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RESPONSE --> ok
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14:40:25 Give the first of your three equations.
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RESPONSE --> c=1
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14:57:32 ** STUDENT SOLUTION CONTINUED: 400a + 20b + c = 2.118034**
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RESPONSE --> For the sake of making my work easier I am changing my first equation to use the point (2.118034,20) therefore 20=4.486068a+2.118034b+c
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14:57:58 Give the second of your three equations.
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RESPONSE --> 40=6.662278a+2.581139b+c
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15:12:41 ** STUDENT SOLUTION CONTINUED: 2500a + 50b + c = 2.767767 **
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RESPONSE --> I got my numbers back wards when entering my info into my equations therefore my first two equations would be 2.118034=400a+20b+c 2.581139=1600a+40b+c and my 3rd one would be 2.936492=3600a+60b+c
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15:13:03 Give the third of your three equations.
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RESPONSE --> 2.936492=3600a+60b+c
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15:13:09 ** STUDENT SOLUTION CONTINUED: 10,000a + 100b + c = 3.5 **
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RESPONSE --> ok
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15:13:31 Give the first of the equations you got when you eliminated c.
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RESPONSE --> .463105=1200a+20b
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15:13:35 ** STUDENT SOLUTION CONTINUED: 7500a + 50b = .732233. **
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RESPONSE --> ok
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15:13:50 Give the second of the equations you got when you eliminated c.
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RESPONSE --> .818458=3200a+40b
.................................................
......!!!!!!!!...................................
15:13:58 ** STUDENT SOLUTION CONTINUED: Subracting the first equation from the third I go 9600a + 80b = 1.381966 **
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RESPONSE --> ok
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15:14:52 Explain how you solved for one of the variables.
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RESPONSE --> To solve for one of the variables i multiplyed each equation by the coefficient for b from the opposite equation. Therefore for a I got a=-.00013469
.................................................
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15:15:00 ** 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 --> ok
.................................................
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15:15:32 What values did you get for a and b?
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RESPONSE --> a=-.00013469 b=.01507385
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15:15:42 ** STUDENT SOLUTION CONTINUED: a = -.0000876638 b = .01727 **
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RESPONSE --> ok
.................................................
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15:15:56 What did you then get for c?
......!!!!!!!!...................................
RESPONSE --> c=1.870433
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......!!!!!!!!...................................
15:16:00 ** STUDENT SOLUTION CONTINUED: c = 1.773. **
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RESPONSE --> ok
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......!!!!!!!!...................................
15:16:55 What is your function model?
......!!!!!!!!...................................
RESPONSE --> y=-.00013469t^2 + .01507385t + 1.870433
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15:17:36 y = (0) x^2 + (.01727)x + 1.773
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RESPONSE --> ok
.................................................
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15:20:02 What is your percent-of-review prediction for the given range of grades (give grade range also)?
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RESPONSE --> I do not understand what this question is asking
.................................................
......!!!!!!!!...................................
15:20:45 ** 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 --> ok
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15:20:59 What grade average corresponds to the given percent of review (give grade average also)?
......!!!!!!!!...................................
RESPONSE --> I understand what was done.
.................................................
......!!!!!!!!...................................
15:21:04 ** 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
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15:25:43 How well does your model fit the data (support your answer)?
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RESPONSE --> I understand this. My model fits well.
.................................................
......!!!!!!!!...................................
15:25:51 ** 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. **
......!!!!!!!!...................................
RESPONSE --> ok
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15:40:12 illumination vs. distance Give your data in the form of illumination vs. distance ordered pairs.
......!!!!!!!!...................................
RESPONSE --> (935.1395,1) (246.4411,2) (105.1209,3) (61.01488.4) (43.06238,5) (25.91537,6) (19.92772,7) (16.27232,8) (11.28082,9) (9.484465,10)
.................................................
......!!!!!!!!...................................
15:40:17 ** 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 --> ok
.................................................
......!!!!!!!!...................................
15:40:54 What three points on your graph did you use as a basis for your model?
......!!!!!!!!...................................
RESPONSE --> (246.4411,2) (43.06238,5) (11.28082,9)
.................................................
......!!!!!!!!...................................
15:40:58 ** STUDENT SOLUTION CONTINUED: (2, 264.4411) (4, 61.01488) (8, 16.27232) **
......!!!!!!!!...................................
RESPONSE --> ok
.................................................
......!!!!!!!!...................................
15:41:21 Give the first of your three equations.
......!!!!!!!!...................................
RESPONSE --> 246.4411 = 4a + 2b + c
.................................................
......!!!!!!!!...................................
15:41:24 ** STUDENT SOLUTION CONTINUED: 4a + 2b + c = 264.4411**
......!!!!!!!!...................................
RESPONSE --> ok
.................................................
......!!!!!!!!...................................
15:41:44 Give the second of your three equations.
......!!!!!!!!...................................
RESPONSE --> 43.06238 = 25a + 5b + c
.................................................
......!!!!!!!!...................................
15:41:48 ** STUDENT SOLUTION CONTINUED: 16a + 4b + c = 61.01488**
......!!!!!!!!...................................
RESPONSE --> ok
.................................................
......!!!!!!!!...................................
15:42:06 Give the third of your three equations.
......!!!!!!!!...................................
RESPONSE --> 11.28082 = 81a + 9b + c
.................................................
......!!!!!!!!...................................
15:42:11 ** STUDENT SOLUTION CONTINUED: 64a + 8b + c = 16.27232**
......!!!!!!!!...................................
RESPONSE --> ok
.................................................
......!!!!!!!!...................................
15:42:31 Give the first of the equations you got when you eliminated c.
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RESPONSE --> 203.37872 = -24a - 3b
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15:42:35 ** STUDENT SOLUTION CONTINUED: 48a + 4b = -44.74256**
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RESPONSE --> ok
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15:42:56 Give the second of the equations you got when you eliminated c.
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RESPONSE --> 235.16028 = -77a - 7b
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15:43:01 ** STUDENT SOLUTION CONTINUED: 60a + 6b = -248.16878**
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RESPONSE --> ok
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15:43:32 Explain how you solved for one of the variables.
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RESPONSE --> I eliminated b by multiplying each equation by the coefficient for b from the opposite equation.
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15:43:37 ** 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 --> ok
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15:43:56 What values did you get for a and b?
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RESPONSE --> a= 11.39 b= -158.91
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15:44:06 ** STUDENT SOLUTION CONTINUED: a = 15.088, b = -192.24 **
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RESPONSE --> ok
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15:44:17 What did you then get for c?
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RESPONSE --> c= 475.64
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15:44:21 ** STUDENT SOLUTION CONTINUED: c = 588.5691**
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RESPONSE --> ok
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15:44:53 What is your function model?
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RESPONSE --> y = 11.39t^2 - 158.91t + 475.64
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15:44:58 ** STUDENT SOLUTION CONTINUED: y = (15.088) x^2 - (192.24)x + 588.5691 **
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RESPONSE --> ok
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15:48:06 What is your illumination prediction for the given distance (give distance also)?
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RESPONSE --> When I figured a distance of 6 I got an output of an illumination = -67.78 which I would assume to be impossible
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15:48:19 ** 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 --> ok
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15:49:10 What distances correspond to the given illumination range (give illumination range also)?
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RESPONSE --> They didnt correspond. I must have obtained a poor model using the points that I did, that is unless I messed up on my math which I have rechecked and did not find an error.
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15:49:14 ** 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 50% - 69% if 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 --> ok
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