course Mth 271 I only got this much done.....I will do the rest of it in the morning.....This is alot of typing to do all at once....Thanks
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21:57:44 `questionNumber 20000 What were temperature and time for the first, third and fifth data points (express as temp vs clock time ordered pairs)?
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RESPONSE --> (0,72) (20,60) (40,46)
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22:00:19 `questionNumber 20000 According to your graph what would be the temperatures at clock times 7, 19 and 31?
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RESPONSE --> (7,68.03) (19,60.65) (31, 52.55)
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22:01:17 `questionNumber 20000 What three points did you use as a basis for your quadratic model (express as ordered pairs)?
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RESPONSE --> (20,60) (40,41) (60,30)
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22:02:30 `questionNumber 20000 ** A good choice of points `spreads' the points out rather than using three adjacent points. For example choosing the t = 10, 20, 30 points would not be a good idea here since the resulting model will fit those points perfectly but by the time we get to t = 60 the fit will probably not be good. Using for example t = 10, 30 and 60 would spread the three points out more and the solution would be more likely to fit the data. The solution to this problem by a former student will be outlined in the remaining `answers'. STUDENT SOLUTION (this student probably used a version different from the one you used; this solution is given here for comparison of the steps) For my quadratic model, I used the three points (10, 75) (20, 60) (60, 30). **
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RESPONSE --> I understand. I think that my points are pretty good and spread apart.
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22:03:13 `questionNumber 20000 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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22:03:57 `questionNumber 20000 What is the second 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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22:04:19 `questionNumber 20000 ** STUDENT SOLUTION CONTINUED: The equation that I got from my second data point was 400a + 20b + c = 60 **
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RESPONSE --> I understand that is what I got for my first one.
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22:04:48 `questionNumber 20000 What is the third equation you got when you substituted into the form of a quadratic?
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RESPONSE --> 30=3600a+60b+c
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22:05:04 `questionNumber 20000 ** STUDENT SOLUTION CONTINUED: The equation that I got from my third data point was 3600a + 60b + c = 30. **
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RESPONSE --> Thats what I got
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22:05:25 `questionNumber 20000 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 1-2 and 1-3
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22:06:11 `questionNumber 20000 ** 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 --> My first equation I got 30=-3200a+-40b
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22:06:50 `questionNumber 20000 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 --> 1-3......I got 19=-1200a=-12b
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22:07:21 `questionNumber 20000 ** STUDENT SOLUTION CONTINUED: This time, I subtracted the first equation from the third equation in order to again eliminate c. I obtained my second new equation: 3500a + 50b = -45**
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RESPONSE --> I didnt get this so I am not sure if mine will be correct.
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22:08:19 `questionNumber 20000 Which variable did you eliminate from these two equations, and what was its value?
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RESPONSE --> I got rid of the b when I mulitplied the 2nd equation by -2 and then I said that a= -.0025
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22:08:43 `questionNumber 20000 ** STUDENT SOLUTION CONTINUED: In order to solve for a and b, I decided to eliminate b because of its smaller value. In order to do this, I multiplied the first new equation by -5 -5 ( 3200a + 40b = -30) and multiplied the second new equation by 4 4 ( 3500a + 50b = -45) making the values of -200 b and 200 b cancel one another out. The resulting equation is -2000 a = -310. **
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RESPONSE --> I should have show more steps
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22:09:34 `questionNumber 20000 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 --> 30=(-.0025)(3200)+-40b......b=-.55
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22:10:12 `questionNumber 20000 ** STUDENT SOLUTION CONTINUED: After eliminating b, I solved a to equal .015 a = .015 I then substituted this value into the equation 3200 (.015) + 40b = -30 and solved to find that b = -1.95. **
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RESPONSE --> I guess this means that I have done somthing wrong, but I am not sure what I have done.
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22:10:57 `questionNumber 20000 What is the value of c obtained from substituting into one of the original equations?
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RESPONSE --> c=72 ......... 60=(-.0025)(400)+20(-.55)+c
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22:11:38 `questionNumber 20000 ** STUDENT SOLUTION CONTINUED: By substituting both a and b into the original equations, I found that c = 93 **
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RESPONSE --> I am not sure that if by using different equation that I would get different answers or what.
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22:12:28 `questionNumber 20000 What is the resulting quadratic model?
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RESPONSE --> y= -.0025 t^2+ -.55 t+ 72
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22:13:00 `questionNumber 20000 ** STUDENT SOLUTION CONTINUED: Therefore, the quadratic model that I obtained was y = (.015) x^2 - (1.95)x + 93. **
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RESPONSE --> I substitued in my values correctly but I am not sure that my values are correct.
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22:14:40 `questionNumber 20000 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 --> (0,72) deviation 23 (10, 66.25) deviation 8.75 (20, 60)deviation 0
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22:15:22 `questionNumber 20000 ** 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 --> My deviations are more off than yours.
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22:15:30 `questionNumber 20000 What was your average deviation?
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RESPONSE --> 3
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22:16:04 `questionNumber 20000 ** STUDENT SOLUTION CONTINUED: My average deviation was .6 **
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RESPONSE --> I added up all 8 and divided I think that this is how you would fined the deviation
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22:16:57 `questionNumber 20000 Is there a pattern to your deviations?
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RESPONSE --> No not really .....I see that I have alot of .75 at the ends of some of my #'s but that is not a pattern
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22:18:26 `questionNumber 20000 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 the steps in the modeling process. I do not know it by heart but I am studing it and understanding it.
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22:18:39 `questionNumber 20000 ** STUDENT SOLUTION CONTINUED: Yes, I do completely understand the process after studying these outlines and explanations. **
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RESPONSE --> I do understand
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22:20:02 `questionNumber 20000 ** STUDENT SOLUTION CONTINUED: Yes, sir, I have memorized the steps of the modeling process at this point. I also printed out an outline of the steps in order to refresh my memory often, so that I will remember them forever!!! INSTRUCTOR COMMENT: OK, I'm convinced. **
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RESPONSE --> I will work on memorizing them and will print this out to look at them alot.
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