course Mth 163 Sorry it took so long. I've been having internet trouble. Thanks. Ì»íã¯sܳ—¥ÔŸ‚íDÙä›Ïˆ°Ü§assignment #002
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18:52:40 `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 --> (95,0) (60,20) (41,40) confidence assessment: 3
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18:52:46 `questionNumber 20000 ** Continue to the next question **
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RESPONSE --> self critique assessment: 3
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18:54:08 `questionNumber 20000 According to your graph what would be the temperatures at clock times 7, 19 and 31?
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RESPONSE --> At clock times 7, 19, and 31 the temperatures are approximately 83 degrees, 62 degrees, and 47 degrees. confidence assessment: 3
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18:54:14 `questionNumber 20000 ** Continue to the next question **
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RESPONSE --> self critique assessment: 3
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18:55:13 `questionNumber 20000 What three points did you use as a basis for your quadratic model (express as ordered pairs)?
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RESPONSE --> (10, 75) (20, 60) (50, 35) I used these b/c they are all divisible by 5. confidence assessment: 3
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18:55:25 `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 --> ok self critique assessment: 3
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18:57:00 `questionNumber 20000 What is the first equation you got when you substituted into the form of a quadratic?
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RESPONSE --> y=at^2+bt+c y=100a+10b+c=75 confidence assessment: 3
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18:57:18 `questionNumber 20000 ** 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 same as me self critique assessment: 3
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18:57:52 `questionNumber 20000 What is the second equation you got when you substituted into the form of a quadratic?
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RESPONSE --> 400a+20b+c=60 confidence assessment: 3
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18:58:03 `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 --> ok self critique assessment: 3
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18:58:31 `questionNumber 20000 What is the third equation you got when you substituted into the form of a quadratic?
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RESPONSE --> 2500a+50b+c=35 confidence assessment: 3
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18:59:05 `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 --> ok self critique assessment: 3
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19:00:45 `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 subtracted equation w/ (20,60) from equation with (50, 35). This gave me 2100a+30b=-25 confidence assessment: 3
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19:00:54 `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 --> ok self critique assessment: 3
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19:02:34 `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 --> I subtracted equation w/ (10, 75) from equation w/ (50, 35). The solution is 2400a+40b=-40 confidence assessment: 3
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19:02:45 `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 --> ok self critique assessment: 3
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19:03:45 `questionNumber 20000 Which variable did you eliminate from these two equations, and what was the value of the variable for which you solved these equations?
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RESPONSE --> I eliminated b b/c it's the smallest. I ended up with the a value of .01 confidence assessment: 3
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19:03:54 `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 --> ok self critique assessment: 3
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19:05:13 `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 --> I plugged in the a value to 2100a+30b=-25 and got the value -1.53 for b. confidence assessment: 3
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19:06:10 `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 --> ok not the exact answer but I see it was just a math error on my part. self critique assessment: 3
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19:07:01 `questionNumber 20000 What is the value of c obtained from substituting into one of the original equations?
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RESPONSE --> My value of c after plugging in a and b into original equation is c=89.3. It will be a little off b/c my b value was a little off. confidence assessment: 2
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19:07:13 `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 --> ok self critique assessment: 2
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19:10:30 `questionNumber 20000 What is the resulting quadratic model?
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RESPONSE --> When I apply the correct answers into the equation it is y=.015(100)^2+(-1.95)(10)+93. y=223.5 confidence assessment: 2
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19:10:49 `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 --> ok self critique assessment: 2
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19:12:21 `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 --> The quadratic model gave me 0, 10, and20 min for the clock times confidence assessment: 2
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19:13:06 `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 --> I see how they got there answers I just didn't read the last half of the question but I do understand. self critique assessment: 2
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19:14:14 `questionNumber 20000 What was your average deviation?
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RESPONSE --> I think my average deviation was approx. 15 confidence assessment: 1
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19:14:41 `questionNumber 20000 ** STUDENT SOLUTION CONTINUED: My average deviation was .6 **
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RESPONSE --> I was totally off and I forgot to put in my percentage decimal. self critique assessment: 2
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19:15:24 `questionNumber 20000 Is there a pattern to your deviations?
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RESPONSE --> Yeah all of mine seem to be off! No seriously I think they slowly elevate at the same rate. confidence assessment: 2
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19:16:45 `questionNumber 20000 ** STUDENT SOLUTION CONTINUED: There was no obvious pattern to my deviations. INSTRUCTOR NOTE: Common patterns include deviations that start positive, go negative in the middle then end up positive again at the end, and deviations that do the opposite, going from negative to positive to negative. **
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RESPONSE --> I think I understand the instructor note. I don't know what I was thinking. self critique assessment: 2
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19:17:46 `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 am still kind of iffy with deviation but I understand the concept of the flow model, and the basics of the modeling process. confidence assessment: 3
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19:17:56 `questionNumber 20000 ** 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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19:18:56 `questionNumber 20000 Have you memorized the steps of the modeling process, and are you gonna remember them forever? Convince me.
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RESPONSE --> I am working on the memorization and hopefully I will be able to retain the information at least for the next few years! confidence assessment:
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19:19:30 `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 actually did print out a copy of the outline. I print out all the notes and outlines and exercises. self critique assessment: 3
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