course mth 163
-Exercise 1-Using the depth vs. clock time data points (5.3, 63.7) (10.6, 54.8) (15.9, 46) (21.2, 37.7) (26.5, 32) (31.8, 26.6) we obtain the model; y = at^2 + bt + c
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-Using the depth vs. clock time data points (10.6, 54.8) (21.2, 37.7) (31.8, 26.6) we obtain the system of equations; 112.36a + 10.6b + c = 54.8 ; 449.44a + 21.2b + c = 37.7 ; 1011.24a + 31.8b + c = 26.6
-From the parameters a = .0267, b = -2.4623 and c = 77.9004 we obtain the function y = .0267t^2 – 2.4623t + 77.9004
-Comparing the predicted depths at clock times t = 5.3, 10.6, 15.9, 21.2, 26.5, 31.8 with the observed depths we see that they are close to one another and the model fits pretty good.
-At 46 seconds the predicted depth is 21.1318cm
-Exercise 2 (data 1)
-Using the Grade Average vs. Percent of Assignments Reviewed data we obtain a model y = at^2 + bt + c
- Using the Grade Average vs. Percent of Assignments Reviewed data points (20, 2.118034) (40, 2.581139) (80, 3.236068) we obtain the system of equations; 400a + 20b + c = 2.118034 ; 1600a + 40b + c = 2.581139 ; 6400a +80b + c = 3.236068
- From the parameters a = -.000130, b = .031634 and c = 1.537354 we obtain the function y = -.000130x^2 + .031634x + 1.537354
-The model fits well for the data.
-80% = 3.236074
-To solve for 3.0 and 4.0 I know to put it in for y in the y = -.000130x^2 + .031634x + 1.537354, but not sure how to solve for x to get the percent. For 4.0 it would not make sense to get any higher than what shows for 100% already because one couldn’t study more than 100% of the assignments.
In each case you get a quadratic equation. You will find that the equation for y = 3.0 actually gives two solutions, but one solution would not correspond to the situation and would be disregarded.
When you solve the equation for y = 4.0 you might well find that the equation has no solution, which would be consistent with your comment.
-Exercise 2 (data 2)
-Obtaining a model from Illumination of Comet vs. Distance from Sun y = at^2 + bt + c
-Using the Illumination of Comet vs. Distance from Sun data points (2, 264.4411) (4, 61.01488) (8, 16.27232) we obtain the system of equations: 4a + 2b + c = 264.4411 ; 16a + 4b + c = 61.01488 ; 64a + 8b + c = 16.27232
-From the parameters a = 15.08791, b = -192.24056 and c = 588.57058 we obtain the function y = 15.08791x^ - 192.24056x + 588.57058
- The model doesn’t seem to fit well for the data except for the three points selected, but as far as the others go it the predicted illumination doesn’t come very close.
There were additional questions related to this problem.
Your problems are well presented. See my notes.
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