course Mth 163
As a followup to your comments on the simulated flow experiment, after going back and looking at how I arrived at a, you were correct in that in the process of eliminating the variable b, I did make an error and a should have been -.9. When I went over my remaining calculations, I got -1.49 (as you suggested it should have been)for b and 84.696 for c. So the revised depth of 46.42 which makes more sense looking at the other numbers. However, since there are no data points beyond the clock time of 16.8, I can only speculate this might be a reasonable prediction. Am I just comparing my calculation to the other data points here or am I ommitting an additional step in evaluating?
Also, when I used the quadratic formula to predict clock time at depth of 14 cm (using a=.0143, b=-1.49, and c=84.96-14), I get a negative discriminant of -1.824. How do I further evaluate? I'm not sure whether the model would be considered a good fit or not. Your comments would be appreciated.I have one last completely unrelated question. In the orientation exercises, you gave a problem in which we were asked to determine how many frogs you would have in a pond if, on a monthly basis, the beginning population of 20 were to increase by 10%. The answer was not given but the formula was, I believe, 20x1.1^300 (we were to estimate what the number would be at 300 months using this formula). This worked on my calculator at three months (26.62) but at 300 I got 5.234021992E13. Can you give actual answer? Thank you.
Your work has been received. Please scroll through the document to see any inserted notes (inserted at the appropriate place in the document, in boldface) and a note at the end. The note at the end of the file will confirm that the file has been reviewed; be sure to read that note. If there is no note at the end, notify the instructor through the Submit Work form, and include the date of the posting to your access page.
100a+10b+c=1.790569
4900a+70b+c=3.09165
10000a+100b+c=3.5
By elimination I arrived at a=.0001512 and by plugging a into my reduced equations, a value of .00960 for b. Using a and b in the 2nd of the original 3 equations, I got c=1.67975.
Using 80 for the t value in the quadratic formula, I get y=3.415 for the predicted GPA which deviates from the given data by .179 and seems likely to be fairly accurate. I would expect the curve I obtained to reflect that of an actual class. Would I be correct in saying the graph is increasing at a steady rate? I think the model fit the data fairly well.
If the graph increases at a constant rate then it's linear, and a model of the form y = a x + b would apply. Since the model here is of the form y = a x^2 + b x + c, with a not equal to 0, there will be some nonlinearity, some curvature to the graph. However the value of a is pretty small, and the curvature might not be obvious over this range of values.
On Data set 2, I used the points (2, 264.441), (5, 43.06238), and (9, 11.280) and obtained the following 3 equations:
4a+2b+c=264.441
25a+5b+c=43.06238
81a+9b+c=11.280
After eliminating c, I was left with 56a+40b=-31.782 and 21a+3b=-221.3786. Eliminating b gave me .1875 for a from which I obtained b (1.0571) and c(261.577).
Using these in the quadratic equation I got y=263.748 as the predicted illumination at 1.6 earth distances. Comparing the earth distances of 1 and 2 and their respective y values, I would expect illumination at 1.6 earth distances to lie closer to the midpoint between 935.1395 and 264.4411 but it did not. With regard to the range of distances which would be comfortable for reading given that this occurs in the range between 25 and 100 watts per square meter, I could not calculate because in this case I got a negative discriminant of -195.116, so I would only be able to estimate roughly that, based on the data given, the answer would at a point slightly above 6 and slightly below 3 earth distances. I would say this model was a poor fit for the data.
You are absolutely correct. The point here is that a quadratic fit works fine for some types of data, and not well at all for others.
Your work on this assignment is good. Let me know if you have questions.