course Math 163
On the completion of the initial flow model in assignment 2, it says
If you have not already done so, obtain your own set of flow depth vs. time data as instructed in the Flow Experiment (either perform the experiment, as recommended, or click on Randomized Problems
In a study of precalculus students, average grades were compared with the percent of classes in which the students took and reviewed class notes. The results were as follows:
Percent of Assignments Reviewed Grade Average
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
Determine from your model the percent of classes reviewed to achieve grades of 3.0 and 4.0.
?? After my eliminations, I came up with a = -.000116 b = 0.031364 c = 1.488529 With this information I determined the grades 3.0 by solving t = [-b +- sqrt(b^2 – 4ac)] / 2a
First, I subtracted 3.0 from both sides of the equation
3.0 = (-.000116)t^2 + (0.031364)t + 1.488529 to get the equation
(-.000116)t^2 + (0.031364)t – 1.511471 = 0
Using that information I get
t = [-0.031364 + -sqrt(.0009837 – 4 * -.000116 * -1.511471)] / 2(-.000116)
t = 8.53732 or t = 11.06517
You're doing great. You completely understand the process.
The only thing I see wrong is that you did your arithmetic incorrectly in
t = [-0.031364 + -sqrt(.0009837 – 4 * -.000116 * -1.511471)] / 2(-.000116)
The expression simplifies to t = 63 or t = 208, approx..
The 63 fits right in with your table. The 208 is on the increasing half of the parabola and can safely be discarded.
Neither of these solutions makes any sense. I must have done the calculations wrong. It should be between 60and 80%
For 4.0 I got an error message saying it was undefined. I am not sure where I went wrong on these problems. I understand how to do them, but my calculations are wrong.??
Determine also the projected grade for someone who reviews notes for 80% of the classes.
??For 80%, the grade average would be approx 3.255249??
Comment on how well the model fits the data. The model may fit or it may not.
??It appears to be a good fit. The average deviance was approximately 0.0256334??
Comment on whether or not the actual curve would look like the one you obtained, for a real class of real students.
??Yes??
You did everything well, except for an arithmetic error. Excellent work.