Mth 163
Your 'question form' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
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Unanswered questions from Assignment 3 Open QU
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iven Solution:
In the preceding problem we approximated the x value at which the function is maximized by averaging 1.48 and 10.16, the x values at which the function is zero. Here we will use x = -b / (2 a) to obtain
x value at which function is maximized: x = -b / (2a) = - 5.33333 / (2 * -0.45833) = 5.81818.
To find corresponding y value we substitute x = 5.81818 into y = - 0.458333 x^2 + 5.33333 x - 6.875 to obtain y = 8.64024.
Thus the vertex of the parabola lies at (5.81818, 8.64024).
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Self-critique (if necessary):
QUESTIONS:
I have a slightly different value for the y coordinate, and I thought the solution provided would be to six significant figures:
(5.81818, 8.64024).
Question asked for five significant figures, which I interpreted as:
(5.8182, 8.6401)
AM I OFF ON THE CONCEPT OF SIGNIFICANT FIGURES?
As long as you have a sufficient number of significant figures to distinguish your results, you have enough. The programs that generate problems, etc., often 'spit out' more significant figures than necessary.
In the case of this problem, three significant figures would be plenty.
Self-critique Rating:2
(2) QUESTION: I am still having slight discrepancies in the values – suspect significant figures issue. Should I be completing the calculations, then reducing the answer to the proper number of significant figures (process I am using) or should I be reducing each factor in the answer to the proper number of significant figures BEFORE doing the calculation?
Self-critique Rating: 2
Usually in this course it doesn't make a lot of difference, but sometimes it does, and when it does you want to use the process you are using. That covers you in just about any instance.
Mth 163
Your 'question form' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
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Regarding Completion of the Initial Flow Model.
I expected matching values here when testing the variables back into the original equations, but they are off – what am I doing wrong here?
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Data Set 1
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/percent reviewed
Grade Average
0
1.000000
10
.0790569
1.790569
20
.0327465
2.118034
30
.0251272
2.369306
40
.0211833
2.581139
50
.0186628
2.767767
60
.0168725
2.936492
70
.0155158
3.091650
80
.0144418
3.236068
90
.0135640
3.371708
100
.0128292
3.500000
GPA
Percent of Assignments Reviewed
Graph as show at right is a fairly smooth curve, beginning on (0, 1) and rising sharply to the next point (10, 1.790569), then increasing at a decreasing rate from that point on up to (100, 3.5)
Evaluation of the points shows a fairly smooth progression, with no obvious outliers except possibly for the first point (0.1)
* Three points selected with a reasonable spread within the data to create the model are:
o 10, 1.790569
o 50, 2.767767
o 100, 3.5
* Use selected points to complete quadratic equation y= ax^2 + bx +c
o 10, 1.790569
* 1.790569 = a 10^2 + b*10 + c
* 1.790569 = 100a + 10b + c
o 50, 2.767767
* 2.767767= a*50^2 + b*50 +c
* 2.767767= 2500a + 50b +c
o 100, 3.5
* 3.5 = a*100^2 + b*100 +c
* 3.5 = 10000a +100 b +c
* Eliminate one variable “c” from first two resulting equations (start with easiest to eliminate)
* (Subtract first equation from second)
* 2.767767= 2500a + 50b +c
* 1.790569 = 100a + 10b + c
o 0.977198 = 2400a + 40b
* Eliminate same variable “c” from second and third equations
* (Subtract second equation from third)
* 3.5 = 10000a +100 b +c
* 2.767767= 2500a + 50b +c
o 0.732233 = 7500a + 50b
* Use factors as needed to create equalities in next variable “b” in order to eliminate it
* 5*(0.977198) = (2400a + 40b)*5
o 4.88599 = 12000a + 200b
* -4*(0.732233) = (7500a + 50b)*-4
o -2.928932 = -30000a – 200b
* Eliminate next variable from revised equations (add) and solve for “a”
* 4.88599 = 12000a + 200b
* -2.928932 = -30000a – 200b
o 1.957058 = -18000a
o -1.08725 e-4 ? a
* Use value provided for last variable, in one of equations to solve for second variable “b”
* 4.8859 ? 12000(-1.08725 e-4) + 200b
* 4.8859 ? -1.3047 + 200b
* 6.1906 ? 200b
o 0.030953 ? b
* Use values of both variables to find the third one by using one of the initial three variable equations.
* 3.5 ? 10000(-1.08725 e-4)+100 0.030953 +c
* 3.5 ? -1.08725 + 3.0953 +c
o 1.49195 ? c
* Test all three values in the three original equations – predictions should match data
* I expected matching values here, but they are off – am I doing something wrong here?
Percent of Assignments Reviewed
? Grade Average/percent reviewed
Grade Average
residual
prediction less data value
y = (-1.08725 e-4) x^2 + 0.030953 x + 1.49195
0
0.62217
1.000000
10
.0790569
1.790569
.00982375
-.00108725 + .30953 + 1.49195 = 1.80039275
20
.0327465
2.118034
30
.0251272
2.369306
40
.0211833
2.581139
50
.0186628
2.767767
.0000205
-.2718125+1.54765 + 1.49195 = 2.7677875
60
.0168725
2.936492
70
.0155158
3.091650
80
.0144418
3.236068
90
.0135640
3.371708
100
.0128292
3.500000
(-1.08725 e-4) x^2 + 0.030953 x + 1.49195
Determine from your model the percent of classes reviewed to achieve grades of 3.0 and 4.0. (
Note: use formula if y = at^2 + bt + c, then y = 0 if and only if t = [-b +- `sqrt(b^2-4ac)] / (2a).
3.0 = (-1.08725 e-4) t^2 + 0.030953 t + 1.49195
0 = (-1.08725 e-4) t^2 + 0.030953 )t -1.50805
t = [-b +- `sqrt(b^2-4ac)] / (2a
t= [-0.030953 +- `sqrt(0.030953 ^2- 4*(-1.08725 e-4)*(-1.50805)] / (2(-1.08725 e-4)
t= [-0.030953 +- `sqrt(.00095800882-.00655850945) / (2(-1.08725 e-4)
t= [-0.030953 +- `sqrt-0.00560050063 / (2(-1.08725 e-4)
I believe this is correct, showing that the equation has no solution. You would conclude that this particular model will never reach a point average of 3.0. This means that the parabola opens downward, and that the vertex lies lower than y = 3.0.
I might have given you a misleading response on a previous question, where I thought you were solving a model very much like this for y = 0 (i.e., solving the equation (-1.08725 e-4) t^2 + 0.030953 t + 1.49195, for which the discriminant would be positive).
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Good questions.
&#See my notes and let me know if you have questions. &#