Asked by Pat
The more you study for a certain exam, the better your performance on it. If you study for 10 hours, your score will be 65%. If you study for 20 hours, your score will be 95%.You can get as close as you want to a perfect score just by studying long enough. Assume your percentage score, p(n), is a function of the number of hours, n, that you study in the form:
p(n)=(an+b)/(cn+d).
If you want a score of 80%, how long do you need to study?
(This problem involves with rational function chapter)
p(n)=(an+b)/(cn+d).
If you want a score of 80%, how long do you need to study?
(This problem involves with rational function chapter)
Answers
Answered by
Steve
Since the limit of p(n) is 1, a = c. So, you can scale things so that a=c=1 and you have
p(n) = (n+b)/(n+c)
Now plug in your numbers:
(10+b)/(10+c) = .65
(20+b)/(20+c) = .95
solve for b and c and you have
p(n) = (n - 107/12)/(n - 100/12)
or, eliminating the fractions,
p(n) = (12n-107)/(12n-100)
So, for a score of 80%,
(12n-107)/(12n-100) = .80
n = 11.25
p(n) = (n+b)/(n+c)
Now plug in your numbers:
(10+b)/(10+c) = .65
(20+b)/(20+c) = .95
solve for b and c and you have
p(n) = (n - 107/12)/(n - 100/12)
or, eliminating the fractions,
p(n) = (12n-107)/(12n-100)
So, for a score of 80%,
(12n-107)/(12n-100) = .80
n = 11.25
Answered by
Ray
Thanks for replying Steve
I just wanted to ask how did you get p(n) = (n - 107/12)/(n - 100/12) after solving for b and c?
I just wanted to ask how did you get p(n) = (n - 107/12)/(n - 100/12) after solving for b and c?
Answered by
Steve
p(n) = (n-b)/(n+c)
Sorry, should have been (n-b)/(n-d)
In any case, when you solve for the two unknowns, you get -107/12 and -25/3
Sorry, should have been (n-b)/(n-d)
In any case, when you solve for the two unknowns, you get -107/12 and -25/3
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