Given:


To find:
The value of
.
Solution:
We have,
![[\because d(t)=80t]](https://tex.z-dn.net/?f=%5B%5Cbecause%20d%28t%29%3D80t%5D)

Now,
![[\because C(d)=0.09d]](https://tex.z-dn.net/?f=%5B%5Cbecause%20C%28d%29%3D0.09d%5D)

Therefore, the value of
is 72.
Assuming the question marks are minus signs
to find max, take derivitive and test 0's and endpoints
take derivitive
f'(x)=18x²-18x-108
it equal 0 at x=-2 and 3
if we make a sign chart to find the change of signs
the sign changes from (+) to (-) at x=-2 and from (-) to (+) at x=3
so a reletive max at x=-2 and a reletive min at x=3
test entpoints
f(-3)=83
f(-2)=134
f(3)=-241
f(4)=-190
the min is at x=3 and max is at x=-2
Yes but it dose depend on the problem
Answer:
The minimum score required for admission is 21.9.
Step-by-step explanation:
Problems of normally distributed samples are solved using the z-score formula.
In a set with mean
and standard deviation
, the zscore of a measure X is given by:

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
In this problem, we have that:

A university plans to admit students whose scores are in the top 40%. What is the minimum score required for admission?
Top 40%, so at least 100-40 = 60th percentile. The 60th percentile is the value of X when Z has a pvalue of 0.6. So it is X when Z = 0.255. So




The minimum score required for admission is 21.9.
Answer: $3,166
Step-by-step explanation: