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1 answer:
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Answer:
The 95% confidence interval for the true proportion of all medical students who plan to work in a rural community is (0.0525, 0.1075).
Step-by-step explanation:
In a sample with a number n of people surveyed with a probability of a success of , and a confidence level of
, we have the following confidence interval of proportions.
In which
z is the zscore that has a pvalue of .
For this problem, we have that:
95% confidence level
So , z is the value of Z that has a pvalue of
, so
.
The lower limit of this interval is:
The upper limit of this interval is:
The 95% confidence interval for the true proportion of all medical students who plan to work in a rural community is (0.0525, 0.1075).
Answer:
18,646
Step-by-step explanation:
>you can graph the equation on a calculator and see that the minimum is at the point (250, 18646) , so the minimum unit cost is 18,646
>you can use the first derivative to find the minimum
c'(x) = (0.2x²-100x+31,146)' = .4x -100
.4x-100= 0 , so x= 250
c(x=250) = 0.2(250)² -100*250 +31,146 =18,646
