What is it you need help with?
The answer is A.) 1/4(n-64) because if you distribute the 1/4 onto the parentheses it becomes n/4 - 16 which is equivalent to 1/4 - 16 if n were to equal 1.
<h2>
Answer with explanation:</h2>
Let
be the average starting salary ( in dollars).
As per given , we have

Since
is left-tailed , so our test is a left-tailed test.
WE assume that the starting salary follows normal distribution .
Since population standard deviation is unknown and sample size is small so we use t-test.
Test statistic :
, where n= sample size ,
= sample mean , s = sample standard deviation.
Here , n= 15 ,
, s= 225
Then, 
Degree of freedom = n-1=14
The critical t-value for significance level α = 0.01 and degree of freedom 14 is 2.62.
Decision : Since the absolute calculated t-value (2.07) is less than the critical t-value., so we cannot reject the null hypothesis.
Conclusion : We do not have sufficient evidence at 1 % level of significance to support the claim that the average starting salary of the graduates is significantly less that $42,000.
Answer:
13
Step-by-step explanation:
Put n as 4

Hope this helps :)