The question is incorrect.
The correct question is:
Three TAs are grading a final exam.
There are a total of 60 exams to grade.
(c) Suppose again that we are counting the ways to distribute exams to TAs and it matters which students' exams go to which TAs. The TAs grade at different rates, so the first TA will grade 25 exams, the second TA will grade 20 exams and the third TA will grade 15 exams. How many ways are there to distribute the exams?
Answer: 60!/(25!20!15!)
Step-by-step explanation:
The number of ways of arranging n unlike objects in a line is n! that is ‘n factorial’
n! = n × (n – 1) × (n – 2) ×…× 3 × 2 × 1
The number of ways of arranging n objects where p of one type are alike, q of a second type are alike, r of a third type are alike is given as:
n!/p! q! r!
Therefore,
The answer is 60!/25!20!15!
The answer is 25.
-1/5n = -5
-n = -25
n = 25
Probably $87 based on the information I was given.
Answer: The average number of hours she danced per day is 1.9 hours (rounded to the nearest tenth)
Step-by-step explanation: We start by calculating how many hours she danced all together which can be derived as follows;
Summation = 3 +2 +2 + 1 + 1.5 + 2 = 11.5
The number of days she danced which is the observed data is 6 days (she did not dance at all on Wednesday).
The average (or mean) hours she danced each day can be calculated as
Average = ∑x ÷ x
Where ∑x is the summation of all data and x is number of observed data
Average = (3+2+2+1+1.5+2) ÷ 6
Average = 11.5 ÷ 6
Average = 1.9166
Approximately, average hours danced is 1.9 hours (to the nearest tenth)
Answer: 
Step-by-step explanation:
According to the given information, we have
Sample size : n= 50
Since population standard deviation is unknown, so we use t-test.
Critical value for 95 percent confidence interval :
Confidence interval : 
Required 95% confidence interval :
