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!
Old
9.99×55
=549.45
New
10.68×55
=587.4
((10.68÷9.99)−1)×100
=6.9%
K is equal to 4 because g(x) is a parrallel line to f(x)
