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!
3 is incorrect because 14.7 + 3 = 17.7
The answer of 15 - 14.7 = 0.3
Since, the number w and 0.8 are additive inverses.
A number 'a' is said to have an additive inverse '-a' if "a+ (-a)= 0".
Since, 'w' and '0.8' are additive inverses of each other such that 
Therefore, the value of 'w' should be '-0.8' so that
.
So, the value of 'w' is =0.8
Now, Refer to the attached image which represents the position of 0.8 , w ( that is -0.8) and the sum of 0.8 and w.
Sum of 0.8 and w = 0.8 + w
= 0.8 +(-0.8)
= 0.
Answer:
P(6) = 0.6217
Step-by-step explanation:
To find P(6), which is the probability of getting a 6 or less, we will need to first calculate two things: the mean of the sample (also known as the "expected value") and the standard deviation of the sample.
Mean = np
Here, "n" is the sample size and "p" is the probability of the outcome of interest, which could be getting a heads when a tossing a coin, for instanc
So, Mean = n × p = (18) ×(0.30) = 5.4
Next we we will find the standard deviation:
Standard Deviation = 
n = 18 and p = 0.3 "q" is simply the probability of the other possible outcome (maybe getting a tails when flipping a coin), so q = 1 - p
Standard Deviation =
= 1.944
Now calculate the Z score for 6 successes.
Z = ( of successes we're interested in - Mean) ÷ (Standard Deviation)
=(6-5.4) ÷ (1.944) = 0.309
we have our Z-score, we look on the normal distribution and find the area of the curve to the left of a Z value of 0.309. This is basically adding up all of the possibilities for getting less than or equal to 6 successes. So, we get 0.6217.