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solong [7]
2 years ago
9

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
Mathematics
1 answer:
abruzzese [7]2 years ago
6 0

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!

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In circle M shown, chords GH and EF intersect at K such that EK  5 and FK  6 . If GK  3 , then which of the following is the
myrzilka [38]

The length of the GH segment is 13  

Step-by-step explanation:

For solving this problem we need to remember some of the circle corollaries-

When two-chord intersects each other, the product of the chord segments are equal

The above corollary can be easily understood by looking at a diagram attached below-

In the figure, EF and GH are two chords intersecting at K

Thus, EK*KF= GK*KH

Values of the EK, KF, GK are given as 5, 6 and 3 respectively

Substituting the values we get

5*6=3*KH

KH= 10

We know that GH= GK+KH

Thus GH= 3+10= 13

6 0
2 years ago
A spinner has four equally sized sectors that are numbered 2, 4, 4, and 5. The spinner is spun twice and the product of the outc
JulsSmile [24]

Answer:

Umm whats the question

Step-by-step explanation:

8 0
2 years ago
!!!! I NEED HELP RIGHT NOW ITS DUE IM 5 MINUTES !!!!
kiruha [24]

Answer:

Step-by-step explanation:

A = h(a+b)/2 for h

2A = h(a + b)

h(a+b) = 2A

h = 2A/(a+b)

3 0
2 years ago
A building across the street casts a shadow that is 24 feet long. Your friend is 6 feet tall and casts a shadow that is 4 feet l
tiny-mole [99]
The height of the building is 16 feet.
5 0
2 years ago
Read 2 more answers
Tickets to a musical cost $30 for adults and $12 for children. At one particular performance 960 people attended and $19 080 was
postnew [5]
In order to solve this, you have to set up a systems of linear equations.

Let's say that children = c and adults = a

30a + 12c = 19,080
a + c = 960

I'm going to show you how to solve this system of linear equations by substitution, the easiest way to solve in my opinion.

   a + c = 960 
       - c    -    c
 ---------------------- ⇒ Step 1: Solve for either a or c in either equation.
   a = 960 - c 



20(960 - c)+ 12c = 19,080
19,200 - 20c + 12c = 19,080
   19,200 - 8c = 19,080
 - 19,200         - 19,200
---------------------------------- ⇒ Step 2: Substitute in the value you got for a or c
         8c = -120                    into the opposite equation.                                                
       ------  ---------
          8         8
 
         c = -15



30a + 12(-15) = 19,080
   30a - 180 = 19,080
          + 180  +     180
 -------------------------------
   30a = 19,260
  -------   -----------
    30          30
     
       a = 642
__________________________________________________________

I just realized that there can't be a negative amount of children, so I'm sorry if these results are all wrong. 


7 0
2 years ago
Read 2 more answers
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