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sergij07 [2.7K]
3 years ago
12

Plzzzzzz someone help I keep on getting a decimal answer that is 100% wrong. Please do number 14

Mathematics
1 answer:
ANEK [815]3 years ago
6 0
Im sorry number 14 of what?
You might be interested in
Determine the explicit equation for the arithmetic sequence given below.-3,-7,-11,-15,-19
liraira [26]

Answer:

-3 + (n-1)(-4)

Step-by-step explanation:

We follow the formula.

a1 + (n-1)d

d is the common difference.

7 0
2 years ago
Read 2 more answers
How many different 5​-letter radio station call letters can be made a. if the first letter must be Upper C comma Upper X comma U
Lady_Fox [76]

Answer:

a) 1,518,000

b) 2,284,880

c) 60,720

Step-by-step explanation:

a) a. if the first letter must be Upper C comma Upper X comma Upper T comma or Upper M and no letter may be​ repeated?

We draw 5 boxes, and based on that we will see the total possible cases. There are 26 alphabets

The first box should have C or X or T or M .No letter may be repeated.

Any                    Any             Any                 Any                    Any

5 alphabets    of the            of the              of the                 of the

C,X , T , M      remaining     remaining       remaining          remaining

                   25 alphabets   24 alphabets   23 alphabets   22 alphabets

Therefore; total possible call letters = 5 × 25 × 24 × 23 × 22 = 1,518,000

b)

The first box should have C or X or T or M Repeats as allowed

Any                    Any             Any                 Any                    Any

5 alphabets    of the            of the              of the                 of the

C,X , T , M      remaining     remaining       remaining          remaining

                   26 alphabets   26 alphabets   26 alphabets   26 alphabets

Therefore Total possible call letters = 5 × 26 × 26 × 26 × 26 = 2,284,880

c)   The first box should have   C,X , T , M  and end with S

So the last place if fixed, and we now have 25 alphabets. The first box can go in 5 ways. The next box then will have only 24 letters to choose from, as the first box has taken a letter and the last box already has S in it. Repetition not allowed

Any                    Any             Any                 Any                       S

5 alphabets    of the            of the              of the                  is fixed

C,X , T , M      remaining     remaining       remaining           here

                   24 alphabets   23 alphabets   22 alphabets  

Therefore Total possible call letters = 5 × 24 × 23 × 22  × 1 =  60,720

3 0
3 years ago
Explain to me how to do this please
nataly862011 [7]
You must first attach the problems with your question.
6 0
3 years ago
2 2/5 x 3/4 <br> plz help me! and plz show your work.
Aneli [31]

Answer:1.8

Step-by-step explanation:

4 0
2 years ago
what is the equation subtracting 7 from the product of 4 and a number results in the number added to 29
vitfil [10]

That would be:-

4x - 7 = x + 29

the solution:-

3x = 29 + 7

3x = 36

x = 12


7 0
3 years ago
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