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SSSSS [86.1K]
2 years ago
15

A group of seven kids line up in a random order. Each ordering of the kids is equally likely. There are three girls and four boy

s in the group. What is the probability that all the girls are ahead of all the boys
Mathematics
1 answer:
jeka57 [31]2 years ago
6 0

Using the arrangements formula, it is found that there is a 0.0286 = 2.86% probability that all the girls are ahead of all the boys.

<h3>What is the arrangements formula?</h3>

The number of possible arrangements of n elements is the factorial of n, that is:

A_n = n!.

The total number of outcomes is the arrangement of 7 elements, hence:

T = 7! = 5040.

The desired number of outcomes is all girls(3!), then all boys(4!), hence:

D = 3! x 4! = 6 x 24 = 144.

Hence, the probability is:

p = D/T = 144/5040 = 0.0286.

There is a 0.0286 = 2.86% probability that all the girls are ahead of all the boys.

More can be learned about the arrangements formula at brainly.com/question/24648661

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R is the interior of angle TUV. The measure of RUV=30degrees, the measure of TUV=3x+16, and the measure of TUR=x+10. Find x and
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Given

R is the interior of ∠ TUV.

m∠ RUV=30degrees, m∠ TUV=3x+16, and m∠ TUR=x+10.

Find the value of x and the m ∠TUV.

To proof

As given in the question

m ∠TUV=3x+16, and m ∠TUR=x+10

thus

m∠ RUV =  m∠ TUV - m∠ TUR

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30 = 2x + 6

24 = 2x

12 = x

put this value in the m ∠TUV= 3x+16

m ∠TUV= 12× 3 +16

            = 52°

Hence proved


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