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2 years ago

Simplify eighth root of y squared

Mathematics

1 answer:

NARA [144]2 years ago

5 0

Answer:

<u>The simplified expression is y raised to the 2/8 power</u>

Step-by-step explanation:

Let's recall that one of the properties of fractional exponents is:

x ^ a/b = ᵇ√x ᵃ

therefore,

⁸√y² = y ^2/8

<u>The simplified expression is y raised to the 2/8 power</u>

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Answer:

0.1108 = 11.08% probability that there are exactly 3 faulty candy bars among the seven.

Step-by-step explanation:

The bars are chosen without replacement, which means that the hypergeometric distribution is used to solve this question.

Hypergeometric distribution:

The probability of x successes is given by the following formula:

$P(X = x) = h(x,N,n,k) = \frac{C_{k,x}*C_{N-k,n-x}}{C_{N,n}}$

In which:

x is the number of successes.

N is the size of the population.

n is the size of the sample.

k is the total number of desired outcomes.

Combinations formula:

$C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

In this question:

60 total candies means that $N = 70$

12 are faulty, which means that $k = 12$

Seven are chosen, so $n = 7$

What is the probability that there are exactly 3 faulty candy bars among the seven?

This is $P(X = 3)$ . So

$P(X = x) = h(x,N,n,k) = \frac{C_{k,x}*C_{N-k,n-x}}{C_{N,n}}$

$P(X = 3) = h(3,70,7,12) = \frac{C_{12,3}*C_{48,4}}{C_{60,7}} = 0.1108$

0.1108 = 11.08% probability that there are exactly 3 faulty candy bars among the seven.

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