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kati45 [8]
3 years ago
14

What is the correct way to write three hundred nine million, fifty-eight thousand, three hundred four? A. 390,580,304 B. 309,58,

34 C. 309,058,304 D. 395,834
Mathematics
2 answers:
vovikov84 [41]3 years ago
5 0
I THINK IT`S C

hope i helped :0
Leni [432]3 years ago
3 0
C. 309,058,304

Three hundred nine million is in the millions bracket; wherein, three is in the hundred million place and nine is in the one million place.

fifty-eight thousand is in the thousands bracket; wherein 0 is placed in the hundred thousand place, five is in the ten thousands place, and eight is in the one thousands place.

three hundred four is placed as such because three should be in the hundreds place, 0 is in the tens place, and 4 is in the ones place.

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Using the hypergeometric distribution, it is found that there is a 0.0273 = 2.73% probability that the third defective bulb is the fifth bulb tested.

In this problem, the bulbs are chosen without replacement, hence the <em>hypergeometric distribution</em> is used to solve this question.

<h3>What is the hypergeometric distribution formula?</h3>

The formula is:

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

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

The parameters are:

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  • N is the size of the population.
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  • k is the total number of desired outcomes.

In this problem:

  • There are 12 bulbs, hence N = 12.
  • 3 are defective, hence k = 3.

The third defective bulb is the fifth bulb if:

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Hence:

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

P(X = 2) = h(2,12,4,3) = \frac{C_{3,2}C_{9,1}}{C_{12,4}} = 0.2182

0.2182 x 1/8 = 0.0273.

0.0273 = 2.73% probability that the third defective bulb is the fifth bulb tested.

More can be learned about the hypergeometric distribution at brainly.com/question/24826394

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A low p-value means a higher chance of the null hypothesis to be true.

It lies between 0 and 1. A small p-value indicates fewer chances of the null hypothesis to be true.

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