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worty [1.4K]
3 years ago
9

In a different plan for area​ codes, the first digit could be any number from 2 through 5​, the second digit was either 2, 3, 4,

or 5​, and the third digit could be any number except 4, 5, or 8. With this​ plan, how many different area codes are​ possible?
Mathematics
1 answer:
GrogVix [38]3 years ago
3 0

Answer:

Step-by-step explanation:

First digit 2,3,4,5 = 4 different digits.

Second digit 2,3,4,5 = 4 different digits.

Third digit = 1,2,3,6,7,9 = 6 different digits.

Total number of possibilities = 4*4*6 = 96

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Using the binomial distribution, it is found that the expected values are given by:

a) 120.

b) 150.

<h3>What is the binomial probability distribution?</h3>

It is the probability of <u>exactly x successes on n repeated trials, with p probability</u> of a success on each trial.

The expected value of the binomial distribution is:

E(X) = np

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The expected value is given by:
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Item b:

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More can be learned about the binomial distribution at brainly.com/question/24863377

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24 + 4(3 + 1)
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