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zzz [600]
3 years ago
6

Simplify. Your answer should contain only positive exponents.

Mathematics
1 answer:
Shkiper50 [21]3 years ago
5 0
<h3>Option A is your answer .</h3>

<u>Solution </u><u>:</u><u>-</u>

\sf \implies 2^{-4} . \big(2^{-3}\big)^4\\\\\sf \implies\dfrac{1}{2^4}.\left(\dfrac{1}{2^{3}}\right)^4\\\\\sf\implies \dfrac{1}{2^4}.\left(\dfrac{1^4}{(2^3)^4}\right)\\\\\sf\implies\dfrac{1}{2^4}.\left(\dfrac{1}{2^{12}}\right)\\\\\sf\implies \dfrac{1}{2^{16}}

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8 0
3 years ago
Assume that when Human Resource managers are randomly selected, 62% say job applicants should follow up within two weeks. If 25
Alenkinab [10]

Using the binomial distribution, it is found that there is a 38% probability that exactly 18 of them say job applicants should follow up within two weeks.

<h3>How to find that a given condition can be modelled by binomial distribution?</h3>

Binomial distributions consists of n independent Bernoulli trials.

Bernoulli trials are those trials that end up randomly either on success (with probability p) or on failures( with probability 1- p = q (say))

The probability that out of n trials, there'd be x successes is given by

P(X =x) = \: ^nC_xp^x(1-p)^{n-x}

Binomial probability distribution  

P(X =x) = \: ^nC_xp^x(1-p)^{n-x}

The parameters are:

n is the number of trials.

x is the number of successes.

p is the probability of success on a single trial.

In this problem:

62% say job applicants should follow up within two weeks, p = 0.62

25 managers are selected, n = 25

The probability that exactly 18 of them say job applicants should follow up within two weeks is P ( X = 18)

P( X > 18) = 1 -  ( X = 18)

= 1 - 0.62

= 0.38

38 % probability that exactly 18 of them say job applicants should follow up within two weeks.

Learn more about binomial distribution here:

brainly.com/question/13609688

#SPJ1

8 0
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