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3 years ago

PLEASE HELP

Mathematics

1 answer:

notka56 [123]3 years ago

8 0

Answer:

D. 28.5 in.

Step-by-step explanation:

Formula: P = 2l + 2w, l = length & w = width.

To solve this problem, just substitute the given value for l & w and simplify.

P = 2l + 2w; l = 8.5 & w = 5.75

P = 2(8.5) + 2(5.75)

P = 17 + 11.5

P = 28.5 in

Hope this helps!

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Answer:

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Step-by-step explanation:

For each student, there are only two possible outcomes. Either they found employment, or they did not. The probability of a student finding employment is independent of any other student, which means that the binomial probability distribution is used to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

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

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

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88% of students who graduate from the university successfully find employment in their chosen field within three months of graduation.

This means that $p = 0.88$

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This means that $n = 9$

What is the probability that of nine randomly selected students who have graduated from this university, at least six of them find employment in their chosen field within three months?

This is:

$P(X \geq 6) = P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9)$

In which

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 6) = C_{9,6}.(0.88)^{6}.(0.12)^{3} = 0.0674$

$P(X = 7) = C_{9,7}.(0.88)^{7}.(0.12)^{2} = 0.2119$

$P(X = 8) = C_{9,8}.(0.88)^{8}.(0.12)^{1} = 0.3884$

$P(X = 9) = C_{9,9}.(0.88)^{9}.(0.12)^{0} = 0.3165$

Then

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0.9842 = 98.42% probability that at least six of them find employment in their chosen field within three months.

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3 years ago

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Answer:

See step-by-step explanation

Step-by-step explanation:

Table

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Answer:

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