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4 years ago

A construction company purchased 387 boxes of nails. Each box costs $87. How much did the nails cost the construction company?

2 answers:

8 0

You multiply the number of boxes purchased by the cost per box.

=387 boxes * $87 per box
=$33,669 how much the nails cost the construction company!

Some expensive nails!

Hope this helps! :)

N76 [4]4 years ago

4 0

Answer:

$33,669

Step-by-step explanation:

In order to calculate this you just need to multiply the number of boxes of nails that the company had to purchase by the cost of each box of nails:

Boxes bought*cost of each box=Total Cost

Total cost= 387*$87

Total cost: $33,669

So the total cost of the nails for the construction company was $33,669.

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Suppose that x is a binomial random variable with n=5, p=. 3,and q=. 7.1. Write the binomial formula for this situation and list

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Answer:

$P(X = x) = C_{5,x}.(0.3)^{x}.(0.7)^{5-x}$

Possible values of x: Any from 0 to 5.

$P(X = 0) = C_{5,0}.(0.3)^{0}.(0.7)^{5-0} = 0.16807$

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Step-by-step explanation:

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.

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

And p is the probability of X happening.

In this question:

$n = 5, p = 0.3, q = 1 - p = 0.7$

$P(X = x) = C_{5,x}.(0.3)^{x}.(0.7)^{5-x}$

Possible values of x: 5 trials, so any value from 0 to 5.

For each value of x calculate p(☓ =x)

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

$P(X = 0) = C_{5,0}.(0.3)^{0}.(0.7)^{5-0} = 0.16807$

$P(X = 1) = C_{5,1}.(0.3)^{1}.(0.7)^{5-1} = 0.36015$

$P(X = 2) = C_{5,2}.(0.3)^{2}.(0.7)^{5-2} = 0.3087$

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$P(X = 5) = C_{5,5}.(0.3)^{5}.(0.7)^{5-5} = 0.00243$

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