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kykrilka [37]
3 years ago
5

The picture shows two sizes of shipping boxes with measurements in inches. The boxes are rectangular prisms. There are two recta

ngular prisms. Prism 1 has a height of 10 inches, length of 20 inches and a width of 7 inches. Prism 2 is smaller and to the right of Prism 1. Prism 2 has a height of 4 inches, length is 20 inches and width is 4 inches. Marcus shipped 1 of the larger boxes and 2 of the smaller boxes. Which equations show the total volume of the boxes Marcus shipped? Select the three equations that apply.
Mathematics
1 answer:
MakcuM [25]3 years ago
7 0

Answer:

6

Step-by-step explanation:

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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
Radda [10]

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

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

P(X = 3) = C_{5,3}.(0.3)^{3}.(0.7)^{5-3} = 0.1323

P(X = 4) = C_{5,4}.(0.3)^{4}.(0.7)^{5-4} = 0.02835

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

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

So

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

P(X = 3) = C_{5,3}.(0.3)^{3}.(0.7)^{5-3} = 0.1323

P(X = 4) = C_{5,4}.(0.3)^{4}.(0.7)^{5-4} = 0.02835

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

8 0
3 years ago
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