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Travka [436]
3 years ago
10

The flower box is 5 ft​ long, and its ends are in the shape of a trapezoid. The upper and lower bases of the trapezoid measure 1

8 in. and 10​ in., respectively, and the height is 9 in. Find the volume of the flower box.

Mathematics
1 answer:
Nady [450]3 years ago
6 0

Answer:

7560 in^3

Step-by-step explanation:

Hello! The first thing that we must take into account to solve this problem is that the volume of any solid is equal to the transverse area by the height, in this case we must find the area of ​​the trapezoid and then multiply it by the height

V=AL

V= Volume

A=area

L=height=9in

finding area of trapezoid

A=H\frac{a+b}{2} \\

To know what each variable corresponds to, please see the attached file

A=(60)\frac{10+18}{2}=840in^2

finally multiply the area by the height of the box(9in)

V=AL

V=(840in^2)(9in)=7560in^3

the volume of the flower box is 7560 in^3

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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

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3 years ago
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