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const2013 [10]
2 years ago
12

Suppose one bag contains 500 cubic inches of confetti. How many bags will Student Government need to fill 100 containers? If the

y need 100 containers, how many TOTAL cubic inches of confetti will they need?
type your answer...
cubic inches How many bags will they need if each bag contains 500 cubic inches of confetti? You cannot buy a partial bag of confetti.

type your answer...
bags
Mathematics
1 answer:
Vanyuwa [196]2 years ago
5 0

The number of bags that they will need if each bag contains 500 cubic inches of confetti is 20,000

<h3>How to find the total volume ?</h3>

One bag contains 500 cubic inches of confetti.

Therefore,

1 bag = 500 inches³

The student need to fill 100 containers with the bag.

Each container can accomodates 100,000 cubic inches.

Therefore,

1 bag = 500 inches³

100,000inches³ = 1 container = 200 bags

Therefore,  

100 containers will contain 20, 000 bags of confetti.

learn more on volume here: brainly.com/question/10224096

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3 0
3 years ago
discrete random variable X has the following probability distribution: x 13 18 20 24 27 P ( x ) 0.22 0.25 0.20 0.17 0.16 Compute
tatuchka [14]

Answer:

(a) P(X = 18) = 0.25

(b) P(X > 18) = 0.53

(c) P(X ≤ 18) = 0.47

(d) Mean = 19.76

(e) Variance = 22.2824

(f) Standard deviation = 4.7204

Step-by-step explanation:

We are given that discrete random variable X has the following probability distribution:

            X                    P (x)             X * P(x)            X^{2}             X^{2} * P(x)

           13                    0.22              2.86              169              37.18

           18                    0.25              4.5                324               81

           20                   0.20               4                  400               80

           24                    0.17              4.08              576              97.92

           27                    0.16              4.32              729             116.64

(a) P ( X = 18) = P(x) corresponding to X = 18 i.e. 0.25

     Therefore, P(X = 18) = 0.25

(b) P(X > 18) = 1 - P(X = 18) - P(X = 13) = 1 - 0.25 - 0.22 = 0.53

(c) P(X <= 18) = P(X = 13) + P(X = 18) = 0.22 + 0.25 = 0.47

(d) Mean of X, \mu = ∑X * P(x) ÷ ∑P(x) = (2.86 + 4.5 + 4 + 4.08 + 4.32) ÷ 1

                                                         = 19.76

(e) Variance of X, \sigma^{2} = ∑X^{2} * P(x) - (\sum X * P(x))^{2}

                                 = 412.74 - 19.76^{2} = 22.2824

(f) Standard deviation of X, \sigma = \sqrt{variance} = \sqrt{22.2824} = 4.7204 .

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