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miv72 [106K]
3 years ago
8

a high school stage collapsed in Fullerton, California, when 250 students got on stage for the finale of a musical production. T

wo dozen students were injured. The stage could support a maximum of 12,750 pounds. If the average weight of a student is assumed to be 140 pounds, what is the maximum number of students who could safely be on the stage?
Mathematics
1 answer:
Lera25 [3.4K]3 years ago
6 0

Answer:

91

Step-by-step explanation:

In this question, the important information we get from this is that the stage can support 12,750 students. We are asked, if the average weight of a student is 140 pounds, what is the maximum number of students that can be on the stage safely?

To solve, we simply divide 12,750 by 140.

x=\frac{12,750}{140}

x= 91.07

We get 91.07 as our answer. Since we are considering the maximum number of students, then we must round 91.07 to the nearest whole number which would be 91.

The maximum number of students who could safely be on the stage is 91.

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3 years ago
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Answer:

a) The expected value is \frac{-1}{15}

b) The variance is  \frac{49}{45}

Step-by-step explanation:

We can assume that both marbles are withdrawn at the same time. We will define the probability as follows

#events of interest/total number of events.

We have 10 marbles in total. The number of different ways in which we can withdrawn 2 marbles out of 10 is \binom{10}{2}.

Consider the case in which we choose two of the same color. That is, out of 5, we pick 2. The different ways of choosing 2 out of 5 is \binom{5}{2}. Since we have 2 colors, we can either choose 2 of them blue or 2 of the red, so the total number of ways of choosing is just the double.

Consider the case in which we choose one of each color. Then, out of 5 we pick 1. So, the total number of ways in which we pick 1 of each color is \binom{5}{1}\cdot \binom{5}{1}. So, we define the following probabilities.

Probability of winning: \frac{2\binom{5}{2}}{\binom{10}{2}}= \frac{4}{9}

Probability of losing \frac{(\binom{5}{1})^2}{\binom{10}{2}}\frac{5}{9}

Let X be the expected value of the amount you can win. Then,

E(X) = 1.10*probability of winning - 1 probability of losing =1.10\cdot  \frac{4}{9}-\frac{5}{9}=\frac{-1}{15}

Consider the expected value of the square of the amount you can win, Then

E(X^2) = (1.10^2)*probability of winning + probability of losing =1.10^2\cdot  \frac{4}{9}+\frac{5}{9}=\frac{82}{75}

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Var(X) = E(X^2)-E(X)^2

Thus

Var(X) = \frac{82}{75}-(\frac{-1}{15})^2 = \frac{49}{45}

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