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creativ13 [48]
2 years ago
7

Mrs. Silas has 80 boys and 70 girls in her Mathematics classes throughout the days. Mrs. Silas took a survey of the students to

determine how each student arrived at school on most days. The table below shows the results of the survey. Students in Mathematics Classes Transportation Number of Students School bus 49 Walk 26 Family Automobile 51 Ride with friends 24 What percent of the students arrive in an automobile, either a family car or riding with friends?
Mathematics
1 answer:
vichka [17]2 years ago
3 0

Answer:

the bring them to my house kidding its 118

Step-by-step explanation:

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At a certain airport, 75% of the flights arrive on time. A sample of 10 flights is studied. Assume each flight is independent of
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Answer:

52.56% probability that eight or more of the flights will arrive on time.

Step-by-step explanation:

For each flight, there are only two possible outcomes. Either it is on time, or it is not. The probability of a flight being on time is independent from other flights. So we use the binomial probability distribution to solve this question.]

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.

At a certain airport, 75% of the flights arrive on time.

This means that p = 0.75

A sample of 10 flights is studied.

This means that n = 10

Find the probability that eight or more of the flights will arrive on time.

P(X \geq 8) = P(X = 8) + P(X = 9) + P(X = 10)

In which

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

P(X = 8) = C_{10,8}.(0.75)^{8}.(0.25)^{2} = 0.2816

P(X = 9) = C_{10,9}.(0.75)^{9}.(0.25)^{1} = 0.1877

P(X = 10) = C_{10,10}.(0.75)^{10}.(0.25)^{0} = 0.0563

P(X \geq 8) = P(X = 8) + P(X = 9) + P(X = 10) = 0.2816 + 0.1877 + 0.0563 = 0.5256

52.56% probability that eight or more of the flights will arrive on time.

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