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GrogVix [38]
3 years ago
11

PLEASE HELP ME IM BAD AT THIS EVEN THOUGH ITS REALLY EASY! SO YOU PROBABLY COULD DO IT!

Mathematics
1 answer:
masha68 [24]3 years ago
5 0

Answer:

<u><em>THE ANSWERS ARE OPTION C and E</em></u>

Step-by-step explanation:

so there are 3 different types of loaves .

the price of french bread is f\\;

from the question , we find that the price of wheat bread is twice of that of the french bread. thus ,

the price of wheat bread is 2f;

the price of cinnamon raisin bread is $2.50 more than the wheat bread. thus ,

the price of cinnamon raisin bread is 2f+2.50;

<em>now the total amount she spent was : $11.25</em>

thus ,

f+2f+(2f+2.50)=11.25

which gives us ,

5f+2.50=11.25\\5f=8.75

<em><u>∴ The correct answers are C and E</u></em>

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based on the simulation, what is the probability that at most 2 of the next 10 callers will have to wait more than 8 minutes to
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Using the binomial distribution, supposing that 0.3 of the callers have to wait more than 8 minutes to have their calls answered, it is found that there is a 0.3828 = 38.28% probability that at most 2 of the next 10 callers will have to wait more than 8 minutes to have their calls answered.

For each caller, there are only two possible outcomes, either they have to wait more than 8 minutes to have their calls answered, or they do not. The probability of a caller having to wait more than 8 minutes is independent of any other caller, which means that the binomial distribution is used to solve this question.

Binomial probability distribution

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

C_{n,x} = \frac{n!}{x!(n-x)!}

The parameters are:

  • x is the number of successes.
  • n is the number of trials.
  • p is the probability of a success on a single trial.

In this problem:

  • 10 callers, hence n = 10
  • Suppose that 0.3 of them have to wait more than 8 minutes, hence p = 0.3

The probability that <u>at most 2</u> of the next 10 callers will have to wait more than 8 minutes is:

P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2)

Then

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P(X = 1) = C_{10,1}.(0.3)^{1}.(0.7)^{9} = 0.1211

P(X = 2) = C_{10,2}.(0.3)^{2}.(0.7)^{8} = 0.2335

Then:

P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) = 0.0282 + 0.1211 + 0.2335 = 0.3828

0.3828 = 38.28% probability that at most 2 of the next 10 callers will have to wait more than 8 minutes to have their calls answered.

A similar problem is given at brainly.com/question/25537909

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