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lesya [120]
3 years ago
8

A firm maintains 22 cars for business purposes. From past experience, it is known that approximately 10% will require major engi

ne service during a one-year period. What is the probability that at most 1 car will require major engine repair next year?
Mathematics
1 answer:
seropon [69]3 years ago
3 0

Answer:

The probability that at most 1 car will require major engine repair next year is P=0.3392.

Step-by-step explanation:

This can be modeled as a binomial random variable, with n=22 and p=0.1.

The probability that exavtly k cars will require major engine repair next year is:

P(x=k) = \dbinom{n}{k} p^{k}(1-p)^{n-k}

Then, the probability that at most 1 car will require major engine repair next year is:

P(x\leq1)=P(x=0)+P(x=1)\\\\\\P(x=0) = \dbinom{22}{0} p^{0}(1-p)^{22}=1*1*0.0985=0.0985\\\\\\P(x=1) = \dbinom{22}{1} p^{1}(1-p)^{21}=22*0.1*0.1094=0.2407\\\\\\P(x\leq1)=0.0985+0.2407\\\\P(x\leq1)=0.3392

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siniylev [52]

Answer:

12 students

Step-by-step explanation:

Because the total is 4 sticks, and each student needs 1/3 of a stick, we would have to see how many 1/3 sticks fit into 4 sticks. Because 3 of the 1/3 sticks fit into 1 stick, then 4 sticks would be 4*3=12 of the 1/3 sticks, which means 12 students can do it.

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2 years ago
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Calculate the mean, median, and mode of the following set of data. Round to the nearest tenth. 7, 3, 2, 1, 13, 8, 1, 5, 14, 11,
zysi [14]

Answer:

Mean = 7.3, median = 7, mode = 1.

Step-by-step explanation:

First arrange the data in order:

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3 years ago
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Verdich [7]

Answer:

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Step-by-step explanation:

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4t is your answer.

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3 years ago
Question 16 of 50
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Does anyone know the answer to the last question? :)
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