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STatiana [176]
3 years ago
8

Suppose we want to choose for letters without replacement from 18 distant lovers how many ways can this be done if the order of

the choices does not matter
Mathematics
1 answer:
Lelu [443]3 years ago
7 0

Answer:

This can be done in 3,060 ways.

Step-by-step explanation:

Letters are chosen without replacement, and the order does not matter, which means that the combinations formula is used.

Combinations formula:

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)!}

In this question:

Four letters from a set of 18. So

C_{18,4} = \frac{18!}{4!14!} = 3060

This can be done in 3,060 ways.

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H_0: \mu_B - mu_A \leq 2

At the alternative hypothesis, it is <u>tested if it outlasts by more than 2 hours</u>, that is:

H_1: \mu_B - \mu_A > 2

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Hence, the standard errors are:

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The distribution of the difference has <u>mean and standard deviation</u> given by:

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The test statistic is given by:

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t = \frac{2.58 - 2}{0.2735}

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A similar problem is given at brainly.com/question/13873630

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