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3 years ago

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A librarian has a set of ten different books, including four different books about Abraham Lincoln. The librarian wants to put t

he ten books on a shelf with the four Lincoln books next to each other, somewhere on the shelf among the other six different books. How many different arrangements of the ten books are possible?

1 answer:

Darya [45]3 years ago

3 0

Answer:

120960

Step-by-step explanation:

Given that a librarian has a set of ten different books, including four different books about Abraham Lincoln.

The librarian wants to put the ten books on a shelf with the four Lincoln books next to each other, somewhere on the shelf among the other six different books.

First arrnage the 6 other books in the shelf.

This can be done in $6! ways$

Now arrange the 4 books of Lincoln

This can be done in $4! ways$

Now we have 7 places to insert Lincoln books including two corners and in between.

Thus this can be done in 7 ways

Total no of ways = $6!(4!)(7) = 120960$

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Answer:

a

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b

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c

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d

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Step-by-step explanation:

From the question we are told that

The sample size is n = 1000

The population proportion is $\r p = 0.25$

Considering question a

The population parameter of interest is the true proportion of Greek who are suffering

While the point estimate of this parameter is proportion of those that would rate their lives poorly enough to be considered "suffering". which is 25%

Considering question b

The condition for constructing a confidence interval is

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$Z_{\frac{\alpha }{2} } = 1.96$

Generally the margin of error is mathematically represented as

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substituting values

$E = 1.96 * \sqrt{ \frac{ 0.25 (1 - 0.25 ) }{ 1000} }$

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The 95% confidence interval is mathematically represented as

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substituting values

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substituting values

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considering d

If the confidence level is increased which will in turn reduce the level of significance but increase the critical value( $Z_{\frac{\alpha }{2} }$ ) and this will increase the margin of error( deduced from the formula for margin of error i.e $E \ \alpha \ Z_{\frac{\alpha }{2} }$ ) which will make the confidence interval wider

considering e

Looking at the formula for margin of error if the we see that if the sample size is increased the margin of error will reduce making the confidence level narrower

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