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denis-greek [22]

3 years ago

12

An english reading list has 9 american novels and 7 english novels. A student must read 5 from the list and at least 3 must be e

nglish novels. In how many different ways can the five books be selected combination

1 answer:

Elenna [48]3 years ago

7 0

$_7C_3\cdot {_{13}C_2}=\dfrac{7!}{3!4!}\cdot\dfrac{13!}{2!11!}=\dfrac{5\cdot6\cdot7}{2\cdot3}\cdot\dfrac{12\cdot13}{2}=2730$

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Minchanka [31]

Answer:

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The confidence interval for the difference of two proportions would be given by this formula

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For this case we have the confidence interval given by: (-0.0313,0.2753). From this we can find the margin of erro on this way:

$ME= \frac{0.2753-(-0.0313)}{2}=0.1533$

And we know that the margin of erro is given by:

$ME=z_{\alpha/2} \sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1} +\frac{\hat p_2 (1-\hat p_2)}{n_2}}$

We have all the values except the value for $z_{\alpha/2}$

So we can find it like this:

$0.1533=z_{\alpha/2} \sqrt{\frac{0.768(1-0.768)}{92} +\frac{0.646 (1-0.646)}{95}}$

And solving for $z_{\alpha/2}$ we got:

$z_{\alpha/2}=2.326$

And we can find the value for $\alpha/2$ with the following excel code:

"=1-NORM.DIST(2.326,0,1,TRUE)"

And we got $\alpha/2 =0.01$ so then the value for $\alpha=0.02$ and then the confidence level is given by: $Conf=1-0.02=0.98$ or 98%

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azamat

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