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Inga [223]
3 years ago
8

What are the steps of factoring (5m-2)^2-(3m-4)^2 ?

Mathematics
1 answer:
Svetlanka [38]3 years ago
6 0
We use the formula a^2 - b^2 = ( a - b )( a + b );
We have a = 5m - 2 and b = 3m - 4;
<span>(5m-2)^2-(3m-4)^2 = (5m - 2 -3m + 4) x (5m-2 + 3m - 4) = (2m + 2)(8m - 6) = 2(m +1) x 2(4m - 3) = 4(m+1)(4m-3);

</span>
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a)

The null hypothesis is H_0: p = 0.91.

The alternate hypothesis is H_1: p < 0.91.

The decision rule is: accept the null hypothesis for z > -1.645, reject the null hypothesis for z < -1.645.

Since z = -1.79 < -1.645, we reject the null hypothesis and accept the alternate hypothesis that that percentage of athletes who graduate is less than for the student population at large.

b)

The p-value for this test is 0.0367. Since this p-value is less than the significance level of \alpha = 0.05, we reject the null hypothesis and accept the alternate hypothesis that that percentage of athletes who graduate is less than for the student population at large.

Step-by-step explanation:

Question a:

Perform the appropriate hypothesis test to determine whether this is significant evidence that the percentage of athletes who graduate is less than for the student population at large:

At the null hypothesis, we test if the proportion is the same as the student population, of 91%. Thus:

H_0: p = 0.91

At the alternate hypothesis, we test that the proportion for athletes is less than 91%, that is:

H_1: p < 0.91

The test statistic is:

z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}

In which X is the sample mean, \mu is the value tested at the null hypothesis, \sigma is the standard deviation and n is the size of the sample.

Test if the proportion is less at the 0.05 level:

The critical value is z with a p-value of 0.05, that is, z = -1.645. Thus, the decision rule is: accept the null hypothesis for z > -1.645, reject the null hypothesis for z < -1.645.

0.91 is tested at the null hypothesis:

This means that \mu = 0.91, \sigma = \sqrt{0.91*0.09}

A sports reporter contacted 152 athletes randomly sampled from that same university and time period and found that 132 of them had graduated within 6 years.

This means that n = 152, X = \frac{132}{152} = 0.8684

Value of the test statistic:

z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}

z = \frac{0.8684 - 0.91}{\frac{\sqrt{0.91*0.09}}{\sqrt{152}}}

z = -1.79

Since z = -1.79 < -1.645, we reject the null hypothesis and accept the alternate hypothesis that that percentage of athletes who graduate is less than for the student population at large.

(b) (3 points) Calculate the P-value for this test. Explain how this P-value can be use to test the hypotheses in part (a).

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The p-value for this test is 0.0367. Since this p-value is less than the significance level of \alpha = 0.05, we reject the null hypothesis and accept the alternate hypothesis that that percentage of athletes who graduate is less than for the student population at large.

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