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4 years ago

HELP ME PLEASEEEE!!!!

1 answer:

4 0

In this question is given to points, with coordinates A(8.5,12) and B(-6.75,7.5). It is asked to find the length between both points. The only way to find the solution is by using the Distance Formula, which is represented by
d=√(x2-x1)^2+(y2-y1)^2.

d=√(x2-x1)^2+(y2-y1)^2
d=√(8.5--6.75)^2+(12-7.5)^2
d=√(15.25)^2+(4.5)^2
d=√(232.5625)+(20.25)
d=√252.8125
d=15.90007862 or 15.90

The distance between the points A(8.5,12) and B(-6.75,7.5) is 15.90 units.

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Answer:

The null and alternative hypothesis are:

$H_0: \mu_X-\mu_Y=0\\\\H_a:\mu_X-\mu_Y> 0$

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The null hypothesis failed to be rejected.

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Step-by-step explanation:

This is a hypothesis test for the difference between populations means.

The claim is that the mean score μX on one-tailed hypothesis test questions is higher than the mean score μY on two-tailed hypothesis test questions .

Then, the null and alternative hypothesis are:

$H_0: \mu_X-\mu_Y=0\\\\H_a:\mu_X-\mu_Y> 0$

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The sample X (one-tailed), of size nX=80 has a mean of 7.8 and a standard deviation of 1.06.

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The difference between sample means is Md=0.16.

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$s_{M_d}=\sqrt{\dfrac{\sigma_X^2+\sigma_Y^2}{n}}=\sqrt{\dfrac{1.06^2+1.32^2}{80}}\\\\\\s_{M_d}=\sqrt{\dfrac{2.866}{80}}=\sqrt{0.036}=0.189$

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The degrees of freedom for this test are:

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This test is a right-tailed test, with 158 degrees of freedom and t=0.8453, so the P-value for this test is calculated as (using a t-table):

$P-value=P(t>0.8453)=0.1996$

As the P-value (0.1996) is greater than the significance level (0.05), the effect is not significant.

The null hypothesis failed to be rejected.

There is not enough evidence to support the claim that the mean score μX on one-tailed hypothesis test questions is higher than the mean score μY on two-tailed hypothesis test questions .

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