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Sphinxa [80]
3 years ago
6

Which system of equations is represented by the graph?

Mathematics
2 answers:
slavikrds [6]3 years ago
5 0

The red line has a y-intercept of -4, so the linear equations of choices A and C can be eliminated.

The numerator of the rational expression of choice B will not be zero when x=4, so that choice can be eliminated. (That is, check to see if the point (4, 0) satisfies the equation. It does not.)

The problem requirements are met by the last choice:

d y = x minus 4, over x plus 2

y = x − 4

Mkey [24]3 years ago
5 0

Answer:

d y = x minus 4, over x plus 2

y = x − 4

Step-by-step explanation:

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miskamm [114]

Answer:

The conditions for use of the normal model to represent the distribution of sample proportion are not met. He should increase the sample size until the conditions are met.

If the test is done anyway, the null hypothesis failed to be rejected.

The conclusion is taht there is not enough evidence to support the claim that the percentage is lower in this district.

Step-by-step explanation:

The conditions for use of the normal model to represent the distribution of sample proportion are not met, as the affirmative responses are less than 10.

np=80*0.1=8

If the test of hypothesis is done as if the conditiones were met, we know that the claim is that  the percentage is lower in this district.

Then, the null and alternative hypothesis are:

H_0: \pi=0.173\\\\H_a:\pi

The significance level is 0.05.

The sample has a size n=80.

The sample proportion is p=0.1.

 

The standard error of the proportion is:

\sigma_p=\sqrt{\dfrac{\pi(1-\pi)}{n}}=\sqrt{\dfrac{0.173*0.827}{80}}\\\\\\ \sigma_p=\sqrt{0.001788}=0.042

Then, we can calculate the z-statistic as:

z=\dfrac{p-\pi+0.5/n}{\sigma_p}=\dfrac{0.1-0.173+0.5/80}{0.042}=\dfrac{-0.067}{0.042}=-1.578

This test is a left-tailed test, so the P-value for this test is calculated as:

P-value=P(z

As the P-value (0.057) 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 percentage is lower in this district.

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3 years ago
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