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NARA [144]
3 years ago
14

(Please Help) Line s ‖ segment RP. m∠3 = m∠__?

Mathematics
1 answer:
goldfiish [28.3K]3 years ago
7 0
<span>before answering this question, we should know the properties of angles and parallel lines. For example, two angles are called interior alternate angles, if each angle is created by the existence of two parallel lines and a an another line called transversal. As for our case, 3 and 5 are alternate angles, so meas 3 = meas 5. It verifies Z test when we look at the figure.</span>
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z = \frac{\bar X -\mu}{\sigma_{\bar X}}= \frac{70.9-71.5}{0.77}=-0.779

From this result we can conclude that the value of 70.9 is 0.78 deviation below the true mean of 71.5 and that can be considered as unusual. If we conduct a hypothesis test or a confidence interval we will see that we have enough evidence to conclude that the true mean is not significantly different from 71.5.

Step-by-step explanation:

For this case from all the population we know that the population mean and deviation are:

\mu = 71.5,\sigma = 4.87

And we take a random sample of size n =40 and we got a sample mean calculated with the following formula:

\bar X =\frac{\sum_{i=1}^n X_i}{n}= \frac{\sum_{i=1}^{40} X_i}{40}=70.9

And we want to test if this value is unusually low.

Since the sample size is large n>30 we can use the central limit theorem who says that the distribution for the sample mean is given by:

\bar X \sim N (\mu , \frac{\sigma}{\sqrt{n}})

And on this case if we replace the values that we have we got:

\bar X \sim N (\mu_{\bar X}=71.5,\sigma_{\bar X}= \frac{\sigma}{\sqrt{n}}=\frac{4.87}{\sqrt{40}}=0.77)

For this case we can calculate how many deviations above or below is our calculated value from the sample of size 40, using the z score given by:

z = \frac{\bar X -\mu}{\sigma_{\bar X}}= \frac{70.9-71.5}{0.77}=-0.779

From this result we can conclude that the value of 70.9 is 0.78 deviation below the true mean of 71.5 and that can be considered as unusual. If we conduct a hypothesis test or a confidence interval we will see that we have enough evidence to conclude that the true mean is not significantly different from 71.5.

3 0
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