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oksian1 [2.3K]
3 years ago
14

Use the diagram. AB is a diameter, and AB is perpendicular to CD. The figure is not drawn to scale.

Mathematics
1 answer:
Westkost [7]3 years ago
7 0

Answer:

The measure of arc BD is 147°

Step-by-step explanation:

we know that

If segment AB is perpendicular to segment CD

then

The measure of the inner angle CPA is a right angle (90 degrees)

Remember that

The measure of the inner angle is the semi-sum of the arcs comprising it and its opposite.

so

m∠CPA=(1/2)[arc AC+arc BD]

substitute the given values

90°=(1/2)[33°+arc BD]

180°=33°+arc BD

arc BD=180°-33°=147°

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A study was conducted to estimate the difference in the mean salaries of elementary school teachers from two neighboring states.
umka21 [38]

Answer:

(28900-30300) -2.07 \sqrt{\frac{2300^2}{10} +\frac{2100^2}{14}} = -3301.70

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The confidence interval would be -3301.70 \leq \mu \leq 501.698 and since the confidence interval contains the value of 0 we don't have enough evidence to conclude that the difference between the two states for the salary of teachers are significantly different.

Step-by-step explanation:

We have the following info given by the problem

\bar X_1 = 28900 the sample mean for the salaries of teachers in Indiana

s_1 = 2300 the sample deviation for the salary of  teachers in Indiana

n_1 =10 the sample size from Indiana

\bar X_2 = 30300 the sample mean for the salaries of teachers in Michigan

s_2 = 2100 the sample deviation for the salary of  teachers in Michigan

n_2 =14 the sample size from Michigan

We want to find a confidence interval for the difference in the two means and the formula for this case is given by;

(\bar X_1 -\bar X_2) \pm t_{\alpha/2}\sqrt{\frac{s^2_1}{n_1} +\frac{s^2_2}{n_2}}

The degrees of freedom for this case are:

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The confidence is 95%so then the significance is \alpha=0.05 and the \alpha/2 =0.025, we need to find a critical value in the t distribution with 22 degrees of freedom who accumulates 0.025 of the area on each tail and we got:

t_{\alpha/2}= 2.07

And now replacing in the formula for the confidence interval we got:

(28900-30300) -2.07 \sqrt{\frac{2300^2}{10} +\frac{2100^2}{14}} = -3301.70

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