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2 years ago

Pleaseeee helpppp answer correctly !!!!!!!!!!!!!! Will mark Brianliest !!!!!!!!!!!!!!

Mathematics

1 answer:

Fed [463]2 years ago

5 0

Answer:

z=44

Step-by-step explanation:

One way to go about this with the given angles is to use the alternate interior angles theorem to find the full measure of ∠C and ∠A. Then, use the same theorem to set ∠BDA as 52°. Subtract all the angle measures from 360 (because the sum of the interior angles of a quadrilateral is = 360 degrees). Then divide the result by 2.

360-84-84-52-52=88

=88/2

=44

So z=44°

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3 years ago

Read 2 more answers

A 90 percent confidence interval is to be created to estimate the proportion of television viewers in a certain area who favor m

Dominik [7]

Answer:

This means that the first interval, with 9000 viewers, is 3 times as narrower as the second interval, with 1000 viewers.

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of $\pi$ , and a confidence level of $1-\alpha$ , we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the zscore that has a pvalue of $1 - \frac{\alpha}{2}$ .

The width of the interval is:

$W = 2z\sqrt{\frac{\pi(1-\pi)}{n}}$

In this sample:

Two 90% intervals, with different lenghts. So both have the same values for z an $\pi$

Interval A:

9000 viewers.

So the width is

$W_{A} = 2z\sqrt{\frac{\pi(1-\pi)}{9000}}$

Interval B:

100 viewers

So the width is

$W_{B} = 2z\sqrt{\frac{\pi(1-\pi)}{1000}}$

Relationship between the widths:

$R = \frac{W_{A}}{W_{B}} = \frac{2z\sqrt{\frac{\pi(1-\pi)}{9000}}}{2z\sqrt{\frac{\pi(1-\pi)}{1000}}} = \frac{\sqrt{1000}}{\sqrt{9000}} = \frac{\sqrt{1}}{\sqrt{9}} = \frac{1}{3}$

This means that the first interval, with 9000 viewers, is 3 times as narrower as the second interval, with 1000 viewers.

8 0

3 years ago

HELP WILL GIVE BRAINLESS Which choice is equivalent to the fraction below? (Hint: Rationalize the

Ad libitum [116K]

Answer:

C. $\sqrt{x + 2} - \sqrt{x}$

Step-by-step explanation:

$\frac{2}{ \sqrt{x} + \sqrt{x + 2} } \\ \\ = \frac{2}{ (\sqrt{x} + \sqrt{x + 2}) } \times \frac{(\sqrt{x} - \sqrt{x + 2})}{(\sqrt{x} - \sqrt{x + 2})} \\ \\ = \frac{2( \sqrt{x} - \sqrt{x + 2}) }{ ({ \sqrt{x} )}^{2} - {( \sqrt{x + 2} )}^{2} } \\ \\ = \frac{2( \sqrt{x} - \sqrt{x + 2}) }{ x - {( x + 2)} } \\ \\ = \frac{2( \sqrt{x} - \sqrt{x + 2}) }{ x - x - 2 } \\ \\ = \frac{2( \sqrt{x} - \sqrt{x + 2}) }{ - 2 } \\ \\ = - ( \sqrt{x} - \sqrt{x + 2}) \\ \\ \frac{2}{ \sqrt{x} + \sqrt{x + 2} } = \purple{ \bold{\sqrt{x + 2} - \sqrt{x} }}$