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3 years ago

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Given: In this figure, `/_AED` and `/_BEC` form a pair of vertical angles. Prove: `/_AED ~= /_ BEC `. Match each statement to it

s corresponding reason. Drag the items on the left to the correct location on the right.

2 answers:

valkas [14]3 years ago

5 0

Answer:

Step-by-step explanation:

weeeeeb [17]3 years ago

4 0

Answer:

Step-by-step explanation:

You might be interested in

The yearbook committee polled 80 randomly selected students from a class of 320 ninth graders to see if they would be willing to

Anon25 [30]

Using the z-distribution, as we are working with a proportion, it is found that the margin of error for the 90% confidence interval is of 0.0524 = 5.24%.

<h3>What is a confidence interval of proportions?</h3>

A confidence interval of proportions is given by:

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

In which:

$\pi$ is the sample proportion.

z is the critical value.

n is the sample size.

The margin of error is given by:

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

In this problem, the critical value is given as z = 1.645, and since 26 out of 80 students said they would be willing to pay extra:

$n = 80, \pi = \frac{26}{80} = 0.325$

Then, the <em>margin of error</em> is of:

$M = 1.645\sqrt{\frac{0.325(0.675)}{80}} = 0.0524$

More can be learned about the z-distribution at brainly.com/question/25890103

5 0

3 years ago

Suppose that we will randomly select a sample of 106 measurements from a population having a mean equal to 19 and a standard dev

svet-max [94.6K]

Answer:

$z =\frac{18.45-19}{\frac{7}{\sqrt{106}}}= -0.809$

And if we use the normal standard distribution or excel we got:

$P(z$

Step-by-step explanation:

For this case we have the following info given:

$\mu = 19$ represent the mean

$\sigma = 7$ represent the standard deviation

$n = 106$ represent the sample size

The distribution for the sample size if we use the central limit theorem (n>30) is given by:

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

And for this case we want to find the following probability:

$P(\bar X< 18.45)$

And for this case we can use the z score formula given by:

$z =\frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}$

And replacing we got:

$z =\frac{18.45-19}{\frac{7}{\sqrt{106}}}= -0.809$

And if we use the normal standard distribution or excel we got:

$P(z$

4 0

4 years ago

Which sequence is geometric and has `1/4` as its fifth term and `1/2` as the common ratio? A. …, `1,` ` 1/2`, `1/4`, `1/8`, … B.

o-na [289]

Looks like its A whose first term would be 4.

6 0

3 years ago

Read 2 more answers

-14+7-9r+13r someone please helpp !

JulsSmile [24]

Answer:

4r-7

Step-by-step explanation:

-14+7-9r+13r

combine like terms

4r-14+7

combine like terms

4r-7

hope this helped :)

4 0

4 years ago

Read 2 more answers

Use complete sentences to describe the transformation that maps ABCD onto its image.

Lapatulllka [165]

General Idea:

When a point or figure on a coordinate plane is moved by sliding it to the right or left or up or down, the movement is called a translation.

Say a point P(x, y) moves up or down ' k ' units, then we can represent that transformation by adding or subtracting respectively 'k' unit to the y-coordinate of the point P.

In the same way if P(x, y) moves right or left ' h ' units, then we can represent that transformation by adding or subtracting respectively 'h' units to the x-coordinate.

P(x, y) becomes $P'(x\pm h, y\pm k)$ . We need to use ' + ' sign for 'up' or 'right' translation and use ' - ' sign for ' down' or 'left' translation.

Applying the concept:

The point A of Pre-image is (0, 0). And the point A' of image after translation is (5, 2). We can notice that all the points from the pre-image moves 'UP' 2 units and 'RIGHT' 5 units.

Conclusion:

The transformation that maps ABCD onto its image is translation given by (x + 5, y + 2),

In other words, we can say ABCD is translated 5 units RIGHT and 2 units UP to get to A'B'C'D'.

6 0

4 years ago