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3 years ago

how do you indicate congruent segments in a diagram? how do you indicate congruent angles in a diagram?

Mathematics

1 answer:

ICE Princess25 [194]3 years ago

8 0

9514 1404 393

Answer:

segments: using the same number of hash marks
angles: using the same number of arcs, or hash marks on an arc

Step-by-step explanation:

The attached diagram shows that segments AC and BD are congruent by using a single hash mark on each of those segments. If other segments are congruent, but not congruent to these two, the "decoration" would be different, probably two hash marks. Segments marked with the same "decoration" are intended to be understood as congruent.

The "decoration" used for congruent angles is an arc of some kind. Here, a single arc is used to signify angle CAB is congruent to angle DBA. Additional arcs could be used for other congruent angles, or hash marks can be put on the arcs.

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In a simple random sample of 1500 young Americans 1305 had earned a high school diploma.

inn [45]

Answer:

a) The standard error for this estimate of the percentage of all young Americans who earned a high school diploma is 0.87%.

b) The margin of error, using a 95% confidence level, for estimating the percentage of all young Americans who earned a high school diploma is of 1.71%.

c) The 95% confidence interval for the percentage of all young Americans who earned a high school diploma is (85.29%, 88.71%).

d) The lower bound of the confidence interval is above 80%, which means that the confidence interval supports the claim that the percentage of young Americans who cam high school diplomas has increased.

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 z-score that has a p-value of $1 - \frac{\alpha}{2}$ .

Standard error:

The standard error is:

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

Margin of error:

The margin of error is:

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

The confidence interval is:

Sample proportion plus/minus margin of error. So

$(\pi - M, \pi + M)$

In a simple random sample of 1500 young Americans 1305 had earned a high school diploma.

This means that $n = 1500, \pi = \frac{1305}{1500} = 0.87$

a. What is the standard error for this estimate of the percentage of all young Americans who earned a high school diploma?

$s = \sqrt{\frac{\pi(1-\pi)}{n}} = \sqrt{\frac{0.87*0.13}{1500}} = 0.0087$

0.0087*100% = 0.87%.

The standard error for this estimate of the percentage of all young Americans who earned a high school diploma is 0.87%.

b. Find the margin of error, using a 95% confidence level, for estimating the percentage of all young Americans who earned a high school diploma.

95% confidence level

So $\alpha = 0.05$ , z is the value of Z that has a p-value of $1 - \frac{0.05}{2} = 0.975$ , so $Z = 1.96$ .

Then

$M = zs = 1.96*0.0087 = 0.0171$

0.0171*100% = 1.71%

The margin of error, using a 95% confidence level, for estimating the percentage of all young Americans who earned a high school diploma is of 1.71%.

c. Report the 95% confidence interval for the percentage of all young Americans who earned a high school diploma.

87% - 1.71% = 85.29%

87% + 1.71% = 88.71%.

The 95% confidence interval for the percentage of all young Americans who earned a high school diploma is (85.29%, 88.71%).

d. Suppose that in the past, 80% of all young Americans earned high school diplomas. Does the confidence interval you found in part c support or refute the claim that the percentage of young Americans who cam high school diplomas has increased? Explain.

The lower bound of the confidence interval is above 80%, which means that the confidence interval supports the claim that the percentage of young Americans who cam high school diplomas has increased.

4 0

3 years ago

If a six-sided die(1-6) and a ten-sided die(1-10) are rolled simultaneously, then what is the probability of rolling both with a

AnnyKZ [126]

Answer:

$Pr = 0.20$