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3 years ago

Prove that angle ABD is congruent to angle CBE with solution!

Mathematics

1 answer:

WINSTONCH [101]3 years ago

8 0

ANSWER:

the conditions are the angle a is equal to angle c and ab = bc . Hence we need to prove that the triangles is congruent to the triangle cbe. ... angle A =angle C and AB=BC.

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What is the measure of angle A in the triangle below?

laiz [17]

The answer is 30.
You need to use sine here
So:

5 0

3 years ago

Find NQ if NP = 13 cm and line l is a segment bisector.

Zielflug [23.3K]

Answer:

NQ = 26 cm

Step-by-step explanation:

A line bisector divides a line into two equal segments.

Given that line l is the bisector of segment NQ, it then means it divides NQ into two equal segments, namely, segment NP and segment PQ.

Thus, NP = ½ of segment NQ

Therefore, if NP = 13 cm,

13 = ½*NQ

multiply both sides by 2

2*13 = NQ

26 = NQ

NQ = 26 cm

7 0

3 years ago

A government report gives a 99% confidence interval for the proportion of welfare recipients who have been receiving welfare ben

juin [17]

Answer:

The estimation for the proportion for this case is $\hat p = 0.21$

And we know that the 99% confidence interval is given by:

$0.21 \pm 0.045$

So then the margin of error at 99% of confidence is 0.045, and the margin of error is given by this formula:

$ME= z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}$

If we decrease the confidence level this margin of error can't be higher than 0.045 so then the correct answer for this case would be:

d. 21% ± 4.8%

Because if the z value decrease from 2.58 to 1.96 not makes sense that the margin of error increase from 0.045(4.5%) to 0.048(4.8%)

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

Solution to the problem

The population proportion have the following distribution

$p \sim N(p,\sqrt{\frac{\hat p(1-\hat p)}{n}})$

The confidence interval would be given by this formula

$\hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}$

For the 95% confidence interval the value of $\alpha=1-0.95=0.05$ and $\alpha/2=0.025$ , with that value we can find the quantile required for the interval in the normal standard distribution.

$z_{\alpha/2}=1.96$

For the 99% confidence interval the value of $\alpha=1-0.99=0.01$ and $\alpha/2=0.005$ , with that value we can find the quantile required for the interval in the normal standard distribution.