Mathematics: For each part below, decide if the triangles are similar. If they are similar, use their similarity to solve for x. If they are not similar, explain why not?

For each part below, decide if the triangles are

Ответ:

Jennifer312332

18.01.2022

The answer to your question is x = 10.7

Step-by-step explanation:

Yes, angles are similar because they share a vertical angle and two lines are parallels so there are two pairs of alternate interior angles.

Use proportions to solve this problem

$\frac{x}{8} = \frac{8}{6}$

Solve for x

x = 8(8/6)

Simplification

x = 64/6

Result

x = 10.7

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Ответ:

keshaayro1618

03.12.2021

Mathematics

We need a sample size of at least 23 for a 99% confidence interval, with a tolerable interval width of 0.4.

We need a sample size of at least 9 for a 95% confidence interval with a tolerable width of 0.5,

Step-by-step explanation:

How large a sample size is required if want a 99% confidence interval, with a tolerable interval width of 0.4?

We have that to find our $\alpha$ level, that is the subtraction of 1 by the confidence interval divided by 2. So:

$\alpha = \frac{1-0.99}{2} = 0.005$

Now, we have to find z in the Ztable as such z has a pvalue of $1-\alpha$ .

So it is z with a pvalue of 1-0.005 = 0.995 , so z = 2.575

Now, find the margin of error(width) as such

$M = z*\frac{\sigma}{\sqrt{n}}$

In which $\sigma$ is the standard deviation of the population and n is the size of the sample.

For this item, we have:

$M = 0.4, \sigma = 0.75$ . So

$M = z*\frac{\sigma}{\sqrt{n}}$

$0.4 = 2.575*\frac{0.75}{\sqrt{n}}$

$0.4\sqrt{n} = 1.93125$

$\sqrt{n} = \frac{1.93125}{0.4}$

$\sqrt{n} = 4.828125$

$\sqrt{n}^{2} = (4.828125)^{2}$

n = 23

We need a sample size of at least 23 for a 99% confidence interval, with a tolerable interval width of 0.4.

How large a sample would we need if were interested in a 95% confidence interval with a tolerable width of 0.5?

We have that to find our $\alpha$ level, that is the subtraction of 1 by the confidence interval divided by 2. So:

$\alpha = \frac{1-0.95}{2} = 0.025$

Now, we have to find z in the Ztable as such z has a pvalue of $1-\alpha$ .

So it is z with a pvalue of 1-0.025 = 0.975 , so z = 1.96

Now, find the margin of error(width) as such

$M = z*\frac{\sigma}{\sqrt{n}}$

In which $\sigma$ is the standard deviation of the population and n is the size of the sample.

For this item, we have:

$M = 0.5, \sigma = 0.75$ . So

$M = z*\frac{\sigma}{\sqrt{n}}$

$0.5 = 1.96*\frac{0.75}{\sqrt{n}}$

$0.5\sqrt{n} = 1.47$

$\sqrt{n} = \frac{1.47}{0.5}$

$\sqrt{n} = 2.94$

$\sqrt{n}^{2} = (2.94)^{2}$

$n \cong 9$

We need a sample size of at least 9 for a 95% confidence interval with a tolerable width of 0.5,

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For each part below, decide if the triangles are similar. If they are similar, use their similarity to solve for x. If they are not similar, explain why not?

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