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3 years ago

Sarah is 8 years older than Tanya. Sarah's age is also 5/4 of Tanya's age. How old is each person?

1 answer:

3 0

S = t + 8
s = 5/4t

t + 8 = 5/4t
8 = 5/4t - t
8 = 5/4t - 4/4t
8 = 1/4t
8 * 4 = t
32 = t <=== tanya's age

s = t + 8
s = 32 + 8
s = 40 <== sarah's age

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bagirrra123 [75]

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3 years ago

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JulsSmile [24]

Answer:

a. The point estimate was of $44,600.

b. The sample size was of 16.

Step-by-step explanation:

Confidence interval concepts:

A confidence interval has two bounds, a lower bound and an upper bound.

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The margin of error is the difference between the two bounds, divided by 2.

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Mean of the bounds, so:

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The point estimate was of $44,600.

b. Determine the sample size used for the analysis.

First we need to find the margin of error, so:

$M = \frac{50463-38737}{2} = 5863$

Relating the margin of error with the sample size:

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

$\alpha = \frac{1 - 0.9}{2} = 0.05$

Now, we have to find z in the Z-table as such z has a p-value of $1 - \alpha$ .

That is z with a pvalue of $1 - 0.05 = 0.95$ , so Z = 1.64.

Now, find the margin of error M 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 problem, we have that $\sigma = 14300, M = 5863$ . So

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

$5863 = 1.645\frac{14300}{\sqrt{n}}$

$5863\sqrt{n} = 1.645*14300$

$\sqrt{n} = \frac{1.645*14300}{5863}$

$(\sqrt{n})^2 = (\frac{1.645*14300}{5863})^2$

$n = 16$

The sample size was of 16.

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3 years ago

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solong [7]

Bse is 10
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2 years ago

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stira [4]

Answer:

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I hope this helped and have a good rest of your day!

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Amanda [17]

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