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allochka39001 [22]

3 years ago

14

How do I do this? Please explain in detail.

2 answers:

umka2103 [35]3 years ago

8 0

22.5/(x-6) + 22.5/(x+6) = 9
multiply by x-6
=> (x-6)22.5/(x-6) + (x-6)22.5/(x+6) = 9(x-6)
=> 22.5 + (x-6)22.5/(x+6) = 9(x-6)
multiply by x+6
=> (x+6)22.5 + (x+6)(x-6)22.5/(x+6) = 9(x-6)(x+6)
=> (x+6)22.5 + (x-6)22.5 = 9(x-6)(x+6)
distribute
=> 22.5x+6(22.5) + 22.5x - 6(22.5) = 9(x^2 - 36)
=> 45x = 9x^2 - 9(36)
=> 0 = 9x^2 - 45x - 9(36)
divide by 9
=> 0 = x^2 - 5x - 36
=> 0 = x^2 - 5x - 36
=> 0 = (x - 9)(x + 4)
x=9 and -4

Neporo4naja [7]3 years ago

7 0

Https://us-static.z-dn.net/files/d39/8be32892e96d7954b568343588bbf6c2.jpeg

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A sociologist wishes to estimate the percentage of the United States population living in poverty. What size sample should be ob

andriy [413]

Answer:

a) We need a sample size of at least 705.

b) We need a sample size of at least 1692.

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 zscore that has a pvalue of $1 - \frac{\alpha}{2}$ .

The margin of error is:

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

90% confidence level

So $\alpha = 0.1$ , z is the value of Z that has a pvalue of $1 - \frac{0.1}{2} = 0.95$ , so $Z = 1.645$ .

a) If she uses 1999 estimate of 11.8% obtained from the Current Population Survey.

We need a sample of size at least n

n is found when $M = 0.02, \pi = 0.118$ .

So

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

$0.02 = 1.645\sqrt{\frac{0.118*0.882}{n}}$

$0.02\sqrt{n} = 1.645\sqrt{0.118*0.882}$

$\sqrt{n} = \frac{1.645\sqrt{0.118*0.882}}{0.02}$

$(\sqrt{n})^{2} = (\frac{1.645\sqrt{0.118*0.882}}{0.02})^{2}$

$n = 704.08$

Rounding up

We need a sample size of at least 705.

b) She does not use any estimate.

Same thing as above, we just use $\pi = 0.5$ when do not use any estimate.

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

$0.02 = 1.645\sqrt{\frac{0.5*0.5}{n}}$

$0.02\sqrt{n} = 1.645\sqrt{0.5*0.5}$

$\sqrt{n} = \frac{1.645\sqrt{0.5*0.5}}{0.02}$

$(\sqrt{n})^{2} = (\frac{1.645\sqrt{0.5*0.5}}{0.02})^{2}$

$n = 1691.2$

Rounding up

We need a sample size of at least 1692.

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