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3 years ago

Please help me on number one I will mark you as the best answer

Mathematics

1 answer:

Fiesta28 [93]3 years ago

4 0

Answer:

-4n - 40 > -120

<=> -4n > -80

<=> 4n < 80

<=> n < 20

=> Option B is correct.

q/(-7) + 1 > 5

<=> q/(-7) > 4

<=> q < 4 x (-7)

<=> q < -28

=> Option C is correct.

Hope this helps!

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2 years ago

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2 years ago

In a sample of 230 teenagers, 202 would like to have smaller class sizes at their school. Find the sample

marishachu [46]

Answer:

The sample proportion is 0.88.

The margin of error is of 0.04.

The interval likely to contain the true population proportion is (0.84,0.92).

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 $1 - \frac{\alpha}{2}$ .

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The margin of error is of:

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

Sample of 230 teenagers, 202 would like to have smaller class sizes at their school.

This means that $n = 230$ , and that the sample proportion is:

$\pi = \frac{202}{230} = 0.88$

The sample proportion is 0.88.

Standard 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 $Z = 1.96$ .

Margin of error:

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

$M = 1.96\sqrt{\frac{0.88*0.12}{230}}$

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The margin of error is of 0.04.

Confidence interval:

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3 years ago

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3 years ago