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3 years ago

2x + 7 + (3 - y) = -4 + 6i

Mathematics

1 answer:

Pavlova-9 [17]3 years ago

5 0

Answer:

2.50 is the answer i hope its right i doubled checked it with my algebra teacher

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A research group conducted an extensive survey of 2968 wage and salaried workers on issues ranging from relationships with their

kodGreya [7K]

Answer:

We need a sample of at least 1797 if we wish to be 95% confident that the sample percentage of those equating success with personal satisfaction is within 2.3% of the population percentage.

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:

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

95% confidence level

So $\alpha = 0.05$ , z is the value of Z that has a pvalue of $1 - \frac{0.05}{2} = 0.975$ , so $Z = 1.96$ .

In this problem, we have that:

$p = 0.55$

How large a sample is needed if we wish to be 95% confident that the sample percentage of those equating success with personal satisfaction is within 2.3% of the population percentage?

We have to find n for which $M = 0.023$ . So

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

$0.023 = 1.96\sqrt{\frac{0.55*0.45}{n}}$

$0.023\sqrt{n} = 0.9751$

$\sqrt{n} = \frac{0.9751}{0.0023}$

$\sqrt{n} = 423.95$

$\sqrt{n}^{2} = (42.395)^{2}$

$n = 1797$

We need a sample of at least 1797 if we wish to be 95% confident that the sample percentage of those equating success with personal satisfaction is within 2.3% of the population percentage.

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3 years ago

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antiseptic1488 [7]

Answer:

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Step-by-step explanation:

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A ferris wheel

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