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3 years ago

Plz answer it fast...

Mathematics

1 answer:

Sveta_85 [38]3 years ago

4 0

Answer:

Rs 2163.20.

Step-by-step explanation:

The formula for this is

A = P(1 + r)^t where P is the amount invested, r = the rate (as a decimal) and t = the number of years.

So for rs 2000 we have:

A = 2000(1 + 0.04)^2 = Rs 2163.20

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3 years ago

An automobile manufacturer would like to know what proportion of its customers are not satisfied by the service provided by the

zheka24 [161]

Answer:

The sample size needed if the margin of error of the confidence interval is to be about 0.04 is 18.

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 $Z = 1.96$ .

Past studies suggest this proportion will be about 0.15

This means that $p = 0.15$

Find the sample size needed if the margin of error of the confidence interval is to be about 0.04

This is n when M = 0.04. So

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

$0.04 = 1.96\sqrt{\frac{0.15*0.85}{n}}$

$0.04\sqrt{n} = 1.96\sqrt{0.15*0.85}$

$\sqrt{n} = \frac{1.96\sqrt{0.15*0.85}}{0.04}$

$(\sqrt{n})^{2} = (\frac{1.96\sqrt{0.15*0.85}}{0.04})^{2}$

$n = 17.5$

Rounding up

The sample size needed if the margin of error of the confidence interval is to be about 0.04 is 18.

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