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3 years ago

Plz give 8th and 9th question

Mathematics

1 answer:

Neko [114]3 years ago

3 0

Answer:

for number 8 it is the multiplictivie inverse because you just do 8/9 x 2 so it would be correct and for number 9 it is not the multiplicative inverse because it is not reversing it also can i pls get brainliest

Step-by-step explanation:

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2 years ago

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

butalik [34]

Answer:

a) A sample size of 5615 is needed.

b) 0.012

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:

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

99.5% confidence level

So $\alpha = 0.005$ , z is the value of Z that has a pvalue of $1 - \frac{0.005}{2} = 0.9975$ , so $Z = 2.81$ .

(a) Past studies suggest that this proportion will be about 0.2. Find the sample size needed if the margin of the error of the confidence interval is to be about 0.015.

This is n for which M = 0.015.

We have that $\pi = 0.2$

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

$0.015 = 2.81\sqrt{\frac{0.2*0.8}{n}}$

$0.015\sqrt{n} = 2.81\sqrt{0.2*0.8}$

$\sqrt{n} = \frac{2.81\sqrt{0.2*0.8}}{0.015}$

$(\sqrt{n})^{2} = (\frac{2.81\sqrt{0.2*0.8}}{0.015})^{2}$

$n = 5615$

A sample size of 5615 is needed.

(b) Using the sample size above, when the sample is actually contacted, 12% of the sample say they are not satisfied. What is the margin of the error of the confidence interval?

Now $\pi = 0.12, n = 5615$ .

We have to find M.

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

$M = 2.81\sqrt{\frac{0.12*0.88}{5615}}$

$M = 0.012$

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3 years ago

Explain in detail how to find the vertex: <br> y=-2x^2 + 2x + 3

Cerrena [4.2K]

$\bf \textit{vertex of a vertical parabola, using coefficients} \\\\ y=\stackrel{\stackrel{a}{\downarrow }}{-2}x^2\stackrel{\stackrel{b}{\downarrow }}{+2}x\stackrel{\stackrel{c}{\downarrow }}{+3} \qquad \qquad \left(-\cfrac{ b}{2 a}~~~~ ,~~~~ c-\cfrac{ b^2}{4 a}\right)$

$\bf \left(-\cfrac{2}{2(-2)}~~,~~3-\cfrac{2^2}{4(-2)} \right)\implies \left( \cfrac{1}{2}~~,~~3+\cfrac{4}{8} \right)\implies \left(\cfrac{1}{2}~~,~~\cfrac{28}{8} \right) \\\\\\ \left(\cfrac{1}{2}~~,~~\cfrac{7}{2} \right)\implies \left(\frac{1}{2}~,~3\frac{1}{2} \right)$

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3 years ago

Plz give 8th and 9th question​

Plz give 8th and 9th question