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3 years ago

What is the simplified form of the following expression? 8ab/-4b

Mathematics

1 answer:

kondaur [170]3 years ago

3 0

-2a

Step-by-step explanation:

steps:1. first write the question again : 8ab/-4b

2.first 8 divide by -4=-2

3.write ' a' with -2because u don't have another 'a' down belowso that u can pair it up and cancel

so: -2a.

4. cancel 'b' on top and 'b' at the bottom.

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Answer:

The minimum sample size needed is 125.

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:

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

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So $\alpha = 0.01$ , z is the value of Z that has a pvalue of $1 - \frac{0.01}{2} = 0.995$ , so $Z = 2.575$ .

What minimum sample size would be necessary in order ensure a margin of error of 10 percentage points (or less) if they use the prior estimate that 25 percent of the pick-axes are in need of repair?

This minimum sample size is n.

n is found when $M = 0.1$

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

$0.1 = 2.575\sqrt{\frac{0.25*0.75}{n}}$

$0.1\sqrt{n} = 2.575{0.25*0.75}$

$\sqrt{n} = \frac{2.575{0.25*0.75}}{0.1}$

$(\sqrt{n})^{2} = (\frac{2.575{0.25*0.75}}{0.1})^{2}$

$n = 124.32$

Rounding up

The minimum sample size needed is 125.

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