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3 years ago

Salma is constructing the inscribed circle for △MNP .

Mathematics

1 answer:

Lena [83]3 years ago

5 0

Construct the angle bisector of angle M.
According to the info of another test :)

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The difference of squares is (a - b)(a + b) = a² - b². Since a = x and b = 2, the answer is x² - 4.

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A line that passes through (-8,4) and has a slope of -5/4

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3 years ago

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3 years ago

A programmer plans to develop a new software system. In planning for the operating system that he will use, he needs to estimat

erma4kov [3.2K]

Answer:

a) $n = 9604$

b) n = 381

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:

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

90% confidence level

So $\alpha = 0.1$ , z is the value of Z that has a pvalue of $1 - \frac{0.1}{2} = 0.95$ , so $Z = 1.645$ .

a. Assume that nothing is known about the percentage of computers with new operating systems. n = ?

When we do not know the proportion, we use $\pi = 0.5$ , which is when we are going to need the largest sample size.

The sample size is n when M = 0.01.

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

$0.01 = 1.96\sqrt{\frac{0.5*0.5}{n}}$

$0.01\sqrt{n} = 1.96*0.5$

$\sqrt{n} = \frac{1.96*0.5}{0.01}$

$(\sqrt{n})^{2} = (\frac{1.96*0.5}{0.01})^{2}$

$n = 9604$

b. Assume that a recent survey suggests that 99% of computers use a new operating system. n = ?

Now we have that $\pi = 0.99$ . So

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

$0.01 = 1.96\sqrt{\frac{0.99*0.01}{n}}$

$0.01\sqrt{n} = 1.96*\sqrt{0.99*0.01}$

$\sqrt{n} = \frac{1.96*\sqrt{0.99*0.01}}{0.01}$

$(\sqrt{n})^{2} = (\frac{1.96*\sqrt{0.99*0.01}}{0.01})^{2}$

$n = 380.3$

Rouding up

n = 381

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3 years ago