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2 years ago

Someone Please help me..What must the value of x be in order for the figures below to be similar?

Mathematics

1 answer:

Ksenya-84 [330]2 years ago

3 0

Answer:

16 cm

Step-by-step explanation:

x+x

8+8

16 cm

therefore 16 cm must be the value of x be in order for the figures below to be similar

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The lifespans of gorillas in a particular zoo are normally distributed. The average gorilla lives 20.8 years; the

Luda [366]

If the mean is 20.8, one standard deviation each way is adding and subtracting 3.1, so 17.7 and 23.9 (68% of values)

Two standard deviations adding and subtracting 3.1*2 = 6.2, or 14.6 and 27.

Three standard deviations is 11.5 and 30.1

So we have
11.5 - 14.6 - 17.7 - 20.8 - 23.9 - 27 - 30.1

Going left to 11.5 is 3 standard deviations out, so 99.7/2 = 49.85%

Going right to 27 is 2 standard deviations out, so 95/2 = 47.5%

Add those two % to get 97.32%

This is hard to do without a picture so I hope that helps!

3 0

3 years ago

A candidate for a US Representative seat from Indiana hires a polling firm to gauge her percentage of support among voters in he

UkoKoshka [18]

Answer:

a) The minimum sample size is 601.

b) The minimum sample size is 2401.

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:

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

For this problem, we have that:

We dont know the true proportion, so we use $\pi = 0.5$ , which is when we are are going to need the largest sample size.

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 $Z = 1.96$ .

a. If a 95% confidence interval with a margin of error of no more than 0.04 is desired, give a close estimate of the minimum sample size that will guarantee that the desired margin of error is achieved. (Remember to round up any result, if necessary.)

This is n for which M = 0.04. So

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

$0.04 = 1.96\sqrt{\frac{0.5*0.5}{n}}$

$0.04\sqrt{n} = 1.96*0.5$

$\sqrt{n} = \frac{1.96*0.5}{0.04}$

$(\sqrt{n})^2 = (\frac{1.96*0.5}{0.04})^2$

$n = 600.25$

Rounding up

The minimum sample size is 601.

b. If a 95% confidence interval with a margin of error of no more than 0.02 is desired, give a close estimate of the minimum sample size necessary to achieve the desired margin of error.

Now we want n for which M = 0.02. So

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

$0.02 = 1.96\sqrt{\frac{0.5*0.5}{n}}$