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2 years ago

7

At a zoo, an Asian elephant is about 3 tons and eats about 300 pounds of food a day what percent of its body weight does the ele

phant eat each day

1 answer:

olchik [2.2K]2 years ago

8 0

3 Tons = 6000 pounds
300 is 5% of 6000 so the answer is 5% :)

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What property is illustrated by the statement? 0 + x = x<br><br>I don't know about this one. Anyone?

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The answer is Identity Property of Addition :D

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2 years ago

8. Larisa buys 3 purses. Each purse costs $126.32. How much did Larisa spend? Draw or use bills and coins to solve. A $126.32 B

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The answer is (c) 368.96

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1 year 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

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

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

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

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

$(\sqrt{n})^2 = (\frac{1.96*0.5}{0.02})^2$

$n = 2401$

The minimum sample size is 2401.

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2 years ago

An auditorium has a seating capacity of 1500. For an upcoming event community event theater play, tickets cost $4 for adults and

hichkok12 [17]

Answer:

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2 years ago

Please help with this problem

GrogVix [38]

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2 years ago