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sleet_krkn [62]

3 years ago

15

Sally’s father baked 8 cookies. Sally and 4 friends (5 people total) decide to share the cookies equally. How much will each per

son get?

2 answers:

Annette [7]3 years ago

7 0

The answer is 1.6

#HopeThatHelps

MariettaO [177]3 years ago

6 0

Each person will get 1.6 parts of the cookies or 1 3/5 cookie.
You do 8/5=1.6

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AfilCa [17]

2 6/7 ÷ 2/3

= 20/7 ÷ 2/3

20/7 × 3/2

= 60/14

reduce to it's lowest term

30/7 = 4 2/7

8 0

2 years ago

The leader board at the Lakeside Golf Tournament shows that Mei's score is 3

weeeeeb [17]

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3 years ago

jillian calculates that see will take 95 minutes to run 7 miles. she runs the distance in 80 minutes. what is jillians percent e

Aneli [31]

Answer:

15.79%

Step-by-step explanation:

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3 years ago

What is the reason for each step in the solution of the inequality?

Stolb23 [73]

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2 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

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

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

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

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

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

$n = 2401$

The minimum sample size is 2401.

4 0

2 years ago