Ryan makes 65 pounds of apple jelly. He put the same amount of jelly into 6 jars how much jelly is in each jar

A certain bean plant grows 15% of its height each day today after 10 days the bean plant is 27 inches tall how tall was it yeste

frez [133]

Given:
15% of its height each day
27 inches on day 10.

How tall was it yesterday?

27 represents 115% of the height yesterday. So we divided 27 by 115% to get yesterdays height.

27 ÷ 115% = 23.48 inches.

The plant was 23.48 inches tall yesterday.

To check:
23.48 x 15% = 3.52 inches
23.48 + 3.52 = 27 inches

3 0

3 years ago

A confidence interval was used to estimate the proportion of statistics students that are female. A random sample of 72 statisti

jeyben [28]

Answer:

We need a sample of size at least 13.

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 interval: (0.438, 0.642).

The proportion estimate is the halfway point of these two bounds. So

$\pi = \frac{0.438 + 0.642}{2} = 0.54$

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 $Z = 1.96$ .

Using the information above, what size sample would be necessary if we wanted to estimate the true proportion to within ±0.08 using 95% confidence?

We need a sample of size at least n.

n is found when M = 0.08. So

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

$0.08 = 1.96\sqrt{\frac{0.54*0.46}{n}}$