The volume of this cylinder is 13,611.9 cubic inches. What is the radius?

Please look at attached images to answer question!!

Zanzabum

Wednesday and bottom one on the left

8 0

3 years ago

Suppose you start an annuity where you invest $2,000 at the beginning of each year and 4% interest is paid at the end of the yea

olga55 [171]

Answer:

$10,400

Step-by-step explanation:

If 2,000 dollars are being invested at the beginning of each year and gets a 4% interest at the end that means 2,000*104%=$2,080

It also continues on for 5 years so 2,080*5=$10,400

5 0

3 years ago

How do you solve for total intrest when you finance $600 for 24 months at 18%

yawa3891 [41]

Multiple 18% by 600 and then multiply the number you get from that by 24 and add that number onto 600 and that’s your answer

4 0

4 years ago

Read 2 more answers

TV advertising agencies face increasing challenges in reaching audience members because viewing TV programs via digital streamin

IRINA_888 [86]

Answer:

a)

[0.5235, 0.5765]

To interpret this result, we could say there is a 99% of probability that the proportion of American adults who have watched digitally streamed TV programming on some type of device is between 52.35% and 57.65%

b) 1,843 American adults

Step-by-step explanation:

The 99% confidence interval is given by

$\bf p\pm z^*\sqrt{p(1-p)/n}$

where

p = the proportion of American adults surveyed who said they have watched digitally streamed TV programming on some type of device = 55% = 0.55

$\bf z^*$ the z-score for a 99% confidence level associated with the Normal distribution N(0,1). We can do this given that the sample size (2,341) is big enough

n = sample size = 2,341

We can find the $\bf z^*$ value either with a table or with a spreadsheet.

In Excel use NORM.INV(0.995,0,1)

In OpenOffice Calc use NORMINV(0.995;0;1)

We get a value of $\bf z*$ = 2.576

and our 99% confidence interval is

$\bf 0.55\pm 2.576\sqrt{0.55*0.45/2341}=0.55\pm 2.576*0.0103=0.55\pm 0.265 = [0.5235, 0.5765]$

To interpret this result, we could say there is a 99% of probability that the proportion of American adults who have watched digitally streamed TV programming on some type of device is between 52.35% and 57.65%

We are 99% confident that this interval contains the true population proportion.

(b) What sample size would be required for the width of a 99% CI to be at most 0.03 irrespective of the value of p?? (Round your answer up to the nearest integer.)

The sample size n in a simple random sampling is given by

$\bf n=\frac{(z^*)^2p(1-p)}{e^2}$

where

e is the error proportion = 0.03

hence

$\bf n=\frac{(2.576)^2p(1-p)}{(0.03)^2}=7373.0844p(1-p)=7373.0844p-7373.044p^2$

taking the derivative with respect to p, we get

n'(p)=7373.0844-2*7373.0844p

and

n'(p) = 0 when p=0.5

By taking the second derivative we see n''(p)<0, so p=0.5 is a maximum of n

This means that if we set p=0.5, we get the maximum sample size for the confidence level required for the proportion error 0.03

Replacing p with 0.5 in the formula for the sample size we get

$\bf n=7373.0844*0.5-7373.044(0.5)^2=1,844$

rounded up to the nearest integer.

6 0

4 years ago

Read 2 more answers

Let Z be the standard normal random variable. Use a probability calculator to answer the following questions: What is the probab

Softa [21]

Answer:

0.6826 = 68.26% probability Z will be within one standard deviation of average.

Step-by-step explanation:

Normal Probability Distribution

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

What is the probability Z will be within one standard deviation of average?

This is the p-value of Z = 1 subtracted by the p-value of Z = -1.

Z = 1 has a p-value of 0.8413.

Z = -1 has a p-value of 0.1587.

0.8413 - 0.1587 = 0.6826

0.6826 = 68.26% probability Z will be within one standard deviation of average.

8 0

3 years ago