What is the distance between Point K and Point L?

In the figure above, circles A and B are tangent to each other and to the large circle. If the radius of circle A

Ne4ueva [31]

Answer:

The answer to your question is A = 200.96 u²

Step-by-step explanation:

Data

Radius A = 6

Radius B = 2

π = 3.14

Area = ?

Formula

Area = πr²

Process

1.- Calculate the radius of the large circle

Radius = radius of circle A + radius of circle B

Radius = 6 + 2

Radius = 8

2.- Substitute in the formula

Area = (3.14)(8)²

Area = 3.14(64)

Area = 200.96 u²

6 0

4 years ago

Tariku was in second grade she read 6 book during each month of the year. She was in third grade she read 8 books during each mo

maks197457 [2]

Answer:

See below

Step-by-step explanation:

Tariku in second grade read 6 books each month during the year. This is equal to 6(12) = 72.

Tariku in third grade read 8 books each month during the year. This is equal to 8(12) = 96.

You can find the total amount with any of the below equations:

6(12) + 8(12)
12(6+8)
12(14)

For a total number of months x the equations would be the same except with x instead of 12.

6x+8x
x(6+8)
14x

4 0

4 years ago

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

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