The length of the chord = 2(sqrt(10^2 - 6^2))
= 2(sqrt (100 - 36))
= 2(sqrt (64))
= 2(8)
= 16cm

The attached pic should put things into perspective hopefully

7 0

4 years ago

Erica a pizza into 9 pieces. What is the angle measure of each piece

Alina [70]

Need more info!!!! !!!!!

6 0

3 years ago

Read 2 more answers

Barron's reported that the average number of weeks an individual is unemployed is 18.5 weeks. Assume that for the population of

Ket [755]

Answer:

A) The sampling distribution for a sample size n=50 has a mean of 18.5 weeks and a standard deviation of 0.849.

B) P = 0.7616

C) P = 0.4441

Step-by-step explanation:

We assume that for the population of all unemployed individuals the population mean length of unemployment is 18.5 weeks and that the population standard deviation is 6 weeks.

A) We take a sample of size n=50.

The mean of the sampling distribution is equal to the population mean:

$\mu_s=\mu=18.5$

The standard deviation of the sampling distribution is:

$\sigma_s=\dfrac{\sigma}{\sqrt{n}}=\dfrac{6}{\sqrt{50}}=0.849$

B) We have to calculate the probability that the sampling distribution gives a value between one week from the mean. That is between 17.5 and 19.5 weeks.

We can calculate this with the z-scores:

$z_1=\dfrac{X_1-\mu}{\sigma/\sqrt{n}}=\dfrac{17.5-18.5}{6/\sqrt{50}}=\dfrac{-1}{0.8485}=-1.179\\\\\\z_2=\dfrac{X_2-\mu}{\sigma/\sqrt{n}}=\dfrac{19.5-18.5}{6/\sqrt{50}}=\dfrac{1}{0.8485}=1.179$

The probability it then:

$P(|X_s-\mu_s|$

C) For half a week (between 18 and 19 weeks), we recalculate the z-scores and the probabilities:

$z=\dfrac{X-\mu}{\sigma/\sqrt{n}}=\dfrac{18-18.5}{6/\sqrt{50}}=\dfrac{-0.5}{0.8485}=-0.589$

$P(|X_s-\mu_s|$

5 0

3 years ago