Suppose the radius of this polygon is 10. What is the perimeter

Can you help me with this???

kvasek [131]

It would look like a diamond with circles around it

3 0

3 years ago

What does x equal in the equation x^1/4=2

Minchanka [31]

Lisa gaff 4 and 2 other

3 0

3 years ago

What is the answer to this?

Vitek1552 [10]

Answer:

each answer is written below

Step-by-step explanation

7 0

3 years ago

One more help please. Thanks!

cricket20 [7]

After factoring this gives

y ( y + 2) y^2 + 2
------------ = --------- y not equal 3 , 0 or -1
y + 1 y + 1

Its B

5 0

4 years ago

Read 2 more answers

The Wall Street Journal reported that the age at first startup for 90% of entrepreneurs was 29 years of age or less and the age

Hitman42 [59]

Answer:

(a) $\hat p\sim N(0.90,\ 0.0212^{2}})$

(b) $\hat q\sim N(0.10,\ 0.0212^{2}})$

(c) Not different.

Step-by-step explanation:

The information provided is:

The age at first startup for 90% of entrepreneurs was 29 years of age or less.
The age at first startup for 10% of entrepreneurs was 30 years of age or more.

According to the Central limit theorem, if from an unknown population large samples of sizes n > 30, are selected and the sample proportion for each sample is computed then the sampling distribution of sample proportion follows a Normal distribution.

The mean of this sampling distribution of sample proportion is:

$\mu_{\hat p}=p$

The standard deviation of this sampling distribution of sample proportion is:

$\sigma_{\hat p}=\sqrt{\frac{p(1-p)}{n}}$

(a)

Let p represent the proportion of entrepreneurs whose first startup was at 29 years of age or less.

A sample of n = 200 entrepreneurs is selected.

As n = 200 > 30, according to the Central limit theorem the sampling distribution of sample proportion can be approximated by the normal distribution.

Compute the mean and standard deviation as follows:

$\mu_{\hat p}=p=0.90\\\\\sigma_{\hat p}=\sqrt{\frac{p(1-p)}{n}}=\sqrt{\frac{0.90(1-0.90)}{200}}=0.0212$

So, $\hat p\sim N(0.90,\ 0.0212^{2}})$ .

(b)

Let q represent the proportion of entrepreneurs whose first startup was at 30 years of age or more.

A sample of n = 200 entrepreneurs is selected.

As n = 200 > 30, according to the Central limit theorem the sampling distribution of sample proportion can be approximated by the normal distribution.

Compute the mean and standard deviation as follows:

$\mu_{\hat q}=q=0.10\\\\\sigma_{\hat q}=\sqrt{\frac{q(1-q)}{n}}=\sqrt{\frac{0.10(1-0.10)}{200}}=0.0212$

So, $\hat q\sim N(0.10,\ 0.0212^{2}})$ .

(c)

The standard deviation of sample proportions is also known as the standard error.

The standard deviation of p is, 0.0212.

The standard deviation of q is, 0.0212.

Thus, the standard errors of the sampling distributions in parts (a) and (b) are same.

5 0

3 years ago