Which of the following illustrates the distributive property?

A machine that produces a major part for an airplane engine is monitored closely. In the past, 10% of the parts produced would b

nata0808 [166]

Answer:

With a .95 probability, the sample size that needs to be taken if the desired margin of error is .04 or less is of at least 216.

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:

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

For this problem, we have that:

$p = 0.1$

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 $Z = 1.96$ .

With a .95 probability, the sample size that needs to be taken if the desired margin of error is .04 or less is

We need a sample size of at least n, in which n is found M = 0.04.

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

$0.04 = 1.96\sqrt{\frac{0.1*0.9}{n}}$

$0.04\sqrt{n} = 0.588$

$\sqrt{n} = \frac{0.588}{0.04}$

$\sqrt{n} = 14.7$

$(\sqrt{n})^{2} = (14.7)^{2}$

$n = 216$

With a .95 probability, the sample size that needs to be taken if the desired margin of error is .04 or less is of at least 216.

7 0

4 years ago

800 grams equals how many pounds?

sattari [20]

1 gram = 0.00220462 pound

Convert:

0.00220462 × 800 = 1.7637

800 grams = 1.7637 pounds

3 0

3 years ago

PLZ HELP FOR BRAINLIEST!!!

natali 33 [55]

Inputs are the first number output is the second. So youre outputs are 2, -1, -1, and 3. so i would say b

5 0

3 years ago

Read 2 more answers

4. Use the process outlined in the lesson to approximate the number 2√3. Use the approximation √3 ≈ 1.732 050 8.

jeka57 [31]

Answer:

sequence of five intervals

(1) 2 < $2^{\sqrt{3} }$ < $2^{2}$

(2) $2^{1.7}$ < $2^{\sqrt{3} }$ < $2^{1.8}$

(3) $2^{1.73}$ < $2^{\sqrt{3} }$ < $2^{1.74}$

(4) $2^{1.732}$ < $2^{\sqrt{3} }$ < $2^{1.733}$

(5) $2^{1.7320}$ < $2^{\sqrt{3} }$ < $2^{1.7321}$

Step-by-step explanation:

as per question given data

√3 ≈ 1.732 050 8

to find out

sequence of five intervals

solution

as we have given that √3 value that is here

√3 ≈ 1.732 050 8 ........................1

so

when we find $2^{\sqrt{3} }$ ................2

put here √3 value in equation number 2

we get $2^{\sqrt{3} }$ that is 3.322

so

sequence of five intervals

(1) 2 < $2^{\sqrt{3} }$ < $2^{2}$

(2) $2^{1.7}$ < $2^{\sqrt{3} }$ < $2^{1.8}$

(3) $2^{1.73}$ < $2^{\sqrt{3} }$ < $2^{1.74}$

(4) $2^{1.732}$ < $2^{\sqrt{3} }$ < $2^{1.733}$

(5) $2^{1.7320}$ < $2^{\sqrt{3} }$ < $2^{1.7321}$

8 0

3 years ago

Please help !! Picture provided !!!

adelina 88 [10]

Method One
Your calculator might be able to do this. Mine does it like this.
6
nCr
2
=
15

Method 2
You could simply set up 6C2

This gives you
$\frac{n!}{(n- r)! r!} = \frac{6!}{(6 - 2)!2!} = \frac{6!}{4! 2!} = \frac{6*5}{2*1} = 15$

Method Three
You only have to do this a couple of times to see how the cancellation works.

$\frac{6!}{4!2!} = \frac{6*5*\st{4*3*2*1}}{4*3*2*1 * 2*1}$
After all the cancellation takes place you have
6*5/2 = 15

7 0

3 years ago