Andrew charges $18 dollars ofr each lawn he mows. Suppose he mows 17 lawns per month. How much money will Andrew make per month?

Translate the sentence into an equation or formula.

AlexFokin [52]

Answer:

Step-by-step explanation:

yo yo yo

= bo obobo ooboob

8 0

2 years ago

Help..... simply this fraction<br><img src="https://tex.z-dn.net/?f=%20%5Cfrac%7B22%7D%7B40%7D%20" id="TexFormula1" title=" \fra

dedylja [7]

Answer:

Reduce the expression, if possible, by cancelling the common factors.

Exact Form:

11/20

Decimal Form:

0.55

22/40

Factor 2 out of 22.

2(11)/40

Cancel the common factors.

Tap for more steps...

11/20

The result can be shown in multiple forms.

Exact Form:

11/20

Decimal Form:0.55

Step-by-step explanation:

Hope it is helpful....

7 0

3 years ago

Manuel ordered pizza for a party. A vegetable pizza has 6 slices per pie. A cheese pizza has 8 slices per pie. Manuel ordered fo

Zanzabum

Answer:

c=4v

8c+6v=152

Step-by-step explanation:

i did it on edge 2020

4 0

3 years ago

Read 2 more answers

We would like to create a confidence interval.

Vlada [557]

Answer:

c.A 90% confidence level and a sample size of 300 subjects.

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 $1-\alpha$ , we have the confidence interval with a margin of error of:

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

In which

z is the zscore that has a pvalue of $1 - \frac{\alpha}{2}$ .

In this problem

The proportions are the same for all the options, so we are going to write our margins of error as functions of $\sqrt{\pi(1-\pi)}$

So

a.A 99% confidence level and a sample size of 50 subjects.

$n = 50$

99% confidence interval

So $\alpha = 0.01$ , z is the value of Z that has a pvalue of $1 - \frac{0.01}{2} = 0.995$ , so $Z = 2.575$ .

The margin of error is

$M = z\sqrt{\frac{\pi(1-\pi)}{n}} = \frac{2.575}{\sqrt{50}}\sqrt{\pi(1-\pi)} = 0.3642\sqrt{\pi(1-\pi)}$

b.A 90% confidence level and a sample size of 50 subjects.

$n = 50$

90% confidence interval

So $\alpha = 0.1$ , z is the value of Z that has a pvalue of $1 - \frac{0.1}{2} = 0.95$ , so $Z = 1.645$ .

The margin of error is

$M = z\sqrt{\frac{\pi(1-\pi)}{n}} = \frac{1.645}{\sqrt{50}}\sqrt{\pi(1-\pi)} = 0.2623\sqrt{\pi(1-\pi)}$

c.A 90% confidence level and a sample size of 300 subjects.

$n = 300$

90% confidence interval

So $\alpha = 0.1$ , z is the value of Z that has a pvalue of $1 - \frac{0.1}{2} = 0.95$ , so $Z = 1.645$ .

The margin of error is

$M = z\sqrt{\frac{\pi(1-\pi)}{n}} = \frac{1.645}{\sqrt{300}}\sqrt{\pi(1-\pi)} = 0.0950\sqrt{\pi(1-\pi)}$

This produces smallest margin of error.

d.A 99% confidence level and a sample size of 300 subjects.

$n = 300$

99% confidence interval

So $\alpha = 0.01$ , z is the value of Z that has a pvalue of $1 - \frac{0.01}{2} = 0.995$ , so $Z = 2.575$ .

The margin of error is

$M = z\sqrt{\frac{\pi(1-\pi)}{n}} = \frac{2.575}{\sqrt{300}}\sqrt{\pi(1-\pi)} = 0.1487\sqrt{\pi(1-\pi)}$

6 0

4 years ago

Solve for C = 2pie.

bonufazy [111]

I don’t think you can do that

5 0

3 years ago