Ask question

Ask question

All categories

2 years ago

7

Madison worked her 40 regular hours last week plus 12 overtime hours at the time and a half rate. Her gross pay was $842.00. Wha

t was her hourly rate? Round to nearest cent, if needed.

1 answer:

GrogVix [38]2 years ago

3 0

she has $13.80 every hour because 842 divided by 40 plus 12 equals 61 so then and you cant round anymore.

You might be interested in

Can someone help me with #3? Thanks!

labwork [276]

I hope this helps you

4 0

2 years ago

Based on the table, which function represents the same relationship? The x column has 0, 1, 2, 3. The y column has 0.0625, 0.25,

Sladkaya [172]

The answer is c. I hope u get it right

8 0

2 years ago

Read 2 more answers

If m\angle A=(6x+10)^{\circ}∠A=(6x+10) ∘ and m\angle B=(4x+30)^{\circ}∠B=(4x+30) ∘ , then find the measure of \angle B∠B.

netineya [11]

A=6 × + 10=88555%__/455

6 0

1 year ago

Complete the equation of the line through (-8,-2) and (-4,6).<br> use exact numbers.<br> y= ?

Travka [436]

Answer:hey

Step-by-step explanation:

3 0

2 years ago

A programmer plans to develop a new software system. In planning for the operating system that he will use, he needs to estimat

Vedmedyk [2.9K]

Using the z-distribution, we have that:

a) A sample of 601 is needed.

b) A sample of 93 is needed.

c) A. Yes, using the additional survey information from part (b) dramatically reduces the sample size.

In a sample with a number n of people surveyed with a probability of a success of $\pi$ , and a confidence level of $\alpha$ , we have the following confidence interval of proportions.

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

In which z is the z-score that has a p-value of $\frac{1+\alpha}{2}$ .

The margin of error is:

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

95% confidence level, hence $\alpha = 0.95$ , z is the value of Z that has a p-value of $\frac{1+0.95}{2} = 0.975$ , so $z = 1.96$ .

For this problem, we consider that we want it to be within 4%.

Item a:

The sample size is <u>n for which M = 0.04.</u>

There is no estimate, hence $\pi = 0.5$

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

$0.04 = 1.96\sqrt{\frac{0.5(0.5)}{n}}$

$0.04\sqrt{n} = 1.96\sqrt{0.5(0.5)}$

$\sqrt{n} = \frac{1.96\sqrt{0.5(0.5)}}{0.04}$

$(\sqrt{n})^2 = \left(\frac{1.96\sqrt{0.5(0.5)}}{0.04}\right)^2$

$n = 600.25$

Rounding up:

A sample of 601 is needed.

Item b:

The estimate is $\pi = 0.96$ , hence:

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

$0.04 = 1.96\sqrt{\frac{0.96(0.04)}{n}}$

$0.04\sqrt{n} = 1.96\sqrt{0.96(0.04)}$

$\sqrt{n} = \frac{1.96\sqrt{0.96(0.04)}}{0.04}$

$(\sqrt{n})^2 = \left(\frac{1.96\sqrt{0.96(0.04)}}{0.04}\right)^2$

$n = 92.2$

Rounding up:

A sample of 93 is needed.

Item c:

The closer the estimate is to $\pi = 0.5$ , the larger the sample size needed, hence, the correct option is A.

For more on the z-distribution, you can check brainly.com/question/25404151

8 0

1 year ago