A notebook has a perimeter of 36 inches. It’s area is 80 square inches. What are the dimensions of the notebook

1 answer:

7 0

Let's call he width of the notebook w and the length l

w*l=80

2(w+l)=36
w+l=18
w=18-l

(18-l)*l=80
-l^2+18*l-80=0
l=8 or 10

Therefore, the notebook is 8 by 10 inches

You might be interested in

I need to know how to solve the question

Ket [755]

I think you divide 1/2 by -2/5 and then add that to -1/4....

4 0

3 years ago

Need Help Asap Please Help

Fittoniya [83]

1/3 that would be it

4 0

3 years ago

Read 2 more answers

In the past, 21% of all homes with a stay-at-home parent, the father is the stay-at-home parent. An independent research firm ha

Afina-wow [57]

Answer:

A sample size of 79 is needed.

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 $1 - \frac{\alpha}{2}$ .

For this problem, we have that:

$\pi = 0.21$

The margin of error is:

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

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 $Z = 1.96$ .

What sample size is needed if the research firm's goal is to estimate the current proportion of homes with a stay-at-home parent in which the father is the stay-at-home parent with a margin of error of 0.09?

A sample size of n is needed.

n is found when M = 0.09. So

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

$0.09 = 1.96\sqrt{\frac{0.21*0.79}{n}}$