Please help. Will mark brainlist if correct

Each page of scrapbook has 144 square inches. In arts and crafts class there will be 13 students. How many pages of scrapbook pa

iogann1982 [59]

Answer:

i would think 1872 square inches

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

Which set of transformations would prove AGFE ~ AGHI?

Georgia [21]

Answer:

Reflect over x=-2 and dilate G'H'I' by a scale factor of 2 from point G

Step-by-step explanation:

First you'd have to reflect from x=-2 because its still on that line but the opposite way.

So we can already narrow it down to the second or last answer.

It wouldn't translate because that doesn't change the size. Its a dilation because the size has changed. Its dilated from point G because as you can see, G is the same point on both.

So the last answer choice is correct.

4 0

3 years ago

3. Ana's bedroom measures 8-feet by 12-feet. She

docker41 [41]

If she wants one fourth of the dimensions of the room for the corner, that means the area will be 1/16 (one fourth squared) of the whole bedroom. The area of the whole bedroom is 8 x 12 = 96. The area of the corner would be 96 / 16 = 6 sq ft.

Or you can do (one fourth of 8) x (one fourth of 12) = 2 x 3 = 6 sq ft.

6 0

3 years ago

In a sample of 330 randomly selected PCC students it was found that 175 of them have a car. Assume this sample comes from a popu

Dvinal [7]

Answer:

a) $0.530 - 1.96\sqrt{\frac{0.530(1-0.530)}{330}}=0.47615$

$0.530 + 1.96\sqrt{\frac{0.530(1-0.530)}{330}}=0.58385$

And if we convert this interval into % we got (47.615%, 58.385%)

b) For this case since the lower value for the confidence interval is a value lower than 0.5 (50%) we don't have enough evidence to conclude at 10% of significance than most students at PCC own a car

Step-by-step explanation:

We can begin calculating the best estimator for the true proportion of students at PCC who own a car:

$\hat p = \frac{175}{330}=0.530$

Part a

The confidence level is 95% , and the significance level would be given by $\alpha=1-0.95=0.05$ and $\alpha/2 =0.025$ . And the critical values would be given by:

$z_{\alpha/2}=-1.96, z_{1-\alpha/2}=1.96$

The confidence interval for the true proportion is given by the following formula:

$\hat p \pm z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}$

Replacing the info given we got:

$0.530 - 1.96\sqrt{\frac{0.530(1-0.530)}{330}}=0.47615$

$0.530 + 1.96\sqrt{\frac{0.530(1-0.530)}{330}}=0.58385$

And if we convert this interval into % we got (47.615%, 58.385%)

Part b

For this case since the lower value for the confidence interval is a value lower than 0.5 (50%) we don't have enough evidence to conclude at 10% of significance than most students at PCC own a car

7 0

3 years ago

This is due in 15 minutes so I need it asap

ser-zykov [4K]

Answer:

B and C

Step-by-step explanation:

$56 - ( {3}^{3} + 13) = 16$