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3 years ago

14

Need help with this problem:

1 answer:

arlik [135]3 years ago

7 0

Answer:

843 days

Step-by-step explanation:

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Find the area of a square if its side length is 3/7 units<br><br> NO LINKS PLEASE

Alex73 [517]

Answer:

A=a2

Step-by-step explanation:

4 0

2 years ago

erik drove 439.92 miles on 15.6 gallons of gas. To the nearest hundredth, how many miles could he drive on 39.7 gallons of gas?

Thepotemich [5.8K]

In the question it is already given that Eric drove for 439.92 miles on 15.6 gallons. It is required to find the distance traveled on 39.7 gallons of gas. Also we have to find the answer to the nearest hundredth.
Then,
In 15.6 gallons Eric can drive for a distance = 439.92 miles
In 39.7 gallons Eric can drive for a distance = [(439.92/15.6) * 39.7] miles
= 1119.54 miles
So the total distance traveled by Eric is 1120 miles with 39.7 gallons of gas. The answer has been calculated to the nearest hundredth.

6 0

3 years ago

If DE = 3x-5 and AC = 5x-3, find x.

deff fn [24]

Step-by-step explanation:

2(3x-5)=5x-3

6x -10 =5x -3

x=7

3 0

2 years ago

Which of the following inequalities are true?

yarga [219]

Answer:

i think its B or D sorry if I'm wrong

3 0

2 years ago

Ten engineering schools in the United States were surveyed. The sample contained 250 electrical engineers, 80 being women; 175 c

steposvetlana [31]

Answer:

There is a significant difference between the two proportions.

Step-by-step explanation:

The (1 - <em>α</em>)% confidence interval for difference between population proportions is:

$CI=(\hat p_{1}-\hat p_{2})\pm z_{\alpha/2}\times\sqrt{\frac{\hat p_{1}(1-\hat p_{1})}{n_{1}}+\frac{\hat p_{2}(1-\hat p_{2})}{n_{2}}}$

Compute the sample proportions as follows:

$\hat p_{1}=\frac{80}{250}=0.32\\\\\hat p_{2}=\frac{40}{175}=0.23$

The critical value of <em>z</em> for 90% confidence interval is:

$z_{0.10/2}=z_{0.05}=1.645$

Compute a 90% confidence interval for the difference between the proportions of women in these two fields of engineering as follows:

$CI=(\hat p_{1}-\hat p_{2})\pm z_{\alpha/2}\times\sqrt{\frac{\hat p_{1}(1-\hat p_{1})}{n_{1}}+\frac{\hat p_{2}(1-\hat p_{2})}{n_{2}}}$

$=(0.32-0.23)\pm 1.645\times\sqrt{\frac{0.32(1-0.32)}{250}+\frac{0.23(1-0.23)}{175}}\\\\=0.09\pm 0.0714\\\\=(0.0186, 0.1614)\\\\\approx (0.02, 0.16)$

There will be no difference between the two proportions if the 90% confidence interval consists of 0.

But the 90% confidence interval does not consists of 0.

Thus, there is a significant difference between the two proportions.

4 0

2 years ago