All categories

3 years ago

What is 1/3 of 6/7. please help very important

Mathematics

1 answer:

SashulF [63]3 years ago

8 0

2 over 7.that is a possible answer.its right

You might be interested in

Find the area of a regular hexagon with the given measurement. 48-inch perimeter A = sq. in. the answer is a square root

svp [43]

The area of a regular hexagon is

... A = (3√3)/2×s² . . . . . where s is the side length

Of course, the 6-sided figure will have a side length that is 1/6 of the perimeter.

... s = (48 in)/6 = 8 in

... A = (3√3)/2×(8 in)² = 96√3 in² . . . area of your regular hexagon

5 0

3 years ago

Identity the zero of the function y= x (x-4)(x+2)

yulyashka [42]

Answer:

- 2, 0 or 4

Step-by-step explanation:

y = x(x - 4)(x + 2)

x' = 0

x - 4 = 0

x" = 4

x + 2 = 0

x"' = - 2

I hope I helped you.

4 0

3 years ago

What is the image of the point (-1,-3)(−1,−3) after a rotation of 90 degrees counterclockwise about the origin?

ycow [4]

Answer:

(3,-1)

Step-by-step explanation:

6 0

3 years ago

Determine the value of x?

Alexxx [7]

hope it will help you..........

6 0

3 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

3 years ago