All categories

3 years ago

HELP VERY MUCH NEEDED PLEASE AND THANKYOUUU!!Drag the tiles to the correct locations on the image. Not all tiles will be used.

Mathematics

1 answer:

WITCHER [35]3 years ago

8 0

Answer:

The perimeter of the square is 16 √ 5

The area of the square is 8√5

Step-by-step explanation:

To solve this you first need to solve 4√5 which is 8.94. This made the perimeter 35.76 which is equaled to 16√5. The area is 17.88 which is the same as 8√5.

You might be interested in

The data below are yields for two different types of corn seed that were used on adjacent plots of land. Assume that the data ar

laiz [17]

Answer:

Step-by-step explanation:

5 0

3 years ago

For what values of x is the expression below defined? Look at the picture(15 points)

Leya [2.2K]

Answer:

D. -5 <= x < 1

Step-by-step explanation:

the values under the square-root radical must not be negative, AND

the value of the denominator must not be 0 or negative

x+5 >=0 or x >= -5

and 1-x > 0 or x < 1

So the answer is -5 <= x < 1

4 0

3 years ago

In a simple random sample of 14001400 young people, 9090% had earned a high school diploma. Complete parts a through d below.

ratelena [41]

Answer:

(a) The standard error is 0.0080.

(b) The margin of error is 1.6%.

(c) The 95% confidence interval for the percentage of all young people who earned a high school diploma is (88.4%, 91.6%).

(d) The percentage of young people who earn high school diplomas has increased.

Step-by-step explanation:

Let p = proportion of young people who had earned a high school diploma.

A sample of n = 1400 young people are selected.

The sample proportion of young people who had earned a high school diploma is:

$\hat p=0.90$

(a)

The standard error for the estimate of the percentage of all young people who earned a high school diploma is given by:

$SE_{\hat p}=\sqrt{\frac{\hat p(1-\hat p)}{n}}$

Compute the standard error value as follows:

$SE_{\hat p}=\sqrt{\frac{\hat p(1-\hat p)}{n}}$

$=\sqrt{\frac{0.90(1-0.90)}{1400}}\\$

$=0.008$

Thus, the standard error for the estimate of the percentage of all young people who earned a high school diploma is 0.0080.

(b)

The margin of error for (1 - α)% confidence interval for population proportion is:

$MOE=z_{\alpha/2}\times SE_{\hat p}$

Compute the critical value of z for 95% confidence level as follows:

$z_{\alpha/2}=z_{0.05/2}=z_{0.025}=1.96$

Compute the margin of error as follows:

$MOE=z_{\alpha/2}\times SE_{\hat p}$

$=1.96\times 0.0080\\=0.01568\\\approx1.6\%$

Thus, the margin of error is 1.6%.

(c)

Compute the 95% confidence interval for population proportion as follows:

$CI=\hat p\pm MOE\\=0.90\pm 0.016\\=(0.884, 0.916)\\\approx (88.4\%,\ 91.6\%)$

Thus, the 95% confidence interval for the percentage of all young people who earned a high school diploma is (88.4%, 91.6%).

(d)

To test whether the percentage of young people who earn high school diplomas has increased, the hypothesis is defined as:

H₀: The percentage of young people who earn high school diplomas has not increased, i.e. p = 0.80.

Hₐ: The percentage of young people who earn high school diplomas has not increased, i.e. p > 0.80.

Decision rule:

If the 95% confidence interval for proportions consists the null value, i.e. 0.80, then the null hypothesis will not be rejected and vice-versa.

The 95% confidence interval for the percentage of all young people who earned a high school diploma is (88.4%, 91.6%).

The confidence interval does not consist the null value of p, i.e. 0.80.

Thus, the null hypothesis is rejected.

Hence, it can be concluded that the percentage of young people who earn high school diplomas has increased.

8 0

3 years ago

3 Select the correct answer. What are the solutions to this equation? 16x² + 9 = 25

Charra [1.4K]

Answer:

Step-by-step explanation:

16x^2 + 9 = 25

16x^2 = 16

x^2 = 1

x = 1, -1

4 0

3 years ago

Who can help me with this for algebra 2

Anuta_ua [19.1K]

The answer to this equation is

x= 1, -3

7 0

3 years ago