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2 years ago

1. How many degrees is x .x 41 82 20.5 o 41 О 10

Mathematics

1 answer:

Alja [10]2 years ago

5 0

Answer:

Option (1)

Step-by-step explanation:

By the inscribed angle theorem inside a circle,

"Measure of an inscribed angle is half the measure of the intercepted arc"

$\frac{1}{2}$ [m(arc AB)] = m(∠ABC)

m(arc AB) = 2[m(∠ABC)]

x = 2(41°)

x = 82°

Option (1) is the correct option.

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Cara has 25$ to buy dry pet food and treats for the animal shelter. A pound of dog food cost $2 and treats are $1 apiece. If she

alina1380 [7]

Let x be the number of treats bought
Let y be the pounds of food bought

25= 2y+x

Given that she has buys 9 pounds of dog food:

25= 2(9)+x
25=18 +x
25-18 =x
7=x

With the money she has left over, she can buy 7 treats.

Hope I helped :)

5 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

Ashanti wants to collect more than 50 food items for the local homeless shelter. If he has already collected 13, how many more i

Marina CMI [18]

The answer is 37 you have to subtract

6 0

3 years ago

Read 2 more answers

What are the solutions to this equation?

ziro4ka [17]

$\displaystyle\\(x-4)^2=49\\\\(x-4)^2-49=0\\\\(x-4)^2-7^2=0\\\\(x-4-7)(x-4+7)=0\\\\(x-11)(x+3)=0\\\\x-11 = 0~~~\text{or}~~~x+3=0\\\\x-11=0~~~\implies~~~\boxed{x_1=11}\\\\x+3=0~~~\implies~~~\boxed{x_2=-3}\\\\\boxed{\bf The~solutions~are:~~x =11~~\text{or}~~x =-3}$