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3 years ago

If the shaded sector is 75m what would the area be of the circle? A. 30 B. 36 C.144 D. 108

Mathematics

1 answer:

fredd [130]3 years ago

8 0

The correct answer is letter B

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Find the value of x (x +40) x

gayaneshka [121]

Answer: x^2 + 40x

Nothing can further be done with the topic.....

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3 years ago

How many times must we toss a coin to ensure that a 0.95-confidence interval for the probability of heads on a single toss has l

musickatia [10]

Answer:

(1) 97

(2) 385

(3) 9604

Step-by-step explanation:

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

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

The margin of error in this interval is:

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

The formula to compute the sample size is:

$\\n=\frac{z_{\alpha/2}^{2}\times \hat p(1-\hat p)}{MOE^{2}}$

(1)

Given:

$\hat p = 0.50\\MOE=0.1\\z_{\alpha/2}=z_{0.05/2}=z_{0.025}=1.96$

*Use the z-table for the critical value.

Compute the value of n as follows:

$\\n=\frac{z_{\alpha/2}^{2}\times \hat p(1-\hat p)}{MOE^{2}}\\=\frac{1.96^{2}\times0.50\times(1-0.50)}{0.1^{2}}\\=96.04\\\approx97$

Thus, the minimum sample size required is 97.

(2)

Given:

$\hat p = 0.50\\MOE=0.05\\z_{\alpha/2}=z_{0.05/2}=z_{0.025}=1.96$

*Use the z-table for the critical value.

Compute the value of n as follows:

$\\n=\frac{z_{\alpha/2}^{2}\times \hat p(1-\hat p)}{MOE^{2}}\\=\frac{1.96^{2}\times0.50\times(1-0.50)}{0.05^{2}}\\=384.16\\\approx385$

Thus, the minimum sample size required is 385.

(3)

Given:

$\hat p = 0.50\\MOE=0.01\\z_{\alpha/2}=z_{0.05/2}=z_{0.025}=1.96$

*Use the z-table for the critical value.

Compute the value of n as follows:

$\\n=\frac{z_{\alpha/2}^{2}\times \hat p(1-\hat p)}{MOE^{2}}\\=\frac{1.96^{2}\times0.50\times(1-0.50)}{0.01^{2}}\\=9604$

Thus, the minimum sample size required is 9604.

8 0

3 years ago

Sara has 3 1/2 feet of ribbon.

Marizza181 [45]

Answer:

5 bows

Step-by-step explanation:

Total feets of ribbon = 3 1/2

Fraction required to make a bow = 2/3

Number of bows that can be made = (total ribbon / Fraction required to make a bow)

3 1/2 ÷ 2/3

7/2 ÷ 2/3

7/2 * 3/2

= 21 / 4

= 5.25

6 0

3 years ago

Hello OwO

Tresset [83]

The prime factorization is 2×5

3 0

3 years ago

Find the circumference of this circle.

finlep [7]

Answer:

81.64

Step-by-step explanation:

2(3.14)=6.28

6.28*13=81.64

:))))

3 0

3 years ago