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Find the circumference and area of a circle with diameter 12 ft. Express your answer in terms of pie

xz_007 [3.2K]

Answer:

Circumference: 12π

Area: 36π

Step-by-step explanation:

The equation for the circumference of a circle is:

C = πd

Where C is the circumference and d is the diameter. Knowing this, we will be able to find the circumference by pugging in 12 for d.

C = 12π

And now we have found the circumference of the circle.

The equation for the area of a circle is:

A = πr²

Where A is the area and r is the radius. Remember that the diameter is double the radius, so we can find the radius:

12 = 2r

r = 6

Now that we know r, we can find the area of the circle:

A = π(6)²

A = 36π

Now we have found the area.

So the circumference is 12π and the area is 36π.

I hope you find my answer and explanation to be helpful. Happy studying.

4 0

4 years ago

A circle has a diameter of 57 in

goldenfox [79]

Answer:

89.49 if is the entire circle and 44.745 if is a semi circle

Step-by-step explanation:

57 / 2 = 28.5 this is the ratio

28.5 * 3.14 = 89.49

7 0

3 years ago

In a large midwestern university (the class of entering freshmen is 6000 or more students), an SRS of 100 entering freshmen in 1

Serga [27]

Answer:

The p-value of the test is 0.0228, which is less than the standard significance level of 0.05, which means that there is evidence that the proportion of freshmen who graduated in the bottom third of their high school class in 2001 has been reduced.

Step-by-step explanation:

Before solving this question, we need to understand the central limit theorem and subtraction of normal variables.

Central Limit Theorem

The Central Limit Theorem establishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean $\mu = p$ and standard deviation $s = \sqrt{\frac{p(1-p)}{n}}$

Subtraction between normal variables:

When two normal variables are subtracted, the mean is the difference of the means, while the standard deviation is the square root of the sum of the variances.

1999:

20 out of 100 in the bottom third, so:

$p_1 = \frac{20}{100} = 0.2$

$s_1 = \sqrt{\frac{0.2*0.8}{100}} = 0.04$

2001:

10 out of 100 in the bottom third, so:

$p_2 = \frac{10}{100} = 0.1$

$s_2 = \sqrt{\frac{0.1*0.9}{100}} = 0.03$

Test if proportion of freshmen who graduated in the bottom third of their high school class in 2001 has been reduced.

At the null hypothesis, we test if the proportion is still the same, that is, the subtraction of the proportions in 1999 and 2001 is 0, so:

$H_0: p_1 - p_2 = 0$

At the alternative hypothesis, we test if the proportion has been reduced, that is, the subtraction of the proportion in 1999 by the proportion in 2001 is positive. So:

$H_1: p_1 - p_2 > 0$

The test statistic is:

$z = \frac{X - \mu}{s}$

In which X is the sample mean, $\mu$ is the value tested at the null hypothesis, and s is the standard error.

0 is tested at the null hypothesis:

This means that $\mu = 0$

From the two samples:

$X = p_1 - p_2 = 0.2 - 0.1 = 0.1$

$s = \sqrt{s_1^2 + s_2^2} = \sqrt{0.04^2 + 0.03^2} = 0.05$

Value of the test statistic:

$z = \frac{X - \mu}{s}$

$z = \frac{0.1 - 0}{0.05}$

$z = 2$

P-value of the test and decision:

The p-value of the test is the probability of finding a difference of at least 0.1, which is the p-value of z = 2.

Looking at the z-table, the p-value of z = 2 is 0.9772.

1 - 0.9772 = 0.0228.

The p-value of the test is 0.0228, which is less than the standard significance level of 0.05, which means that there is evidence that the proportion of freshmen who graduated in the bottom third of their high school class in 2001 has been reduced.

5 0

3 years ago

9/6 × 4/5 1 ) 36/11 ) 6/5 ) 32/30 ) 13/15

erik [133]

Answer:

6/5

Step-by-step explanation:

9×4=36

6×5=30

36/30÷2

18/15÷3

6/5

6 0

3 years ago

Consider the points A(5, 3t+2, 2), B(1, 3t, 2), and C(1, 4t, 3). Find the angle ∠ABC given that the dot product of the vectors B

Vilka [71]

Answer:

66.42°

Step-by-step explanation:

Given:

A(5, 3t+2, 2)

B(1, 3t, 2)

C(1, 4t, 3)

BA • BC = 4

Step 1: Find t.

First we have to find vectors BA and BC. We do that by subtracting the coordinates of the initial point from the coordinates of the terminal point.

In vector BA B is the initial point and A is the terminal point.

BA = OA - OB = (5-1, 3t+2-3t, 2-2) = (4, 2, 0)

BC = OC - OB = (1-1, 4t-3t, 3-2) = (0, t, 1)

Now we can find t because we know that BA • BC = 4

BA • BC = 4

To find dot product we calculate the sum of the produts of the corresponding components.

BA • BC = (4)(0) + (2)(t) + (0)(1)

4 = (4)(0) + (2)(t) + (0)(1)

4 = 0 + 2t + 0

4 = 2t

2 = t

t = 2

Now we know that:

BA = (4, 2, 0)

BC = (0, 2, 1)

Step 2: Find the angle ∠ABC.

Dot product: a • b = |a| |b| cos(angle)

BA • BC = 4

|BA| |BC| cos(angle) = 4

To get magnitudes we square each compoment of the vector and sum them together. Then square root.

$|BA| = \sqrt{4^2 + 2^2 + 0^2} = \sqrt{20} = 2\sqrt{5}$

$|BC| = \sqrt{0^2 + 2^2 + 1^2} = \sqrt{5}$

$2\sqrt{5}\sqrt{5}\cos{(m\angle{ABC})} = 4$

$10\cos{(m\angle{ABC})} = 4$

$\cos(m\angle{ABC}) = \frac{4}{10}=\frac{2}{5}$

$m\angle{ABC} = cos^{-1}{(\frac{2}{5})}$

$m\angle{ABC} = 66.4218^{\circ}$

Rounded to two decimal places:

$m\angle{ABC} = 66.42^\circ$

3 0

2 years ago