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3 years ago

What is the arc length if the radius of the circle is 15 and the angle og the sector id 90 degrees

Mathematics

1 answer:

Sergio039 [100]3 years ago

5 0

Answer:

P = 53.6 square cm

Step-by-step explanation:

We know that the measure of an arc of a sector is given by: $P=\theta r+2r$

Where $\theta$ is the angle of the arc of the sector (radian) while $r$ is the radius of the circle.

Here, $\theta =$ 90° $=\frac{\pi }{2}$ .

Substituing the given values to get:

$P=\frac{\pi}{2}(15)+2\times 15$

P = 53.6 square cm

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Answer:

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Step-by-step explanation:

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kvv77 [185]

Answer:

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Step-by-step explanation:

The area (A) of a trapezoid is calculated as

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3 years ago

Devaughn is 15 years younger than Sydney. The sum of their ages is 73. What is Sydney’s age?

Olegator [25]

Answer:

Sydney is 44 years old

Step-by-step explanation:

Step 1: Make a system of equations

d = s - 15

d + s = 73

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3 years ago

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Step-by-step explanation:

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3 years ago

Exercise 6.13 presents the results of a poll evaluating support for the health care public option in 2009, reporting that 52% of

sleet_krkn [62]

Answer:

A sample size of 6755 or higher would be appropriate.

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of $\pi$ , and a confidence level of $1-\alpha$ , we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the zscore that has a pvalue of $1 - \frac{\alpha}{2}$ .

The margin of error M is given by:

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

90% confidence level

So $\alpha = 0.1$ , z is the value of Z that has a pvalue of $1 - \frac{0.1}{2} = 0.95$ , so $Z = 1.645$ .

52% of Independents in the sample opposed the public option.

This means that $p = 0.52$

If we wanted to estimate this number to within 1% with 90% confidence, what would be an appropriate sample size?

Sample size of size n or higher when $M = 0.01$ . So

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

$0.01 = 1.645\sqrt{\frac{0.52*0.48}{n}}$

$0.01\sqrt{n} = 0.8218$

$\sqrt{n} = \frac{0.8218}{0.01}$

$\sqrt{n} = 82.18$

$\sqrt{n}^{2} = (82.18)^{2}$

$n = 6754.2$

A sample size of 6755 or higher would be appropriate.

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3 years ago