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3 years ago

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PLZ HELP ASAP This partial circle has a radius of 11 inches. What is the area of this figure? Use 3.14 pi. Round your answer to

the nearest hundredth. 379.94 in2 25.91 in2 284.962

1 answer:

valina [46]3 years ago

4 0

<h3><u>Complete Question:</u></h3>

The figure for the question is attached below

This partial circle has a radius of 11 inches. What is the area of this figure? Use 3.14 pi. Round your answer to the nearest hundredth. 379.94 in2 25.91 in2 284.962

<h3><u>Answer:</u></h3>

Area of circle is 284.955 square inches

<h3><u>Solution:</u></h3>

Given that,

Radius of this partial circle = 11 inches

To find: Area

Given is 3/4 of a circle

<em><u>Therefore, area of 3/4 of a circle is given as:</u></em>

$Area = \frac{3}{4} \times \pi \times r^2$

Substitute r = 11

$Area = \frac{3}{4} \times 3.14 \times 11^2\\\\Area = \frac{3}{4} \times 3.14 \times 121\\\\Area = \frac{1139.82}{4}\\\\Area = 284.955$

Thus area of circle is 284.955 square inches

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The probability that your call to a service line is answered in less than 30 seconds is 0.75. Assume that your calls are indepen

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a) 0.2581

b) 0.4148

c) 17

Step-by-step explanation:

For each call, there are only two possible outcomes. Either they are answered in less than 30 seconds. Or they are not. The probabilities for each call are independent. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

In this problem we have that:

$p = 0.75$

a. If you call 12 times, what is the probability that exactly 9 of your calls are answered within 30 seconds? Round your answer to four decimal places (e.g. 98.7654).

This is $P(X = 9)$ when $n = 12$ . So

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 9) = C_{12,9}.(0.75)^{9}.(0.25)^{3} = 0.2581$

b. If you call 20 times, what is the probability that at least 16 calls are answered in less than 30 seconds? Round your answer to four decimal places (e.g. 98.7654).

This is $P(X \geq 16)$ when $n = 20$

So

$P(X \geq 16) = P(X = 16) + P(X = 17) + P(X = 18) + P(X = 19) + P(X = 20)$

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 16) = C_{20,16}.(0.75)^{16}.(0.25)^{4} = 0.1897$

$P(X = 17) = C_{20,17}.(0.75)^{17}.(0.25)^{3} = 0.1339$

$P(X = 18) = C_{20,18}.(0.75)^{18}.(0.25)^{2} = 0.0669$

$P(X = 19) = C_{20,19}.(0.75)^{19}.(0.25)^{1} = 0.0211$

$P(X = 20) = C_{20,20}.(0.75)^{20}.(0.25)^{0} = 0.0032$

So

$P(X \geq 16) = P(X = 16) + P(X = 17) + P(X = 18) + P(X = 19) + P(X = 20) = 0.1897 + 0.1339 + 0.0669 + 0.0211 + 0.0032 = 0.4148$

c. If you call 22 times, what is the mean number of calls that are answered in less than 30 seconds? Round your answer to the nearest integer.

The expected value of the binomial distribution is:

$E(X) = np$

In this question, we have $n = 22$

So

$E(X) = 22*0.75 = 16.5$

The closest integer to 16.5 is 17.

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3 years ago