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2 years ago

Follow the steps to find the

Mathematics

1 answer:

KIM [24]2 years ago

6 0

The area of the sector rounded to 4 decimal place is 78.6396 cm²

<h3 /><h3>How to find the area of a sector?</h3>

area of sector = ∅/ 360 × πr²

Therefore,

r = radius

∅ = 46 degrees.

Hence,

area of a sector = 46 / 360 × 3.14 × 14²

area of a sector = 46 / 360 × 3.14 × 196

area of a sector = 46 / 360 × 615.44

area of a sector = 28310.24 / 360

area of a sector = 78.6395555556

Hence,

area of a sector = 78.6396 cm²

learn more on sector here: brainly.com/question/27946869

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What is the volume (in cubic units) of a sphere with a radius of 18 units? Assume that π = 3.14 and round your answer to the nea

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Answer:

The volume of sphere is $24416.64\ cm^3$ .

Step-by-step explanation:

We have,

Radius of a sphere is 18 units.

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3 years ago

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Answer:

840 ways.

Step-by-step explanation:

For the first jar, you have 7 marbles to put in.

For the second, you now have 6 marbles to put in.

For the third, you have 5 marbles.

For the fourth, you have 4 marbles.

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3 years ago

Read 2 more answers

The arrivals of clients at a service firm in Santa Clara is a random variable from Poisson distribution with rate 2 arrivals per

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Answer:

1.76% probability that in one hour more than 5 clients arrive

Step-by-step explanation:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

In which

x is the number of sucesses

e = 2.71828 is the Euler number

$\mu$ is the mean in the given time interval.

The arrivals of clients at a service firm in Santa Clara is a random variable from Poisson distribution with rate 2 arrivals per hour.

This means that $\mu = 2$

What is the probability that in one hour more than 5 clients arrive

Either 5 or less clients arrive, or more than 5 do. The sum of the probabilities of these events is decimal 1. So

$P(X \leq 5) + P(X > 5) = 1$

We want P(X > 5). So

$P(X > 5) = 1 - P(X \leq 5)$

In which

$P(X \leq 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5)$

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

$P(X = 0) = \frac{e^{-2}*2^{0}}{(0)!} = 0.1353$

$P(X = 1) = \frac{e^{-2}*2^{1}}{(1)!} = 0.2707$

$P(X = 2) = \frac{e^{-2}*2^{2}}{(2)!} = 0.2707$

$P(X = 3) = \frac{e^{-2}*2^{3}}{(3)!} = 0.1804$

$P(X = 4) = \frac{e^{-2}*2^{4}}{(4)!} = 0.0902$

$P(X = 5) = \frac{e^{-2}*2^{5}}{(5)!} = 0.0361$

$P(X \leq 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) = 0.1353 + 0.2702 + 0.2702 + 0.1804 + 0.0902 + 0.0361 = 0.9824$

$P(X > 5) = 1 - P(X \leq 5) = 1 - 0.9824 = 0.0176$

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