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3 years ago

Draw to show how to take apart 5

Mathematics

1 answer:

ELEN [110]3 years ago

3 0

You can take part by eating show dn dkdrnn

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yarga [219]

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2 years ago

Area of a semicircle is 308cm. Find the perimeter

fgiga [73]

Answer:

72 cm

Step-by-step explanation:

Find the radius using the given information

308=1/2(πr^2)

r^2=616/π

r=14

The perimeter is πr + 2r so plug in 14

π(14) + 2(14)

43.96+28=71.96

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2 years ago

True or false? it's not possible to build a triangle with side lengths of 3 3 and 9

Anit [1.1K]

True

The two shorter lengths do not add up to more than the longest length. 3+3 is less than 9. Therefore, even if the two shorter lengths lay on top of the longer side, the two ends cannot meet to form a closed 3 sided figure

6 0

2 years ago

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Perry spent $22.50 on material to make necklaces. He used all of the material and made 15 necklaces. How much should Perry charg

Vika [28.1K]

Answer:

Guys his answer is wrong. It’s 5.50 in edg

Step-by-step explanation:

5 0

2 years ago

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

ICE Princess25 [194]

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

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

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

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

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

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

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

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

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

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

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

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

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2 years ago

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