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2 years ago

(50 points really need help on this it's the last question)

Mathematics

2 answers:

larisa86 [58]2 years ago

8 0

Answer:

Pretty sure it's C

Step-by-step explanation:

trUst mE

erik [133]2 years ago

8 0

A,C,E

Step-by-step explanation:

I got it right on my test

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easy algebra question below first correct answer gets brainliest, if you put a link i will report and block

boyakko [2]

Answer:

6x²

Step-by-step explanation:

3x×2x=6...... then an x²

Therefore answer is 6x²

3 0

2 years ago

Use the arc length formula and the given information to find r. s = 12 cm, 0 = 360; r = ?

hjlf

use s=r0

but first convert 360 degree to radian
you will get 6.284 radian

then you substitute and get the answer!

12= r (6.284)
r= 1.91 cm

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2 years ago

What do you get when you multiply 3x(8x-11)

Masteriza [31]

2.18 that will be ur answer

3 0

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

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

8 0

2 years ago

Read 2 more answers

what is the largest and smallest number of the people that could have been there and what is it rounded to the nearest thousand

Luda [366]

I really don't know that sweetly try someone else

4 0

3 years ago