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3 years ago

5

BOTS ARE LEGIT EVERYWHERE LIKE I CANT EVEN FIND A SPOT FOR PPL TO POST THEIR REAL ANSWER PLS ANYONE, BRAINLY NEEDS TO BE BETTER

AT THIS

1 answer:

Dvinal [7]3 years ago

5 0

Answer:

y=-5x-3

Step-by-step explanation:

Slope: -5/1 =-5

Y2-Y1/X2-X1=2-7/1-2=-5/1

Y-intercept: -3

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What do the diagram look like

qaws [65]

Answer:

sorry bro but i cant see the digram it did not upload

Step-by-step explanation:

3 0

2 years ago

The number of people arriving for treatment at an emergency room can be modeled by a Poisson Distribution with a rate parameter

AlekseyPX

Answer:

a) 0.052 = 5.2% probability that exactly three arrivals occur during a particular hour

b) 0.971 = 97.1% probability that at least three people arrive during a particular hour

c) 5.25 people are expected to arrive during a 45-min period

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 interval.

Poisson Distribution with a rate parameter of seven per hour.

This means that $\mu = 7$

(a) What is the probability that exactly three arrivals occur during a particular hour?

This is P(X = 3).

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

$P(X = 3) = \frac{e^{-7}*7^{3}}{(3)!} = 0.052$

0.052 = 5.2% probability that exactly three arrivals occur during a particular hour.

(b) What is the probability that at least three people arrive during a particular hour? (Round your answer to three decimal places.)

Either less than three people arrive, or at least three does. The sum of the probabilities of these events is 1. So

$P(X < 3) + P(X \geq 3) = 1$

We want $P(X \geq 3)$ , which is

$P(X \geq 3) = 1 - P(X < 3)$

In which

$P(X \geq 3) = P(X = 0) + P(X = 1) + P(X = 2)$

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

$P(X = 0) = \frac{e^{-7}*7^{0}}{(0)!} = 0.001$

$P(X = 1) = \frac{e^{-7}*7^{1}}{(1)!} = 0.006$

$P(X = 2) = \frac{e^{-7}*7^{2}}{(2)!} = 0.022$

So

$P(X \geq 3) = P(X = 0) + P(X = 1) + P(X = 2) = 0.001 + 0.006 + 0.022 + 0.029$

$P(X \geq 3) = 1 - P(X < 3) = 1 - 0.029 = 0.971$

0.971 = 97.1% probability that at least three people arrive during a particular hour

(c) How many people do you expect to arrive during a 45-min period?

During an hour(60 minutes), 7 people are expected to arrive. So, using proportions:

$\frac{45*7}{60} = \frac{3*7}{4} = \frac{21}{4} = 5.25$

5.25 people are expected to arrive during a 45-min period

5 0

3 years ago

Index and laws<br><br> evaluate <br> A) 55^0<br> B)100^1

lisabon 2012 [21]

Answer:

Step-by-step explanation:

Hey there!

so answer for (A is 1)

Answer for (B is 100)

Thank you have fun!

5 0

3 years ago

Read 2 more answers

Is 2,925 more than 2,9250

k0ka [10]

Answer: 32,175

Step-by-step explanation:

2,925 + 29,250 = 32,175

hope this helps

plz mark brainleist

3 0

3 years ago

I need help fast i will mark brainliest

almond37 [142]

Answer:

50.24

Step-by-step explanation:

Area=R^2*PI

OR
Area-R*R*PI

So, Area=4*4*3.41

=50.24

7 0

2 years ago

Read 2 more answers