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2 years ago

Which equation represents the graph

Mathematics

1 answer:

cricket20 [7]2 years ago

3 0

Answer:

y300

Step-by-step explanation:

line graph start at y300

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What is a horseshoe crab?

Yanka [14]

Horseshoe crabs are marine and brackish water arthropods. They live primarily in and around shallow coastal waters on soft sands.

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4 years ago

PLEASEE HELP THIS IS DUE TODAY

dangina [55]

Answer:

The first one is 360 degrees. The second one is 1080.

Step-by-step explanation:

The sum of the interior angles of a quadrilateral are always 360. If you subtract 90 from 360 to get 270. Multiply 270 by 4 to get 1080.

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3 years ago

Evgesh-ka [11]

Answer:

He bought 20 shirts

Step-by-step explanation:

shirts -- x

pants -- 3/4 of x

Therefore,

He bought --- 28(3/4 of x) pants and 40(x) shirts.

So, 28(3/4 of x) + 40(x) = 1220

=> 21x + 40x = 1220

=> 61x = 1220

=> x = 20

He bought 20 shirts and 15 pants.

Thenks and meak me brainliest :))

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3 years ago

Which is greater 7/8 or 5/10

Sloan [31]

Find a common denominator 7/8 = 35/40 and 5/10 = 20/40. Also 7/8 is almost a whole and 5/10 is half. 7/8 is greater.

7 0

3 years ago

Read 2 more answers

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

OverLord2011 [107]

Answer:

a) P(x=3)=0.089

b) P(x≥3)=0.938

c) 1.5 arrivals

Step-by-step explanation:

Let t be the time (in hours), then random variable X is the number of people arriving for treatment at an emergency room.

The variable X is modeled by a Poisson process with a rate parameter of λ=6.

The probability of exactly k arrivals in a particular hour can be written as:

$P(x=k)=\lambda^{k} \cdot e^{-\lambda}/k!\\\\P(x=k)=6^k\cdot e^{-6}/k!$

a) The probability that exactly 3 arrivals occur during a particular hour is:

$P(x=3)=6^{3} \cdot e^{-6}/3!=216*0.0025/6=0.089\\\\$

b) The probability that <em>at least</em> 3 people arrive during a particular hour is:

$P(x\geq3)=1-[P(x=0)+P(x=1)+P(x=2)]\\\\\\P(0)=6^{0} \cdot e^{-6}/0!=1*0.0025/1=0.002\\\\P(1)=6^{1} \cdot e^{-6}/1!=6*0.0025/1=0.015\\\\P(2)=6^{2} \cdot e^{-6}/2!=36*0.0025/2=0.045\\\\\\P(x\geq3)=1-[0.002+0.015+0.045]=1-0.062=0.938$

c) In this case, t=0.25, so we recalculate the parameter as:

$\lambda =r\cdot t=6\;h^{-1}\cdot 0.25 h=1.5$

The expected value for a Poisson distribution is equal to its parameter λ, so in this case we expect 1.5 arrivals in a period of 15 minutes.

$E(x)=\lambda=1.5$

3 0

3 years ago