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3 years ago

Someone help please i dont know how to do this

Mathematics

1 answer:

Olenka [21]3 years ago

3 0

Area of the parallelogram = 14 * PQ

Area of the triangle

= 1/2 * PQ*MQ = 1/6 * 14 * PQ

Divide through by PQ

1/2 MQ = 1/6 * 14

MQ = 1/6 * 14 * 2 = 4 2/3 cm^3 ( or 4.67 cm^3 to nearest hundredth.)

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What issue did the great compromise resolve

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4 years ago

John bought three times as many Cokes as Dr. Peppers for his party. If he bought a total of 144 Cokes and

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3 years ago

Read 2 more answers

If a change dispenser contains 32 coins consisting of

swat32

<h3>Answer: 19 dimes</h3>

=================================================

Work Shown:

d = number of dimes

q = 32-d = number of quarters

$5.15 = 515 cents

10d+25q = 515

10d+25(32-d) = 515

10d+800-25d = 515

-15d+800 = 515

-15d = 515-800

-15d = -285

d = -285/(-15)

d = 19

There are 19 dimes. You can stop here if you want.

q = 32-d = 32-19 = 13

There are 13 quarters

-------------------------------

Check:

1 dime = 10 cents

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1 quarter = 25 cents

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total value = 190 cents + 325 cents = 515 cents = $5.15

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6 0

3 years ago

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