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pashok25 [27]
2 years ago
6

(Y^3 - 125) divided (y -5)

Mathematics
1 answer:
dimaraw [331]2 years ago
5 0

Answer:

(y^2+5y+25)

Step-by-step explanation:

(y^3 - 125) ÷ (y -5)

Factor the first term.  This is the difference of cubes

a^3 – b^3 = (a – b)(a^2 + ab + b^2)

Let a=y    and b = 5  (5^3 =125)

(y^3-125) = (y-5) (y^2+5y+25)

Substitute this back into the original equation

(y-5) (y^2+5y+25)

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

(y-5)

The y-5 term cancels

(y^2+5y+25)

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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:

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

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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

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2 years ago
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