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3 years ago

Determine the product:

1 answer:

4 0

$800.5\cdot(2\cdot10^6)=(8.005\cdot10^2)\cdot(2\cdot10^6)\\\\(8.005\cdot2)(10^2\cdot10^6)=16.01\cdot10^{2+6}=1.601\cdot10\cdot10^8\\\\=\boxed{1.601\cdot10^9}\to\boxed{D.}$

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4 years ago

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3 years ago

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2 years ago

Read 2 more answers

Changes in airport procedures require considerable planning. Arrival rates of aircrft are important factors that muct be taken i

Irina18 [472]

Answer:

a) 0.1558

b) 0.7983

c) 0.1478

Step-by-step explanation:

If we suppose that small aircraft arrive at the airport according to a Poisson process at the rate of 5.5 per hour and if X is the random variable that measures the number of arrivals in one hour, then the probability of k arrivals in one hour is given by:

$\bf P(X=k)=\displaystyle\frac{(5.5)^ke^{-5.5}}{k!}$

(a) What is the probability that exactly 4 small aircraft arrive during a 1-hour period?

$\bf P(X=4)=\displaystyle\frac{(5.5)^4e^{-5.5}}{4!}=0.1558$

(b) What is the probability that at least 4 arrive during a 1-hour period?

$\bf P(X\geq4)=1-P(X$

(c) If we define a working day as 12 hours, what is the probability that at least 75 small aircraft arrive during a working day?

If we redefine the time interval as 12 hours instead of one hour, then the rate changes from 5.5 per hour to 12*5.5 = 66 per working day, and the pdf is now

$\bf P(X=k)=\displaystyle\frac{(66)^ke^{-66}}{k!}$

and we want P(X ≥ 75) = 1-P(X<75). But

$\bf P(X$

hence

P(X ≥ 75) = 1-0.852 = 0.1478

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4 years ago