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Mathematics

1 answer:

siniylev [52]3 years ago

5 0

a)3-2=1 b)4-2=2 c)8-2=6 d)7-4=3

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Assume that during each second, a job arrives at a webserver with probability 0.03. Use the Poisson distribution to estimate the

Savatey [412]

Answer:

The probability that at lest one job will be missed in 57 second is $=1- e^{-1.71}$ =0.819134

Step-by-step explanation:

Poisson distribution:

A discrete random variable X having the enumerable set {0,1,2,....} as the spectrum, is said to be Poisson distribution.

$P(X=x)=\frac{e^{-\lambda t}({-\lambda t})^x}{x!}$ for x=0,1,2...

λ is the average per unit time

Given that, a job arrives at a web server with the probability 0.03.

Here λ=0.03, t=57 second.

The probability that at lest one job will be missed in 57 second is

=P(X≥1)

=1- P(X<1)

=1- P(X=0)

$=1-\frac{e^{-1.71}(1.71)^0}{0!}$

$=1- e^{-1.71}$

=0.819134

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Explain truth tables and how they work

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I think the answer is D

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In the figure below, FH= 14 and GH=6. Find FG.

Katyanochek1 [597]

Answer:

Let's see what to do buddy....