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Charra [1.4K]
3 years ago
15

256899 in extended form

Mathematics
1 answer:
Daniel [21]3 years ago
6 0

200000 + 50000 + 6000 + 800 + 90 + 9

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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
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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!}

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What fraction is equivalent to 30/360?
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Given that

30/360 to lowest fraction equivalent

Now, we can write 30 as 3×10

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knowing the values of h, k, and r we can substitute them into the formula

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