What fraction is 17 out of 50 percentage?

Mathematics

1 answer:

Morgarella [4.7K]3 years ago

3 0

Your answer would be 34%,i hope this helps

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2 years ago

Suppose the number of hits a webpage receives follows a Poisson distribution. The average number of hits per minute is 2.4. What

REY [17]

Answer:

The probability that the page will get at least one hit during any given minute is 0.9093.

Step-by-step explanation:

Let X = number of hits a web page receives per minute.

The random variable X follows a Poisson distribution with parameter,

λ = 2.4.

The probability function of a Poisson distribution is:

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

Compute the probability that the page will get at least one hit during any given minute as follows:

P (X ≥ 1) = 1 - P (X < 1)

= 1 - P (X = 0)

$=1-\frac{e^{-2.4}(2.4)^{0}}{0!}\\=1-\frac{0.09072\times1}{1} \\=1-0.09072\\=0.90928\\\approx0.9093$

Thus, the probability that the page will get at least one hit during any given minute is 0.9093.

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2 years ago

Determine whether each integral is convergent or divergent. If it is convergent evaluate it. (a) integral from 1^(infinity) e^(-

AVprozaik [17]

Answer:

a) So, this integral is convergent.

b) So, this integral is divergent.

c) So, this integral is divergent.

Step-by-step explanation:

We calculate the next integrals:

$\int_1^{\infty} e^{-2x} dx=\left[-\frac{e^{-2x}}{2}\right]_1^{\infty}\\\\\int_1^{\infty} e^{-2x} dx=-\frac{e^{-\infty}}{2}+\frac{e^{-2}}{2}\\\\\int_1^{\infty} e^{-2x} dx=\frac{e^{-2}}{2}\\$

So, this integral is convergent.

$\int_1^{2}\frac{dz}{(z-1)^2}=\left[-\frac{1}{z-1}\right]_1^2\\\\\int_1^{2}\frac{dz}{(z-1)^2}=-\frac{1}{1-1}+\frac{1}{2-1}\\\\\int_1^{2}\frac{dz}{(z-1)^2}=-\infty\\$

So, this integral is divergent.

$\int_1^{\infty} \frac{dx}{\sqrt{x}}=\left[2\sqrt{x}\right]_1^{\infty}\\\\\int_1^{\infty} \frac{dx}{\sqrt{x}}=2\sqrt{\infty}-2\sqrt{1}\\\\\int_1^{\infty} \frac{dx}{\sqrt{x}}=\infty\\$