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Montano1993 [528]
3 years ago
12

If f(x)=x/2+8what if f(x) when x=10

Mathematics
2 answers:
Zinaida [17]3 years ago
5 0

Answer:

f(10)=13

Step-by-step explanation:

f(x)=x/2+8

Let x = 10

f(10)=10/2 +8

      = 5 + 8

      = 13

vovangra [49]3 years ago
4 0
F(10)=13

f(10)= 10/2+8=5+8=13
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Suppose that X has a Poisson distribution with a mean of 64. Approximate the following probabilities. Round the answers to 4 dec
o-na [289]

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(a) The probability of the event (<em>X</em> > 84) is 0.007.

(b) The probability of the event (<em>X</em> < 64) is 0.483.

Step-by-step explanation:

The random variable <em>X</em> follows a Poisson distribution with parameter <em>λ</em> = 64.

The probability mass function of a Poisson distribution is:

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

(a)

Compute the probability of the event (<em>X</em> > 84) as follows:

P (X > 84) = 1 - P (X ≤ 84)

                =1-\sum _{x=0}^{x=84}\frac{e^{-64}(64)^{x}}{x!}\\=1-[e^{-64}\sum _{x=0}^{x=84}\frac{(64)^{x}}{x!}]\\=1-[e^{-64}[\frac{(64)^{0}}{0!}+\frac{(64)^{1}}{1!}+\frac{(64)^{2}}{2!}+...+\frac{(64)^{84}}{84!}]]\\=1-0.99308\\=0.00692\\\approx0.007

Thus, the probability of the event (<em>X</em> > 84) is 0.007.

(b)

Compute the probability of the event (<em>X</em> < 64) as follows:

P (X < 64) = P (X = 0) + P (X = 1) + P (X = 2) + ... + P (X = 63)

                =\sum _{x=0}^{x=63}\frac{e^{-64}(64)^{x}}{x!}\\=e^{-64}\sum _{x=0}^{x=63}\frac{(64)^{x}}{x!}\\=e^{-64}[\frac{(64)^{0}}{0!}+\frac{(64)^{1}}{1!}+\frac{(64)^{2}}{2!}+...+\frac{(64)^{63}}{63!}]\\=0.48338\\\approx0.483

Thus, the probability of the event (<em>X</em> < 64) is 0.483.

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3 years ago
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