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3 years ago

Can someone help me expand this polynominal?

Mathematics

2 answers:

DIA [1.3K]3 years ago

4 0

Answer:

Step-by-step explanation:

( 5 - x) ( 4-x )

=5(4-x)-x(4-x)

=20-5x-4x+x^2

=x^2-9x+20

hope it helps...

dalvyx [7]3 years ago

4 0

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Multiply first two values in brackets:

5 x 4 = 20

Multiply the outside values in the brackets:

-x * 5 = -5x

Multiply the inside terms:

-x * 4 = -4x

Multiply the last terms:

-x * -x = x^2

Place all values together:

x^2 - 5x - 4x + 20

Simplify it:

x^2 -9x + 20

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Answer:

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

Step-by-step explanation:

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The probability mass function of a Poisson distribution is:

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(a)

Compute the probability of the event (X > 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 (X > 84) is 0.007.

(b)

Compute the probability of the event (X < 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 (X < 64) is 0.483.

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Answer:

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