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4 years ago

Find BD???????????????

Mathematics

2 answers:

Rom4ik [11]4 years ago

8 0

Hey.....
In a kite each angle =90degree
=>and BI=IC=12
=>in triangleDIB
By Pythagoras theorem,
(DB)^2=(DI)^2+(BI)^2
(BD)^2=(5)^2+(12)^2
(BD)^2=25+144
(BD)^2=169
=>BD=root of 169
=>BD=13
Hope this helps u ......

Trava [24]4 years ago

5 0

See the attached picture:

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3 years ago

Consider a Poisson distribution with μ = 6.

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e = 2.71828 is the Euler number

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a. Write the appropriate Poisson probability function.

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$P(X = x) = \frac{e^{-6}*6^{x}}{(x)!}$

b. Compute f (2).

This is P(X = 2). So

$P(X = x) = \frac{e^{-6}*6^{x}}{(x)!}$

$P(X = 2) = \frac{e^{-6}*6^{2}}{(2)!} = 0.04462$

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c. Compute f (1).

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$P(X = x) = \frac{e^{-6}*6^{x}}{(x)!}$

$P(X = 1) = \frac{e^{-6}*6^{1}}{(1)!} = 0.01487$

So f(1) = 0.01487.

d. Compute P(x≥2)

This is:

$P(X \geq 2) = 1 - P(X < 2)$

In which:

$P(X < 2) = P(X = 0) + P(X = 1) + P(X = 2)$

$P(X = x) = \frac{e^{-6}*6^{x}}{(x)!}$

$P(X = 0) = \frac{e^{-6}*6^{0}}{(0)!} = 0.00248$

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$P(X = 2) = \frac{e^{-6}*6^{2}}{(2)!} = 0.04462$

Then

$P(X < 2) = P(X = 0) + P(X = 1) + P(X = 2) = 0.00248 + 0.01487 + 0.04462 = 0.06197$

$P(X \geq 2) = 1 - P(X < 2) = 1 - 0.06197 = 0.93803$

$P(X \geq 2) = 0.93803$

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