PLEASE ..... HELP THANK YOU

Mathematics

2 answers:

andrezito [222]3 years ago

8 0

5x-10=3x+40

This is because of the vertical angles theorem (vertical angles are always equal).

Now, subtract 3x from each side. Do this because you cannot have the same variable on two sides of the equation.

2x-10=40

(This is the subtraction property of equality)

Now, add 10 to each side

2x=50
(This is the addition property of equality)

Finally, divide each side by two.

x=25
(This is the division property of equality)

5(25)-10
125-10
115

m

lara [203]3 years ago

4 0

5x-10=3x+40
5x=3x+50
2x=50
x=25
5*25=125
125-10=115
angle AEB=115

Hope this helps :)

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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]

Answer:

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

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

Step-by-step explanation:

The random variable X follows a Poisson distribution with parameter λ = 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 (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)

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