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3 years ago

WILL MARK BRAINLIEST HELP

Mathematics

2 answers:

denis-greek [22]3 years ago

7 0

We start by solving the triangles. Where the base is 3 and the height is 8.
1/2 • 8 • 3 = 12 • 2 = 24 (because there’s 2 triangles)

And now we solve the rectangle:
6•8= 48

Now we add up the areas:
48+24= 72

72 squared units.

anastassius [24]3 years ago

6 0

Answer:

<h2>72 square units</h2>

Step-by-step explanation:

First let's find the area of the middle rectangle.

We know the width of this rectangle is 6 and the length is 8 so we will multiply these two numbers.

6 × 8 = 48

Now let's find the area of the triangles

The formula is ((length × width) ÷ 2)

The width is 3, as it is not part of the rectangle, and the length is 8

And there are two triangles so we will do this twice

((8 × 3) ÷ 2) = 12

Now add up all of the areas

48 + 12 + 12 = 72

So 72 is our final answer

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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)

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

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

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