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3 years ago

8

Could u guys help me plz?

1 answer:

Harman [31]3 years ago

7 0

9514 1404 393

Answer:

84.7 units²
118.125 in³

Step-by-step explanation:

The surface area of the triangular prism is the sum of its base areas and the lateral area. The total area of the two triangular bases is ...

A = bh = 7(6.1) = 42.7 . . . square units

The lateral area is the product of the perimeter of the base and the length of the prism. The height given for the prism is consistent with it having triangular bases that are equilateral triangles. That is, the unmarked dimension is likely 7 units, so the perimeter is ...

P = 7+7+7 = 21

and the lateral surface area is ...

LA = PL = 21·2 = 42

Then the total surface area of the prism is ...

total area = A + LA = 42.7 +42 = 84.7 . . . square units

__

The volume of a rectangular prism is given by ...

V = Bh

where B is the area of the base, and h is the height. Your prism has a volume of ...

V = (26.25 in²)(4.5 in) = 118.125 in³

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Answer:

In the long run, ou expect to lose $4 per game

Step-by-step explanation:

Suppose we play the following game based on tosses of a fair coin. You pay me $10, and I agree to pay you $n^2 if heads comes up first on the nth toss.

Assuming X be the toss on which the first head appears.

then the geometric distribution of X is:

X $\sim$ geom(p = 1/2)

the probability function P can be computed as:

$P (X = n) = p(1-p)^{n-1}$

where

n = 1,2,3 ...

If I agree to pay you $n^2 if heads comes up first on the nth toss.

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$\sum \limits ^{n}_{i=1} n^2 P(X=n) = E(X^2)$

$\sum \limits ^{n}_{i=1} n^2 P(X=n) =Var (X) + [E(X)]^2$

$\sum \limits ^{n}_{i=1} n^2 P(X=n) = \dfrac{1-p}{p^2}+(\dfrac{1}{p})^2$ ∵ X $\sim$ geom(p = 1/2)

$\sum \limits ^{n}_{i=1} n^2 P(X=n) = \dfrac{1-p}{p^2}+\dfrac{1}{p^2}$

$\sum \limits ^{n}_{i=1} n^2 P(X=n) = \dfrac{1-p+1}{p^2}$

$\sum \limits ^{n}_{i=1} n^2 P(X=n) = \dfrac{2-p}{p^2}$

$\sum \limits ^{n}_{i=1} n^2 P(X=n) = \dfrac{2-\dfrac{1}{2}}{(\dfrac{1}{2})^2}$

$\sum \limits ^{n}_{i=1} n^2 P(X=n) =\dfrac{ \dfrac{4-1}{2} }{{\dfrac{1}{4}}}$

$\sum \limits ^{n}_{i=1} n^2 P(X=n) =\dfrac{ \dfrac{3}{2} }{{\dfrac{1}{4}}}$

$\sum \limits ^{n}_{i=1} n^2 P(X=n) =\dfrac{ 1.5}{{0.25}}$

$\sum \limits ^{n}_{i=1} n^2 P(X=n) =6$

Given that during the game play, You pay me $10 , the calculated expected loss = $10 - $6

= $4

∴

In the long run, you expect to lose $4 per game

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