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deff fn [24]
4 years ago
11

A boat’s sail is a right triangle. The length of one side of the sail is 7 feet more than the other side. The hypotenuse is 13 f

eet. Find the lengths of the two sides of the sail.
Mathematics
1 answer:
Svetach [21]4 years ago
8 0

Answer: the other two sides are 5 feet and 12 feet

Step-by-step explanation:

Length of one side be x , that means the other side is x + 7

by Pythagoras theorem:

x^{2} + (x+7)^{2} = 13^{2}

expanding , we have:

x^{2} + x^{2}+ 14x + 49 = 169

2x^{2} + 14x + 49 - 169 = 0

2x^{2}  + 14x - 120 = 0

divide through by 2 , we have

x^{2}+ 7x - 60 = 0

By factorizing , we have

(x - 5)(x+12) = 0

Therefore:

x = 5 or x = -12

Since , the side can not be negative , so

x = 5

That means the other side will be 5+ 7 = 12

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Let X and Y be discrete random variables. Let E[X] and var[X] be the expected value and variance, respectively, of a random vari
Ulleksa [173]

Answer:

(a)E[X+Y]=E[X]+E[Y]

(b)Var(X+Y)=Var(X)+Var(Y)

Step-by-step explanation:

Let X and Y be discrete random variables and E(X) and Var(X) are the Expected Values and Variance of X respectively.

(a)We want to show that E[X + Y ] = E[X] + E[Y ].

When we have two random variables instead of one, we consider their joint distribution function.

For a function f(X,Y) of discrete variables X and Y, we can define

E[f(X,Y)]=\sum_{x,y}f(x,y)\cdot P(X=x, Y=y).

Since f(X,Y)=X+Y

E[X+Y]=\sum_{x,y}(x+y)P(X=x,Y=y)\\=\sum_{x,y}xP(X=x,Y=y)+\sum_{x,y}yP(X=x,Y=y).

Let us look at the first of these sums.

\sum_{x,y}xP(X=x,Y=y)\\=\sum_{x}x\sum_{y}P(X=x,Y=y)\\\text{Taking Marginal distribution of x}\\=\sum_{x}xP(X=x)=E[X].

Similarly,

\sum_{x,y}yP(X=x,Y=y)\\=\sum_{y}y\sum_{x}P(X=x,Y=y)\\\text{Taking Marginal distribution of y}\\=\sum_{y}yP(Y=y)=E[Y].

Combining these two gives the formula:

\sum_{x,y}xP(X=x,Y=y)+\sum_{x,y}yP(X=x,Y=y) =E(X)+E(Y)

Therefore:

E[X+Y]=E[X]+E[Y] \text{  as required.}

(b)We  want to show that if X and Y are independent random variables, then:

Var(X+Y)=Var(X)+Var(Y)

By definition of Variance, we have that:

Var(X+Y)=E(X+Y-E[X+Y]^2)

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Since X and Y are independent, Cov(X,Y)=0

=Var(X)+Var(Y)

Therefore as required:

Var(X+Y)=Var(X)+Var(Y)

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