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Orlov [11]
3 years ago
8

6-45. THE TORTOISE AND THE HARE The tortoise and the hare are having a race. In this tale, the tortoise moves at 50 feet per hou

r while the hare moves at 250 feet per hour. The tortoise takes 8 hours longer than the hare to finish. What was the distance of the race? Let t represent the unknown time in hours that it took the hare to complete the race. Write an expression (not an equation!) representing the time it took the tortoise to finish the race. Write two equations that represent the race, one for the hare and one for the tortoise. What is the distance of the race?
Mathematics
1 answer:
REY [17]3 years ago
6 0
It 7 hours and 5 minutes to complete the race because it would be 8 times 50 which equals 400 and 400 plus 250 which equals 650 and 650 divided by 60 is 7.5 which is 7 hours and 5 minutes I believe so 7.5 is the answer.............
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(a)E[X+Y]=E[X]+E[Y]

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Step-by-step explanation:

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

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

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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:

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

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

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Therefore as required:

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

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