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krok68 [10]
3 years ago
7

There are three highways in the county. the number of daily accidents that occur on these highways are poisson random variables

with respective parameters .3, .5, and .7. find the expected number of accidents that will happen on any of these highways today.
Mathematics
1 answer:
Romashka-Z-Leto [24]3 years ago
8 0
There is a theorem that we can use to find the expected value of a random variable that is a sum of other random variables as follows.

If X=X_1 + X_2 + ...+X_k, then E(X)=E(X_1)+E(X_2)+...+E(X_k).

In this case, let X = the number of accidents that will happen on any of those highways today,
X_1,X_2,X_3 are the numbers of accidents on each highway, respectively.

Then  X = X_1+X_2+X_3.

Since X_1,X_2,X_3 are Poisson variates, their expected values are the parameters given, .3, .5 and .7.

So E(X_1)=.3, E(X_2) = .5, E(X_3) = .7
Thus, E(X)=E(X_1)+E(X_2)+E(X_3) = .3 + .5 + .7 = 1.5

The answer is 1.5.
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The sum of the squares of two consecutive positive even integers is 52.Find the integers
ankoles [38]

Answer:

The two consecutive positive integers are 4 and 6.

Step-by-step explanation:

Let x be an even integer, then the next even integer is (x+2).


The sum of the squares of these even integers is 52.


We can write the following equation and solve for x.


x^2+(x+2)^2=52.

We expand to get,

x^2+x^2+4x+4=52.

We simplify to get,

2x^2+4x+4-52=0.


This gives us,

2x^2+4x-48=0


We divide through by 2 to get,

x^2+2x-24=0.

This is now a quadratic trinomial

We split the middle term to get,


x^2+6x-4x-24=0.


We factor to obtain,

x(x+6)-4(x+6)=0


\Rightarrow (x+6)(x-4)=0


\Rightarrow (x+6)=0 or (x-4)=0


\Rightarrow x=-6 or x=4


We discard x=-6 since it is not a positive integer.

\therefore x=4


Hence the two consecutive positive even integers are,

4

and

4+2=6








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