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krok68 [10]
2 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]2 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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Pls help pls help me<br><br><br><br><br><br> Compare the following 2/-3, -4/5, 21/-32, -15/24
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Step-by-step explanation:

to compare them, we should all bring to the same denominator.

the lcm (lowest common multiplier) is the that denominator.

let's use the approach of the prime factors :

3 : 3

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32 : 2×2×2×2×2

24 : 2×2×2×3

the lcm is the combination of the longest "streaks" of the prime factors.

so,

2×2×2×2×2 × 3 × 5 = 32×15 = 480

2/-3 = -2/3 = -320/480 (as 3×160 = 480)

-4/5 = -4×96 / 5×96 = -384/480 (as 5×96 = 480)

21/-32 = -21/32 = -315/480 (as 32×15 = 480)

-15/24 = -15×20 / 24×20 = -300/480 (24×20 = 480)

as

-384/480 < -320/480 < -315/480 < -300/480

we can say

-4/5 < -2/3 < -21/32 < -15/24

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Check all of the ordered pairs that satisfy the equation below y equals 2 / 3 x
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where are the ordered pairs

Step-by-step explanation:

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