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 [tex]X=X_1 + X_2 + ...+X_k[/tex], then [tex]E(X)=E(X_1)+E(X_2)+...+E(X_k)[/tex].
In this case, let X = the number of accidents that will happen on any of those highways today,
[tex]X_1,X_2,X_3[/tex] are the numbers of accidents on each highway, respectively.
Then [tex]X = X_1+X_2+X_3[/tex].
Since [tex]X_1,X_2,X_3[/tex] are Poisson variates, their expected values are the parameters given, .3, .5 and .7.
So [tex]E(X_1)=.3, E(X_2) = .5, E(X_3) = .7[/tex]
Thus, [tex]E(X)=E(X_1)+E(X_2)+E(X_3) = .3 + .5 + .7 = 1.5[/tex]
The answer is 1.5.