The number of accidents that occur each day at a certain intersection along
with the corresponding probabilities are
shown.

0.12 accidents.

Step-by-step explanation:

The expected value is:

EV = ∑pₙ*xₙ

where xₙ is the event number n, pₙ is the probability of event xₙ

in this case we have:

x₀ = 0 accidents and p₀ = 0.935

x₁ = 1 accident, and p₁ = 0.03

x₂ = 2 accidents, and p₂ = 0.02

x₃ = 3 accidents, and p₃ = 0.01

x₄ = 4 accidents, and p₄ = 0.005

Then the expected value is:

EV = (0*0.935 + 1*0.03 + 2*0.02 + 3*0.01 + 4*0.05) = 0.12 accidents.

x=-7

step-by-step explanation:

x+3=-10

-10+3=-7

x=-7