For there to be 1 car, we consider two possible outcomes:
The first door opened has a car or the second door opened has a car.
P(1 car) = 2/6 x 4/5 + 4/6 x 2/5
P(1 car) = 8/15
For there to be no car in either door
P(no car) = 4/6 x 3/5
P(no car) = 2/5
Probability of at least one car is the sum of the probability of one car and probability of two cars:
P(2 cars) = 2/6 x 1/5
= 1/15
P(1 car) + P(2 cars) = 8/15 + 1/15
= 3/5
10 1/18...........................................
...................................................
The total amount of angles in quadrilateral is 360.so:
z=360-(122+79+90)
z=360-291
z=69
Answer:
p(x)=x^3-4x^2+5x-2
=(2)^3-4(2)^2+5(2)-2
=8-4(4)+10-2
=8-16+10-2
=0
hope it helps mark as brainliest
