48
+14
-------
62
If you need more work:
Ones place: 8+4=12
Tens place: 40+10=50
Add up: 50+12=62
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
Answer:
Step-by-step explanation:
If I read this correctly...
5 1/5^3= 5.20^3=140.608
5(5)^3=625
If this isn't the correct problems, sorry. it was kind of hard to decipher the last one.
Y² + 10z - 10y - yz = (y - 10)(y - z)
