In developing their warranty policy, an automobile company estimates that over a 1-year period 17% of their new cars will need to be repaired once, 7% will need repairs twice, and 4% will require three or more repairs. If you bought two cars from this company, what is the probability that neither will need repair over a 1-year period

Step-by-step explanation:

OVER A 1-YEAR PERIOD, it is estimated that

17 + 7 + 4 = 28% of new cars will need repairs (ranging from one repair to infinity number of repairs)

This means that the remaining percentage of new cars will NOT need any repairs over the 1-year period.

This percentage is 100 - 28 = 72%

The question is reframed thus: If you bought 2 cars from this company, what is the probability that none of them will need repair over the 1-year period?

What is the probability that none of the 2 cars falls within the 28% of cars which will need repair in the 1-year period?

72% = 72/100 = 0.72

this is the probability that a single car will not need repairs

the probability that 2 cars from this company will not need repair over a year is  0.72 ÷ 2 = 0.36

v = 1/3 a^2h

step-by-step explanation:

the volume is 1/3 a^2h.

look at the picture i have filled out a table for it. hope this .