A used car dealer sells SUVs and cars. Of all the vehicles, 90% are cars. Of all the vehicles 40% are red cars. What is the probability that a car chosen at random is red? Round your answer to the nearest tenth of a percent.

44.4%

Step-by-step explanation:

To calculate this, we proceed as follows.

we use the probability equation below;

P(A|B) = P(A and B) / P(B)

Applying the above to the scenario at hand;

P(red | car) = P(red and car) / P(car)

P(red and car) = 40% or simply 40/100 = 0.4

P(car) = 90% = 90/100 = 0.9

P(red | car) = 0.4/0.9

P(red | car) = 0.4444 which is = 44.44% ; to the nearest tenth of a percent = 44.4%

step-by-step explanation:

given perimeter: 28cm

length: 9cm

atp,

2(l+w)=p of the reactangle

2(9+x)=28

18+2x=28

2x=28-18

x=10/2

x=5

therefore width is 5 cm