A used car dealer sells SUVs and cars. Of all the vehicles, 70% are cars. Of all the vehicles, 20% 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.

28.5%

Step-by-step explanation:

We know that out of all the vehicles, 20% are red cars, and we have a total that out of all the vehicles 70% are cars, then the probabilty will be equal to the quotient between the percentage of red cars divided by the percentage of cars, this is:

20/70 = 0.285

This means that the probability is 0.285 or 28.5%

You can prove this because if the total number of vehicles is 100, then we have 70 cars and 20 of these cars are red, then the relative frequency is 20/70 = 0.285.

The probability that a car chosen at random is red in percent is 28.6%.

Step-by-step explanation:

Let the total number of vehicles be 100. So that:

Of all the vehicles, 70% are cars = × 100

                                                      = 70

Therefore, 70 of the vehicles are cars and 30 are SUVs.

Of all the vehicles, 20% are red cars = × 100

                                                            = 20

Therefore, 20 of the cars are red and 50 had other colors.

Pr(a car chosen at random is red) =  

                                                       =

                                                       = 0.285714

                                                      = 0.29

The probability of a car chosen at random in percent is;

                       = × 100

                      = 28.5714

                      = 28.6%

The probability that a car chosen at random is red  in percent is  28.6%.

you would use (d.) the law of cosines or a^2 = b^2 + c^2 – 2bc(cos a)

, b^2 = a^2 + c^2 – 2ac(cos b), or c^2 = a^2 + b^2 - 2ab(cos c). you wouldn't use b. the law of sines as it is mostly effective when 1side and 2angles are given. a and c aren't the answer for they are the improper versions of the law of cosines. hope this is .

536

step-by-step explanation: