Suppose that a recent poll of american households about car ownership found that for households with a car, 39% owned a sedan, 33% owned a van, and 7% owned a sports car. suppose that three households are selected randomly and withreplacement.35)what is the probability that none of the three randomly selected households own a van? (round to the nearest thousandth) a)0.060 b)0.699 c)0.036 d)0.301

Let the total number of cars = 100

Total of sedan, van and sports cars = 39+33+7 = 79

Number of other cars = 100 - 79 = 21.

Let A be the event of owning a sedan,

B be the event of owning a van and

C be the event of owning a sports car.

D be the event of owning other cars.

Then,

and

Now,

p(non selection of van) = p(A ∪ C ∪ D)

= p(A) + p(C) + p(D)

= 0.67

The probability of none of the three selections would be a van is:

0.67 × 0.67 × 0.67

= 0.300763

=0.301

Hence, the correct option is (D).

dc=6 square root of 3 and ac is 12 square root 3

c and d

step-by-step explanation:

square root of 3/2 = 18/c

cross multiply to get 18*2=√3*c

36/√3 * √3/√3 = c

c= 36*√3/3

c=12√3