The probability that a student owns a car is 74%. The probability that a student owns a car or drives to school is 88%. The probability that a student owns a car and drives to school is 56%. Given that the student drives to school, what is the probability that the student owns a car?

The probability that the student owns a car given that the student drives to school is 0.8

Step-by-step explanation:

Lets put names to each variable

X = The student owns a car

Y = The student drives to school

We alredy know:

P(X) = 0.74

P(X ∪ Y) = 0.88

P(X ∩ Y) = 0.56

And we want to know P(X | Y)

Recall that P(X ∩ Y) = P(X|Y)*P(Y), and also

P(X ∪ Y) = P(X) + P(Y) - P(X∩Y)

Thus,

P(Y) = P(X ∪ Y) - P(X) + P(X ∩ Y) = 0.88-0.74+0.56 = 0.70

And

P(X|Y) = P(X ∩ Y)/P(Y) = 0.56/0.70 = 0.8

Therefore, the probability that the student owns a car given that the student drives to school is 0.8

(1, –1), (16, –4), (36, –6)

step-by-step explanation:

given that the function is

this can also be written as

we have to graph this function and check which list of points lie on the graph

see attachment for graph:

we find that the points lying are (1,-1) (16,-4) and (36,-6)

since square root is there we find that x can only be posiitve but y can be since square root both positive and negative.

answer:   expression= p+p*d%

price reduction + sale price = $625

sale price: $p+$p*d%

where d% is the percentage of discount

step-by-step explanation:

original price = p

price went down by $150 price reduction $p*d%

sale price= $ 475

therefore the price before sale is: $625

and the expression that shows the price before sale is: p+p*d%