According to the associated press report, 44 percent of parents who have purchased tv sets after v-chips became standard equipment in january 2000 are aware that their sets have v-chips, and of those who are aware of the option, 36 percent have programmed their v-chips. using these results, find an estimate of the probability that a randomly selected parent who has bought a tv set since january 2000 has programmed the v-chip. (round your answers to 4 decimal places.)

0.1584

Step-by-step explanation:

The relationship between the two events can be estimated by using the conditional probability. This is used to show that the occurrence of an event is dependent on another one.

P (purchase) is the probability of those that bought television before V-chips installation = 44/100 = 0.44

P(V-chips | purchase) probability of those that have installed after purchase = 36/100 = 0.36

Therefore:

P(V-chip ∩ purchase) is the probability of a random person that bought and have installed V-chips = 0.44*0.36 = 0.1584

y < -3

step-by-step explanation:

-3y - 2 > 7

-3y -2 + 2 > 7+2

-3y > 9

-3y/-3 < 9/-3   (change the inequality sign when multiplying or dividing by a negative number)

y < -3

7,24

step-by-step explanation: