(1 point) In a random sample of 1,000 people, it is found that 6.3% have a liver ailment. Of those who have a liver ailment, 7% are heavy drinkers, 55% are moderate drinkers, and 38% are nondrinkers. Of those who do not have a liver ailment, 14% are heavy drinkers, 44% are moderate drinkers, and 42% are nondrinkers. If a person is chosen at random, and he or she is a heavy drinker, what is the empirical probability of that person having a liver ailment? 0.00854217

P(liver ailment | heavy drinker) = 0.087*0.08 = 0.00696

Ans: # = 0.00696*1000 = 6.96 or approximately 7

Step-by-step explanation:

the simplified form is

step-by-step explanation:

we want to simplify the expression,

let us change the middle bar to a normal division sign to obtain,

we multiply the first fraction by the reciprocal of the second fraction to obtain,

we factor to obtain,

we cancel the common factors to get,

we simplify to get,

the correct answer is b

hi how are you doing today, if you need just ask me.

step-by-step explanation: