in a survey of 330 employees , 104 of them are uninsured , 54 work part time , and 12.5 percent of employees who are uninsured work part time . if a person is to be randomly selected from those surveyed , what is the probability that the person will neither work part time nor be uninsured ?
"- - - - - - - - - ui - - - - - - - - - - - - - - - - nui - - - - - - - total pt - - - - ( 12.5 / 100 ) * 104 = 13 - - - - - - - - - - - - - 54 npt - - - 104 - 13 - - - - - - - - - - - - - - x - - - - - - - - 276 total - - 104 - - - - - - - - - - - - - - - - - - - - - - - - - - - - 330 we have to find not part time and not uninsured . in other words not part time and insured = x / 330 = ( 276 - 104 + 13 ) / 330 = 37 / 66 answer is a ."
divide(subtract(330, add(subtract(54, divide(multiply(12.5, 104), multiply(const_1, const_100))), 104)), 330)