Suppose that 39% of a town's population have type o blood, 15% are rh-negative, and 7% have type o
blood and are rh-negative. what is the probability that a randomly selected individual in the town will
neither have type o blood nor be rh-negative?

53 %

Step-by-step explanation:

Let O represents the event of Type O blood and R represents the type Rh-negative blood,

According to the question,

P(O) = 39% = 0.39,

P(R) = 15% = 0.15,

P(O ∩ R) = 7% = 0.07,

We know that,

P(O ∪ R) = P(O) + P(R) - P(O ∩ R)

= 0.39 + 0.15 - 0.07

= 0.47

Hence, the probability of an individual who is neither have type O blood nor be Rh-negative = 1 - P(O ∪ R)

= 1 - 0.47

= 0.53

= 53 %

a = 3.14

b = 18

c = 18/5 or 3.6 or 3 3/5

step-by-step explanation:

$3,200 x 7.25% = $3,200 x .0725 = $232 interest monthly

$232 monthly x 12 annual = $2, 784

step-by-step explanation: