A= P(1 + r/n) ^nt
=8000(1+0.08/2)^8x2
=8000(1.04)^16
=$14983.85
Literally just divide are you slow
Answer:
51,111%
Step-by-step explanation:
We know that 45% have a dog but no cat. While 23% have both a dog and a cat.
Then Pr(X has cat|X has dog) = Pr(X has dog and cat)/Pr(X has dog)
= Pr(X has dog and cat)/Pr(X has dog) = 0.23/0.45 = 0.51 ≈ 51,111%
The probability that a student’s family owns a cat if the family owns a dog is 51,111%
