Answer: (a) 0.006
(b) 0.027
Step-by-step explanation:
Given : P(AA) = 0.3 and P(AAA) = 0.70
Let event that a bulb is defective be denoted by D and not defective be D';
Conditional probabilities given are :
P(D/AA) = 0.02 and P(D/AAA) = 0.03
Thus P(D'/AA) = 1 - 0.02 = 0.98
and P(D'/AAA) = 1 - 0.03 = 0.97
(a) P(bulb from AA and defective) = P ( AA and D)
= P(AA) x P(D/AA)
= 0.3 x 0.02 = 0.006
(b) P(Defective) = P(from AA and defective) + P( from AAA and defective)
= P(AA) x P(D/AA) + P(AAA) x P(D/AAA)
= 0.3(0.02) + 0.70(0.03)
= 0.027
Answer:
B. Yes, because 0.004<0.010.004<0.010, point, 004, is less than, 0, point, 01.
Step-by-step explanation:
Khan Academy
Simplify the radical by breaking the radicand up into a product of known factors, assuming positive real numbers.
1.5
i
1.5i
Answer:
56 + 8x
Step-by-step explanation:
8(7) + 8(x)
= 56 + 8x
