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
Variable, because its a letter<span />
Answer:
Answer
Step-by-step explanation:
3 is the constant since there is no variable beside it, and 7 is the coefficient because there is a variable beside it.
D)5,400,000 Is your answer....
A is additive equality, where -x and x adds to 0
b is distributive, where x(y+z)=xy+xz
c is multiplicative equality, where x*1=x
d is subtracting both sides by 3.
