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
$20.70
Eighteen dollars
Fifteen percent of eighteen is 2.7
(.15 X 18)+18
minus 4 both sides
square both sides
x+2=4
minus 2
x=2
<span>the answer is
x = 2; not extraneous</span>
Distribute:
(12n-12)5
60n-60 = A(n)
You can't really find what n is I don't think, because you have 2 unknown variables n and A(n).
Prime numbers from 1 to 50:
2, 3, 5, 7, 11, 13<span>, 17, 19, 23, 29, 31, </span>37,41<span>, </span>43<span>, </span><span>47</span>
