The answer is 58 I think 166-8= 158
17/10-(2/5x2/2)
17/10-4/10
=13/10
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
Step-by-step explanation:
(1,5) (5,8)
x,y. x,y
Formula: <u>y2-y1</u>
x2-x1
so...
<u>8-5</u><u>=</u> <u>3</u>
5-1 = 4
so... <u>3</u>
4
