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
B. represents a dialation
Answer:
y = -4/3x + 6
Step-by-step explanation:
1. 3y - 4x + 3y = 18 - 3y
2. 4x = -3y + 18
3. 18 - 4x = -3y + 18 - 18
4. <u>-</u><u>3</u><u>/</u><u> </u><u>-3y = 4x - 18</u><u> </u><u>/</u><u>3</u>
5. y = -4/3x + 6
Answer:
1.3, 2 1/3, 1.34 is the order from least to greatest
Answer:
46.67
Step-by-step explanation: