N3Y+7 can have infinate solutions
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
23) 17/30
41) 3/10 because 3/10 + 7/10 = 10/10 or 1.
Answer:
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Answer:
A.
Step-by-step explanation:
The formula for circumference is:
We know that the circumference is 8π inches. So, substitute 8π for C:
Solve for r. Divide both sides by 2π. So:
Therefore, the radius is 4 inches.
Hence, our answer is A.
