Answer:
6, 5, 1, 7, 9, 2, 4, 8, 3.
Step-by-step explanation:
(from top to bottom, this should be right)
(x, y)
The domain are all the x-values, the range are all the y-values.
R={(19,96),(20,101),(21,106),(22,111)}
The domain is: 19, 20, 21, and 22
The range is: 96, 101, 106, and 111
Answer:
The solutions are 
and 
Step-by-step explanation:
we have
Divide by 
both sides
------> 
we know that
The formula to solve a quadratic equation of the form 
is equal to
in this problem we have
so
substitute
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
