Answer:
Infinite solutions
Step-by-step explanation:
1) First, you can solve this easily by elimination. Multiply the first equation by -2 in order to cancel out terms when adding to the second equation.
2) Then, add the new set of equations together. However, everything cancels out, bringing us to 0 = 0. This means that the lines the equations make must be the same. Thus, all real numbers must make this equation true, meaning that there are infinite 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
Answer:
6/5 x (x-6)
Step-by-step explanation:
convert the decimal into a fraction
1.2x -7.2
6/5x - 36/5
factor 6/5 from the expression
6/5x (x-6)
3x^2 + 8x - 4 - (6x^2 - 5x + 3)....distribute thru the parenthesis
3x^2 + 8x - 4 - 6x^2 + 5x - 3 .....combine like terms
-3x^2 + 13x - 7 <==
You would run 2.5/3 miles or .833 miles
