(i) The product of the two expressions is equal to the product of their factors. (ii) The product of the two expressions is equal to the product of their H.C.F. and L.C.M. 2.
Answer:
probability of an electric power outage shipboard is 0.075 or 7.5 %
Step-by-step explanation:
Given the data in the question;
Probability of engine malfunction = 25% = 0.25
Probability of generator malfunction = 10% = 0.1
Probability of generator being repaired = 50% = 0.5
probability of an electric power outage shipboard = ?
To get the probability of an electric power outage shipboard, we use the expression;
probability of an electric power outage = 2
× p( Probability of generator malfunction ) × p( Probability of generator being repaired ) × ( 1 - p( Probability of engine malfunction ) )
so we substitute in our given values;
probability of an electric power outage = 2 × 0.1 × 0.5 × ( 1 - 0.25 )
probability of an electric power outage = 2 × 0.1 × 0.5 × 0.75
probability of an electric power outage = 0.075 or 7.5 %
A) 6
b) x/f
c) 5/f
Mark brainliest please
Hope this helps you
Sure. What do you need help with exactly?
Answer:
Step-by-step explanation:
8/3=3(c+5/3)
multiply both sides by 3
8=9(c+5/3)
divide both sides by 9
8/9=c+5/3
subtract 5/3 from both sides
c=8/9-5/3
change 5/3 to 15/9
c=8/9-15/9
c=-7/9