Suppose LCM(a,b) = 20160, a=60. Find the smallest possible value of b.
2 answers:
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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 %
Answer:
129.996 cubic feet
Step-by-step explanation:
Assuming that the barrel is a perfect cylinder, we can use the formula to find the volume:
First, we have to find the radius, which is:
D = 2r
So the radius is 3 feet, since it is half of the diameter. Then, we plug in the values.
Now, we solve. Exponents are first...
3.14(9)(4.6)
Now multiply left to right.
28.26(4.6)
129.996 cubic feet
Therefore, the barrel can hold 129.996 cubic feet of oil. Hope this helps you!