Answer:
Step-by-step explanation:
Discussion.
Not directly. But the quadratic formula can do it. But that's not your question.
the factors you get must contain the factors for 1 which are 1 and - 1
These factors must add to - 5. There's no way that will happen with 1 and - 1 and you would be creative math if you tried to say that you could make one of the factors (x -1-1-1-1-1-1). That creates a whole new question.
There are C(5,2) = 10 ways to choose 2 contraband shipments from the 5. There are C(11, 1) = 11 ways to choose a non-contraband shipment from the 11 that are not contraband. Hence there are 10*11 = 110 ways to choose 3 shipments that have 2 contraband shipments among them.
There are C(16,3) = 560 ways to choose 3 shipments from 16. The probability that 2 of those 3 will contain contraband is
110/560 = 11/56 ≈ 19.6%
_____
C(n,k) = n!/(k!(n-k)!)
Answer:
4 mugs
Step-by-step explanation:
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