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%
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C(n,k) = n!/(k!(n-k)!)
Answer:
Domain: {0, 1, 2, 3, 4, 5, 6}
Range: {4, 4.5, 5, 5.5, 6, 6.5, 7}
Answer:6.6 y
Step-by-step explanation:
Answer:
(0,-2)
(2,-1)
(4,0)
Step-by-step explanation:
The points are the above ones.
9514 1404 393
Answer:
x = 7
Step-by-step explanation:
The product of segment lengths is the same for the two chords.
9(21) = x(27)
x = 189/27 . . . . . divide by 27
x = 7
