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)!)
Lets solve the equalities first.
NP>MN
Cancel out the N
P>M
Next
MP<MN
Cancel out the M
P<N
M<P<N
The answer is C.
Answer:
225
Step-by-step explanation:
3240k = 2×2×2×3×3×3×3×5×k
Every factor should be in groups of 3
(3×5)² = k
k = 225
Answer:
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