You need as much honey as you do lime juice. So the answer is ( 1/2 )
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)!)
Im confused on what your trying to ask...
Answer:
21
Step-by-step explanation:
Add all the numbers together
6+8+2.5+2.5+2=21
Answer:
(3,-3)
Step-by-step explanation:
