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:
3t - 16
Step-by-step explanation:
(18/12 t-8)*2
First simplify inside the parentheses
(3/2 t -8)*2
Then multiply
3/2t *2 -8*2
3t - 16
1. y = -1/2 + 2.
2. y= 2/5x - 4/5
Answer: -30
Step-by-step explanation:
Answer:
7.52
Step-by-step explanation:
- Write it out: 3.76 × 2
- 3.76 × 2 = 7.52
I hope this helps!
