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:
56,500,000.
Step-by-step explanation:
56,477,812 rounded to the nearest hundred thousand:
The 4 is in the hundred thousands place, so we'll look at the next digit to the right of that, which is the 7:
56,<u>4</u>77,812
Since 7 is more than 5, we'll have to go up a number, which will be the 4. Afterwards, we'll have to replace all the digits after the 4 with zeros.
56,500,000.
Answer:
its gonna be the second one down on the left!! i can give an explanation if ur having trouble
Step-by-step explanation:
Answer:
Step-by-step explanation:
the equation could model a scenario determining the number of players on each team in a league of five teams, if 2 players have to be taken out of the possible players to be referees and to keep score. there will be 11 on each team
