Get all your variables on same side and solve x=-2
Y=-6-12x-2
y=-6-12x-2
y=-8-12x
y=-8-12x
and these are real numbers
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)!)
The answers would be A D and E because if you were to flip them onto each other they would be the exact same
A nonmember would have to go skating 11 times a year and a member would have to go skating 12 times a year.
They each paid $110.
I hope this is correct!
Good luck:)
