Total number of friends = 3
Let the fare be = $ J
As it is equally divided, each friend has to pay 
amount
Angela paid $5 additional as tip so her share is 
Hence,Angela paid $ in total.
920 because 15 is closer to 20 than to 10
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%
_____
C(n,k) = n!/(k!(n-k)!)
HI THERE!!!
The answer you chose is correct: Face
Have great Saturday!!!!!!!!
~TRUE BOSS
