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:
D)
Step-by-step explanation:
Every tenth person will get it no matter what, every time. The order will not change.
Answer:
You can regroup it 3 times and a remainder of 43. :) hope it helped!
Step-by-step explanation:
Answer:
y=2/7x-4
Step-by-step explanation:
Write in Slope intercept form
y=mx+b
m = slope
b = y intercept
x = x intercept
y = (0,-4)
So,
y = 2/7x+(-4) substitute
y = 2/7x-4
