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:
Step-by-step explanation:
We are given that

Function f decreases from quadrant 2 to quadrant 1 and approaches y=0
It cut the y- axis at (0,6) and passing through the point (1,2).
Function g(x) approaches y=0 in quadrant 2 and increases into quadrant 1.
It passing through the point (-1,2) and cut the y-axis at point (0,6).
Reflection across y- axis:
Rule of transformation is given by

Using the rule then we get

By using

Substitute x=-1

Substitute x=0

Therefore,
is true.
60 divided by 4 is 15. so 2*15= 30 // 6*15= 90 // 8*15=120
Add 16.937 and 7.5. It's basically doing the problem backwards.