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)!)
Answer:
Step-by-step explanation:
When x=0, y = ab⁰ = a
The y-intercept of the graph is 3, so a=3.
5 to 1 because they lost 10 games in the time they lost 2 games so it would in smallest form at 5 to 1
Equivalent lines for sure
Answer: The length of the bottom left side is 14.
To find the length of this side, we need to know the relationship between the different sections when chords cross each other.
The relationship is that if you multiply the 2 parts of each chord together, it will equal the product of the two sides in the other chord.
Our equation can be written as:
7(8) = 4(x)
56 = 4x
14 = x
