0.000038 = 3.8e-5 = 3.8×10⁻⁵
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)!)
256 is the value
REMEMBER PEMDAS
No, because it has a constant rate of change
By looking at this table
The y value changes at a constant multiple of 2
6 4
7 2 (4-2 is a difference of 2)
8 0 (2-0 is a difference of 2)
9 -2 (0-2 is a difference of 2)
This means it is Linear
Answer:
29 and 35
Step-by-step explanation:
x+y=64
x-y=6
Solve by substitution.
x=6+y
6+y+y=64
6+2y=64
y=29
Then plug in 29 for y.
x+29=64 ......35
