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)!)
The answer is 370 because you need to work it out.
Basically what I did was look at the first number, 12, and the answer to the equation and saw that 12 x 10 would be 120. And 360 + 10 would be 370.
It fits in perfectly.
12(370 - 360) = 120
370 - 360 = 10
12(10) = 120
So x = 370
Answer:
It's easy to figure it out. The equation we have is 8 = y - 9. We need to isolate y. Simply add 9 to each side of the equation to get this: 17 = y.
ANSWER: 17 = y
Answer:
B. (3, –1)
Step-by-step explanation:
Reflection in the line y=-x is described by the transformation ...
(x, y) ⇒ (-y, -x)
Then the point of interest becomes ...
F(1, -3) ⇒ F'(3, -1) . . . . matches choice B
