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)!)
If you were trying to simplify it, it's not posible because 23 is a composite number so I just went ahead and did division for you:
23/46=0.5
I hope this helped!
Answer:
See below.
Step-by-step explanation:
2b + 8 − 5b + 3 = −13 + 8b − 5
Reorder like terms.
2b-5b+8+3=8b-13-5
Combine those like terms. Then solve.
-3b+11=8b-18
-11 -11
-3b=8b-29
-8b -8b
-11b=-29
/-11 /-11
b=2.64
-hope it helps
Answer:
C
Step-by-step explanation:
No, she didn’t use the entire fraction as the base, so wrote the expanded form wrongly.
and I know this cuz I took the test
The 2 angles at a vertex are supplementary (one interior and one exterior) the exterior angle is = the the sum of the 2 remote interior angles..