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:
200-225
Step-by-step explanation:
trust me haha
Answer:
a.b=[-1]
Step-by-step explanation:
This is a dot product of a and b
a.b =[6,-2,3].[2,5, -1] = [12-10-3] =[-1]
Answer:
C. 5
> 15
Step-by-step explanation:
In the given series of inequalities, the inequality sign determines the final outcome.
A. 4 - 2 < 10
4 < 12
< 3
Since < 3, then the inequality is false
B. + 9 < 12
< 3
Since < 3, then the inequality is false
c. 5 > 15
> 3
Since > 3, then the inequality is true
D. 6 > 2
3 >
Since < 3, then the inequality is false
Thus, the only inequality that is true is option C.
Answer:
0.75
Step-by-step explanation:
