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 31 and 1/3 yards
There are 5 data in this question. To answer these question it is best to make the list based on price in ascending order.
The list would be: $650, $650, $740, $820, $ 1650
Mean: (650+650+740+820+1650) /5= $902
Median: $740 since there is 5 data, the median should be (5+1)/2= 3rd data.
Mode= $650 because it only the number with 2 frequency
Answer:
11.4
Step-by-step explanation: