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)!)
Answer:
the ability to objectively analyze information and draw a rational conclusion
Step-by-step explanation:
Answer:
46656
Step-by-step explanation:
Well, since 3/15 can be simplified to 1/5 by dividing the top and the bottom by 3, you would just need to find the decimal of 1/5 which is much more commonly recognized. it's .2 (becasue 1/10 would be .1 and 1/5=2/10 so .1x2=.2)
Answer:
y=x+4
Step-by-step explanation:
So, let's start with the fact that we have two numbers. Let's call one of them "x" and the other one "y." Let's also assume that "y" is the larger number and "x" is the smaller number.
We know that both of them added together (the sum) gives us 74. Mathematically, that means:
x+y=74.
We also know that the larger number ("y") is 4 more than the smaller number ("x"). Another way to say this is, if you have the smaller number and add 4 more to it, you'll end up with the larger number. Mathematically, this means:
y=x+4
Since we know that y=x+4, we can put "x+4" into the first equation for "y", giving us:
x+(x+4)=74
Simplifying gives us:
2x+4=74
-4 -4
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2x=70
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2 2
x=35
Since we now know that x=35, and we know that y=x+4, we now know that:
y=(35)+4
y=39
Now that we have both of the numbers, we should check our work to ensure it's correct:
Does 39+35=74? Yes it does.
Is 39 (the larger of the two numbers) four more than the smallest number (35)? Yes it is.
