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:
- (9.5, 6.5) and (-4.5, -7.5)
Step-by-step explanation:
Let the extended points be A' and B' and add the point M as midpoint of AB
<u>Coordinates of M are:</u>
- ((6 - 1)/2, (3-4)/2) = (2.5, -0.5)
Now point A is midpoint of A'M and point B is midpoint of MB'
<u>Finding the coordinates using midpoint formula:</u>
- A' = ((2*6 - 2.5),(2*3 - (-0.5)) = (9.5, 6.5)
- B' = ((2*(-1) - 2.5), (2*(-4) - (-0.5)) = (-4.5, -7.5)
The correct should be X=10
Answer:x = 164686.592598
Step-by-step explanation:
219379.01−x/x = 0.3321
−x+219379.01/x = 0.3321
Step 1: Multiply both sides by x.
−x+219379.01≈0.3321x
−x+219379.01 − 0.3321x = 0.3321x−0.3321x (Subtract 0.3321x from both sides)
−1.3321x + 219379.01 = 0
−1.3321x + 219379.01 − 219379.01 = 0 − 219379.01 (Subtract 219379.01 from both sides)
−1.3321x = −219379.01
−1.3321x/−1.3321 = −219379.01/−1.3321
(Divide both sides by -1.3321)
x = 164686.592598
