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:
hard
Step-by-step explanation:
I have never seen this before
Answer:
I'm pretty sure A is the correct answer :)
(x-2)(x^4+3) is a polynomial
Let x=ab=ac, and y=bc, and z=ad.
Since the perimeter of the triangle abc is 36, you have:
Perimeter of abc = 36
ab + ac + bc = 36
x + x + y = 36
(eq. 1) 2x + y = 36
The triangle is isosceles (it has two sides with equal length: ab and ac). The line perpendicular to the third side (bc) from the opposite vertex (a), divides that third side into two equal halves: the point d is the middle point of bc. This is a property of isosceles triangles, which is easily shown by similarity.
Hence, we have that bd = dc = bc/2 = y/2 (remember we called bc = y).
The perimeter of the triangle abd is 30:
Permiter of abd = 30
ab + bd + ad = 30
x + y/2 + z =30
(eq. 2) 2x + y + 2z = 60
So, we have two equations on x, y and z:
(eq.1) 2x + y = 36
(eq.2) 2x + y + 2z = 60
Substitute 2x + y by 36 from (eq.1) in (eq.2):
(eq.2') 36 + 2z = 60
And solve for z:
36 + 2z = 60 => 2z = 60 - 36 => 2z = 24 => z = 12
The measure of ad is 12.
If you prefer a less algebraic reasoning:
- The perimeter of abd is half the perimeter of abc plus the length of ad (since you have "cut" the triangle abc in two halves to obtain the triangle abd).
- Then, ad is the difference between the perimeter of abd and half the perimeter of abc:
ad = 30 - (36/2) = 30 - 18 = 12
Answer:
1920 cubic inches
Step-by-step explanation:
Volume = Base Area × Height
= 240 × 8
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