It's a conditional probability
P( getting a given number AND getting again the same number)
P(of getting any number) =1/6
P(of getting once again the same number) 1/6 x 1/6 = 1/36
Answer:
61.84%
Step-by-step explanation:
Let the cost of the box be x. Since the price of the box and the pen is Rs 80, the pen's price can be represented as 80 - x. The box is sold at a ten percent profit, and an added ten percent is equal to 1.1. Therefore, the price the box sells at is 1.1(x). A 20% loss is the same a keeping 80% or multiplying by 0.8. This means the pen sold at 0.8(80 - x). Now, we are given the box went for Rs 28 more than the pen, so we can create an equation:
1.1x = 0.8(80 - x) + 28
We can simplify and solve:
1.1x = 64 - 0.8x + 28
1.9x = 92
x = 92/1.9
x = 920/19
The cost of the box after the increase would be 1.1(920/19) and the pen would be 0.8(80 - 920/19).
The sum of these two can be written as a percent x of 80.
80x = 0.8(80 - 920/19) + 1.1(920/19)
80x = 64 - 0.8(920/19) + 1.1(920/9)
80x = 64 - 0.3(920/19)
80x = 64 - 276/19
80x = 940/19
x = 940/1520
x = 0.6184
This is 61.84%
Check the picture below.
so the area of the hexagon is really just the area of two isosceles trapezoids.
![\textit{area of a trapezoid}\\\\ A=\cfrac{h(a+b)}{2}~~ \begin{cases} h=height\\ a,b=\stackrel{parallel~sides}{bases}\\[-0.5em] \hrulefill\\ a=2\\ b=4\\ h=2 \end{cases}\implies \begin{array}{llll} A=\cfrac{2(2+4)}{2}\implies A=6 \\\\\\ \stackrel{\textit{twice that much}}{2A = 12} \end{array}](https://tex.z-dn.net/?f=%5Ctextit%7Barea%20of%20a%20trapezoid%7D%5C%5C%5C%5C%20A%3D%5Ccfrac%7Bh%28a%2Bb%29%7D%7B2%7D~~%20%5Cbegin%7Bcases%7D%20h%3Dheight%5C%5C%20a%2Cb%3D%5Cstackrel%7Bparallel~sides%7D%7Bbases%7D%5C%5C%5B-0.5em%5D%20%5Chrulefill%5C%5C%20a%3D2%5C%5C%20b%3D4%5C%5C%20h%3D2%20%5Cend%7Bcases%7D%5Cimplies%20%5Cbegin%7Barray%7D%7Bllll%7D%20A%3D%5Ccfrac%7B2%282%2B4%29%7D%7B2%7D%5Cimplies%20A%3D6%20%5C%5C%5C%5C%5C%5C%20%5Cstackrel%7B%5Ctextit%7Btwice%20that%20much%7D%7D%7B2A%20%3D%2012%7D%20%5Cend%7Barray%7D)
8 pt = 4 qt
7c = 56 fl oz
