What is the least common denominator of 1/8 , 7/12 , and 1/6?
2 answers:
You might be interested in
Solution
To solve division problems you can make equal groups
Example
It is
Answer:
(a)$67
(b)You are expected to win 56 Times
(c)You are expected to lose 44 Times
Step-by-step explanation:
The sample space for the event of rolling two dice is presented below
Total number of outcomes =36
The event of rolling a 5 or a 6 are:
Number of outcomes =20
Therefore:
P(rolling a 5 or a 6)
The probability distribution of this event is given as follows.
First, we determine the expected Value of this event.
Expected Value
Therefore, if the game is played 100 times,
Expected Profit =$0.67 X 100 =$67
If you play the game 100 times, you can expect to win $67.
(b)
Probability of Winning
If the game is played 100 times
Number of times expected to win
Therefore, number of times expected to loose
= 100-56
=44 times
Check the picture below.
so.. simply, use the distance formula, to get their length an add them up, and that's the perimeter of the polygon.
![\bf -------------------------------\\\\ d=\sqrt{[2-(-1)]^2+(4-2)^2}\implies d=\sqrt{(2+1)^2+(2)^2} \\\\\\ d=\sqrt{3^2+2^2}\implies \boxed{d=\sqrt{13}}\\\\ -------------------------------\\\\ d=\sqrt{(3-2)^2+(-2-4)^2}\implies d=\sqrt{1^2+(-6)^2}\implies \boxed{d=\sqrt{37}}\\\\ -------------------------------\\\\ d=\sqrt{(-2-3)^2+[-3-(-2)]^2}\implies d=\sqrt{(-5)^2+(-3+2)^2} \\\\\\ d=\sqrt{(-5)^2+(-1)^2}\implies \boxed{d=\sqrt{26}}](https://tex.z-dn.net/?f=%5Cbf%20-------------------------------%5C%5C%5C%5C%0Ad%3D%5Csqrt%7B%5B2-%28-1%29%5D%5E2%2B%284-2%29%5E2%7D%5Cimplies%20d%3D%5Csqrt%7B%282%2B1%29%5E2%2B%282%29%5E2%7D%0A%5C%5C%5C%5C%5C%5C%0Ad%3D%5Csqrt%7B3%5E2%2B2%5E2%7D%5Cimplies%20%5Cboxed%7Bd%3D%5Csqrt%7B13%7D%7D%5C%5C%5C%5C%0A-------------------------------%5C%5C%5C%5C%0Ad%3D%5Csqrt%7B%283-2%29%5E2%2B%28-2-4%29%5E2%7D%5Cimplies%20d%3D%5Csqrt%7B1%5E2%2B%28-6%29%5E2%7D%5Cimplies%20%5Cboxed%7Bd%3D%5Csqrt%7B37%7D%7D%5C%5C%5C%5C%0A-------------------------------%5C%5C%5C%5C%0Ad%3D%5Csqrt%7B%28-2-3%29%5E2%2B%5B-3-%28-2%29%5D%5E2%7D%5Cimplies%20d%3D%5Csqrt%7B%28-5%29%5E2%2B%28-3%2B2%29%5E2%7D%0A%5C%5C%5C%5C%5C%5C%0Ad%3D%5Csqrt%7B%28-5%29%5E2%2B%28-1%29%5E2%7D%5Cimplies%20%5Cboxed%7Bd%3D%5Csqrt%7B26%7D%7D)
![\\\\ -------------------------------\\\\ d=\sqrt{[-1-(-2)]^2+[2-(-3)]^2}\implies d=\sqrt{(-1+2)^2+(2+3)^2} \\\\\\ d=\sqrt{(1)^2+(5)^2}\implies \boxed{d=\sqrt{26}}](https://tex.z-dn.net/?f=%5C%5C%5C%5C%0A-------------------------------%5C%5C%5C%5C%0Ad%3D%5Csqrt%7B%5B-1-%28-2%29%5D%5E2%2B%5B2-%28-3%29%5D%5E2%7D%5Cimplies%20d%3D%5Csqrt%7B%28-1%2B2%29%5E2%2B%282%2B3%29%5E2%7D%0A%5C%5C%5C%5C%5C%5C%0Ad%3D%5Csqrt%7B%281%29%5E2%2B%285%29%5E2%7D%5Cimplies%20%5Cboxed%7Bd%3D%5Csqrt%7B26%7D%7D)
so, those are their lengths, sum them all up, that's the polygon's perimeter.
so.. simply, use the distance formula, to get their length an add them up, and that's the perimeter of the polygon.
so, those are their lengths, sum them all up, that's the polygon's perimeter.