<span>1.) The probability that the traveler flew on East-West Air is 4/(4 + 3 + 6) = 4/13 = 31%
2.) </span>T<span>he probability that the traveler flew on Air One is 3/13 = 23%
3.) </span>T<span>he probability that the traveler flew on International Air is 6/13 = 46%</span>
Answer:
![y=\frac{11}{5}](https://tex.z-dn.net/?f=y%3D%5Cfrac%7B11%7D%7B5%7D)
Step-by-step explanation:
Given expression
![\frac{1}{3}(y-2)-\frac{5}{6}(y+1)=\frac{3}{4}(y-3)-2](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B3%7D%28y-2%29-%5Cfrac%7B5%7D%7B6%7D%28y%2B1%29%3D%5Cfrac%7B3%7D%7B4%7D%28y-3%29-2)
To solve for
for the given expression.
Solution:
We multiply each term with the least common multiple of the denominators of the fraction in order to remove fractions.
The multiples of the denominators are:
3 = 3,6,9,<u>12</u>,15
6 = 6,<u>12</u>
4 = 4,8,<u>12</u>
The least common multiple = 12.
Multiplying each term with 12.
![12.\frac{1}{3}(y-2)-12.\frac{5}{6}(y+1)=12.\frac{3}{4}(y-3)-2(12)](https://tex.z-dn.net/?f=12.%5Cfrac%7B1%7D%7B3%7D%28y-2%29-12.%5Cfrac%7B5%7D%7B6%7D%28y%2B1%29%3D12.%5Cfrac%7B3%7D%7B4%7D%28y-3%29-2%2812%29)
![4(y-2)-10(y+1)=9(y-3)-24](https://tex.z-dn.net/?f=4%28y-2%29-10%28y%2B1%29%3D9%28y-3%29-24)
Using distribution.
![4y-8-10y-10=9y-27-24](https://tex.z-dn.net/?f=4y-8-10y-10%3D9y-27-24)
Simplifying.
![-6y-18=9y-51](https://tex.z-dn.net/?f=-6y-18%3D9y-51)
Adding
both sides.
![-6y+6y-18=9y+6y-51](https://tex.z-dn.net/?f=-6y%2B6y-18%3D9y%2B6y-51)
![-18=15y-51](https://tex.z-dn.net/?f=-18%3D15y-51)
Adding 51 both sides.
![-18+51=15y-51+51](https://tex.z-dn.net/?f=-18%2B51%3D15y-51%2B51)
![33=15y](https://tex.z-dn.net/?f=33%3D15y)
Dividing both sides by 15.
![\frac{33}{15}=\frac{15y}{15}](https://tex.z-dn.net/?f=%5Cfrac%7B33%7D%7B15%7D%3D%5Cfrac%7B15y%7D%7B15%7D)
![\frac{33}{15}=y](https://tex.z-dn.net/?f=%5Cfrac%7B33%7D%7B15%7D%3Dy)
Simplifying fractions.
![\frac{11}{5}=y](https://tex.z-dn.net/?f=%5Cfrac%7B11%7D%7B5%7D%3Dy)
∴
(Answer)
Answer:
30
Step-by-step explanation: