Respuesta :
Solution :
It is given that :
The car of Mike's can hold = [tex]$14 \frac{2}{5}$[/tex] g of gas.
[tex]$=\frac{72}{5}$[/tex] g of gas
∴ [tex]$=\frac{1}{4} $[/tex] th of [tex]$\frac{72}{5}$[/tex] g of gas is [tex]$=\frac{1}{4} \times \frac{72}{5}$[/tex]
[tex]$=\frac{18}{5}$[/tex] g of gas
The average distance that the car can travel when the tank is full is = 20 mpg
∴ In [tex]$\frac{72}{5}$[/tex] g of gas, the car travels = 20 mpg
In 1 g of gas, the car travels = [tex]$20 \times \frac{5}{72} $[/tex] mpg
So in [tex]$\frac{18}{5}$[/tex] g of gas, the car travels = [tex]$20 \times \frac{5}{72} \times \frac{18}{5} $[/tex] mpg
= 5 mpg
