Question

A 30% cranberry juice drink is mixed with a 100% cranberry juice drink. The function f(x)=(6)(1.0)+x(0.3)6+x models the concentration of cranberry juice in the drink after x gallons of the 30% drink are added to 6 gallons of pure juice. What will be the concentration of cranberry juice in the drink if 2 gallons of 30% drink are added? Give the answer as a percent.

Answer 1

82.5%

Explanation:

It helps to start with the correct formula:

f(x) = ((6)(1.0) +x(0.3))/(6 +x) . . . . parentheses are required

Then f(2) is ...

f(2) = (6 +.3(2))/(6+2) = 6.6/8

f(2) = 82.5%