Answer
See below.
Work Step by Step
Let $y=ax^b$.
Then two of our equations are: $0.5=a5^b$ and $0.75=a11^b$.
If we divide the second equation by the first one we get:
$1.5=2.2^b=\\b=\ln_{2.2}{1.5}\approx0.514$.
Then $0.5=a5^{0.514}\\a=\frac{0.5}{5^{0.514}}\approx0.219$.
Thus $y=0.219\cdot x^{0.514}$
Then plugging in $x=120$ we get: $y=0.219\cdot (120)^{0.514}\approx2.5$ is required.
