Question 61
What percentage of
$$\frac{(\frac12)^2+(\frac12)^3}{\frac14\cdot(1-\frac13)}-\frac1{1+\frac12}$$
is
$$\frac1{12}\cdot\frac{(0{,}5-\frac14)^2\cdot8}{0{,}008\cdot(2+\frac12)^4}$$
? Round the result to hundredths. Solve without using a calculator.
$$\frac{(\frac12)^2+(\frac12)^3}{\frac14\cdot(1-\frac13)}-\frac1{1+\frac12}$$
is
$$\frac1{12}\cdot\frac{(0{,}5-\frac14)^2\cdot8}{0{,}008\cdot(2+\frac12)^4}$$
? Round the result to hundredths. Solve without using a calculator.