Answer
$0.58m$
Work Step by Step
Use equation 9-14 $X_{\mathrm{c}\mathrm{m}}=\displaystyle \frac{\sum mx}{M}$
Let $X_{\mathrm{c}\mathrm{m}}=0$, the middle of length of the basket $(L=0.71m)$. Let right be +x.
Cereal: $m_{1}=2\displaystyle \cdot 0.56kg, \qquad x_{1}=-L/2=-\frac{0.71m}{2}$
Milk:$\quad m_{2}=1.8kg,\qquad x_{2}=?$
$X_{\mathrm{c}\mathrm{m}}=\displaystyle \frac{m_{1}x_{1}+m_{2}x_{2}}{m_{1}+m_{2}}=0$
$m_{1}x_{1}+m_{2}x_{2}=0$
$m_{2}x_{2}=-m_{1}x_{1}$
$x_{2}=-\displaystyle \frac{m_{1}x_{1}}{m_{2}}=-\frac{(2\cdot 0.56kg)(-\frac{0.71m}{2})}{1.8kg}=+0.22m$
that is, $0.22m$ right of center.
Relative to the left end, this is $\displaystyle \frac{0.71m}{2}+0.22m=0.58m$
