## Thinking Mathematically (6th Edition)

The length of the femur is given by$f=0.432h-10.44$. It is also provided that $f=16\text{ inches}$. Now substituting $f=16$ in the above equation and solve for $h$: $16=0.432h-10.44$ Add $10.44$ on both sides; \begin{align} & 16+10.44=0.432h-10.44+10.44 \\ & 26.44=0.432h \end{align} Divide $0.432$on both sides. \begin{align} & \frac{26.44}{0.432}=\frac{0.432h}{0.432} \\ & 61.20=h \end{align} So,$h=61.20\text{ inches}$. As it is provided that the skeleton found is $5$feet tall. Since, \begin{align} & 1\text{ feet}=12\text{ inches} \\ & \Rightarrow 5\text{ feet}=12\times 5 \\ & =60\text{ inches} \end{align} which is approximately equal to, $61.20\ \text{inches}$. Yes, the partial skeleton height is slightly over 5 feet tall.