## Elementary Technical Mathematics

a. $\frac{1}{16}\ mi$ b. $\frac{1}{2}\cdot \frac{1}{16}\ mi = \frac{1}{32}\ mi.$
For the measurement $10\frac{3}{16}\ mi$, we have a. The precision is $\frac{1}{16}\ mi$ because the last fraction $\frac{3}{16}$ has an accuracy of $\frac{1}{16}$. b. The greatest possible error is one-half the precision: $$\frac{1}{2}\cdot \frac{1}{16}\ mi = \frac{1}{32}\ mi.$$