Elementary Technical Mathematics

Published by Brooks Cole
ISBN 10: 1285199197
ISBN 13: 978-1-28519-919-1

Chapter 4 - Review - Page 207: 16


a. $\frac{1}{16}\ mi$ b. $\frac{1}{2}\cdot \frac{1}{16}\ mi = \frac{1}{32}\ mi.$

Work Step by Step

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: $$\frac{1}{2}\cdot \frac{1}{16}\ mi = \frac{1}{32}\ mi.$$
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