Answer
relative error: $0.057$
percent error: $0.57\%$
Work Step by Step
For the measurement $5\frac{7}{16}~in$, we have
The precision is $\frac{1}{16}~in$ because the last fraction $7/16$ has a precision of $1/16$.
The greatest possible error is one-half the precision: $\frac{1}{2}\cdot\frac{1}{16}~in=\frac{1}{32}~in$
The relative error is
$relative~error=\dfrac{greatest~possible~error}{measurement}$
$\frac{\frac{1}{32}~in}{5\frac{7}{16}~in}=0.0057$
The percent of error is: $0.57\%$
