Answer
Range: $59^{\mathrm{o}}F $ to $95^{\mathrm{o}}F $, inclusive, or $[59^{\mathrm{o}}F,95^{\mathrm{o}}F].$
Work Step by Step
Given:
$ 15\leq C\leq 35\qquad $
Replace C with the expression for F
$ 15\displaystyle \leq\frac{5}{9}(F-32)\leq 35\qquad $
Solve for F. Begin by multiplying all parts with $\displaystyle \frac{9}{5}$
$15\displaystyle \cdot\frac{9}{5}\leq\frac{9}{5}\cdot\frac{5}{9}(F-32)\leq\frac{9}{5}\cdot 35$
$ 27\leq F-32\leq 63\qquad $ ... add 32
$59\leq F\leq 95$
Range: $\ \ 59^{\mathrm{o}}F $ to $95^{\mathrm{o}}F $, inclusive, or $[59^{\mathrm{o}}F,95^{\mathrm{o}}F].$
