a) Please see the graph. b) 10 miles: $s=100$, 1 mile: $s=1600$, .1 mile: $s=16000$ c) The closer a radio is to a transmitter, the stronger the frequency. Since you are closer to one transmitter than to the other, then the weaker frequency would be muffled by the stronger frequency.
Work Step by Step
a) $d \geq 40$ when $s \leq 1$ b) $s=1600/d^2$ $s = 1600/10^2$ $s=1600/100$ $s=160$ $s=1600/d^2$ $s=1600/1^2$ $s=1600/1$ $s=1600$ $s=1600/d^2$ $s=1600/.1^2$ $s=1600/.01$ $s=16000$