From the question;
The number of voters randomly polled n is inversely proportional to the square of the desired margin of error E
This implies
[tex]n\text{ }\propto\frac{1}{E^2}[/tex]that is
[tex]\begin{gathered} n\text{ = }\frac{k}{E^2} \\ \text{Where k = constant of proportionality} \end{gathered}[/tex]we are given
when, E = 0.1 (10%), n = 58 voters
we get
[tex]\begin{gathered} 58\text{ = }\frac{k}{(0.1)^2} \\ 58\text{ }\times(0.1)^2\text{ = k} \\ 58\text{ }\times\text{ 0.01 = k} \\ k\text{ = 0.58} \end{gathered}[/tex]Therefore the connection between n and E is
[tex]n\text{ = }\frac{0.58}{E^2}[/tex]