#### Answer

a. $C=\frac{kP_1P_2}{d^2}$, where $k$ is a constant.
b. $k\approx0.02$, $C=\frac{0.2P_1P_2}{d^2}$
c. $39813$ calls.

#### Work Step by Step

a. Based on the given conditions, we can write the relation as $C=\frac{kP_1P_2}{d^2}$, where $k$ is a constant.
b. Given
$P_1=777,000, P_2=3,695,000, d=420\ mi$ and $C=326,000$
we have
$C=\frac{777000(3695000)k}{(420)^2}=326000$
which gives $k\approx0.02$ and the variation equation becomes
$C=\frac{0.2P_1P_2}{d^2}$
c. Given
$P_1=650,000, P_2=490,000, d=400\ mi$
we have
$C=\frac{0.2(650000)(490000)}{400^2}\approx39813$ calls.