= Solution
With the stated ratio $r=v_A/v_B=2$, the time difference equals the <travel time> of A: $\Delta t=R/v_A$. Plotting the squared differences against squared epicentral distances therefore gives a straight line of slope $M=1/v_A^2$ and intercept $B=(\Delta t_0)^2$.
Using the fitted vertex from (a), an ordinary least-squares line gives
$$
M=0.0156088\,\mathrm{s^2\,km^{-2}},\qquad B=221.904\,\mathrm{s^2}.
$$
Consequently
$$
\boxed{v_A=M^{-1/2}\simeq8.0\,\mathrm{km\,s^{-1}},\quad
v_B\simeq4.0\,\mathrm{km\,s^{-1}},\quad
z_0=\sqrt{B/M}\simeq119\,\mathrm{km}.}
$$
For graph-level uncertainties, taking $v_A=8.0\pm0.2\,\mathrm{km\,s^{-1}}$ and $\Delta t_0=15.0\pm0.5\,\mathrm{s}$ gives
$$
\delta z_0\simeq\sqrt{(v_A\delta\Delta t_0)^2+(\Delta t_0\delta v_A)^2}\simeq5\,\mathrm{km}.
$$
Thus \b[a defensible graphical depth estimate is $z_0\simeq120\pm5\,\mathrm{km}$]. It is not a calibrated <confidence interval>. If each difference has independent uniform <rounding error> within $\pm0.5\,\mathrm{s}$, a direct nonlinear fit to the unsquared differences gives $x_0=140.23\,\mathrm{km}$, $\Delta t_0=14.896\,\mathrm{s}$, and $v_A=8.005\,\mathrm{km\,s^{-1}}$; the fitted depth is again $119.24\,\mathrm{km}$. A <covariance> calculation must include <correlations> between fitted slope, intercept, and vertex, rather than treating all three as independent precise measurements. Unmodelled velocity heterogeneity would add uncertainty beyond rounding.
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