 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 givesConsequentlyFor graph-level uncertainties, taking $v_A=8.0\pm 0.2\mathrm{km}{\mathrm{s}}^{-1}$ and $\Delta t_0=15.0\pm 0.5\mathrm{s}$ givesThus a defensible graphical depth estimate is $z_0\simeq 120\pm 5\mathrm{km}$. It is not a calibrated confidence interval. If each difference has independent uniform rounding error within $\pm 0.5\mathrm{s}$, a direct nonlinear fit to the unsquared differences gives $x_0=140.23\mathrm{km}$, $\Delta t_0=14.896\mathrm{s}$, and $v_A=8.005\mathrm{km}{\mathrm{s}}^{-1}$; 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.