The numbers are tractable, though. The circumference of the earth is very close to 40Mm (useful fact to know -- it was one of the original design points for the meter). At the equator (where the rotational acceleration is highest, even though coriolis itself is zero) we have a velocity of 40e6 / (24 x 60 x 60) == 463m/s. So the acceleration (v^2/r) of the ground the target is standing on relative to the free-flying bullet is 0.034 m/s^2. Taking the 3-second shot example from above, the target will move 0.5 x 0.034 x 3^2 == 15.3cm during the bullet's flight.

And yeah, a six-inch error is just about what I'd consider the maximum tolerance if you're trying to hit someone's torso. So it seems that yes, the numbers back it up.

Yay, physics is fun!