| It would depend on the speed at which the woodchuck chucks wood and the woodchucking time allotted to the aforementioned woodchuck to perform his woodchucking task.
Our first approximation then would be:
W = S.T
where
W = amount of wood chucked
S = woodchucking speed of our trusty woodchuck
T = total woodchucking time.
This uses the assumption that the woodchuck chucks wood at a constant rate. However it is likely that as the woodchuck gets tired of chucking wood his woodchucking rate will decrease over time. We will assume a constant drop of woodchucking speed over time of D, ie
D = dS/dt
Therefore S will start at a maximum value Smax at t=0 and drop to zero at time t'. The total amount of chucked wood is represented by the area under the graph of S against t. The equation of the graph of S against T is given by
S = Dt + Smax
(remember that D will be a negative number)
The amount of chucked wood therefore is given by the integral of this equation between t=0 and t=t', which can be calculated from the above equation by substituting 0 for S, ie
0 = Dt' + Smax
t' = -Smax/D
Now the general integral of our equation is given by:
W = D.(t exp2) + Smax.t
so if we put our limits in:
W = D.(-Smax/D)exp2 + Smax.(-Smax/D) - 0
W = D.(-Smax exp2/D exp2) - Smax exp2/D
W = -Smax exp2/D - Smax exp2/D
W = -2Smax exp2/D
(BTW, I'm not a math head, I got that somewhere on alltheweb)
My Question
How many left handed people a year are killed from using products made for right handed people? | 
