Yes you can compute the total number of nuclei from such a graph.
Basically, if you plot # of decays per second on the Y axis, and time on the X axis, then the area under the curve (if your curve ran all the way to infinite time) would be "all the nuclei" since they all decay exactly once.
If you have just a short time segment, you can estimate the half life (for example, by plotting as a lin-log plot and fitting a straight line). When you have the half life (or the 1/e time) it is trivial to compute the total number of nuclei from the initial activity.
If we describe the activity at time $t$ as
$$A_t = A_0 e^{-t/\tau}$$
then we can find $A_0$ directly from the plot; we find $\tau$ as the slope from the lin-log plot; and the total area (total number of atoms that will eventually decay) is
$$\int_0^\infty A(t) dt = \frac{A_0}{\tau}$$
Note than in the above, $\tau$ is the 1/e time. It is related to the half life by
$$\tau = \frac{t_{\frac12}}{\ln 2}$$
What do you mean by your question:
doesn't each nucleus take an infinite amount of time to decay?
As far as I know, this is not true. A nucleus will start in one state, and end in another "decayed state" + radiation ($\alpha^{2+}, \beta^\pm, \gamma$ or whatever), and this is not an infinitely long process.
A nucleus has a probability of decaying within the next time interval, say $\delta t$, or not. Thanks to how statistics and probability work, if we have a large number of these nuclei, they will collectively exhibit a "mean lifetime" (i.e. we are able to obtain an average time it takes for one nucleus to decay).
Perhaps you're getting confused by this formula:
$$N = N_0e^{-\lambda t} = N_0e^{-t/\tau}$$
where $N$ is the number of non-decayed nuclei present in your sample, and $N_0$ is the number of initial non-decayed nuclei.
In this case, yes it takes (in theory) an infinite amount of time for $N$ to reach $0$, though this assumes $N$ can vary continuously (such as taking values like $N=0.01$, which is non-physical - $N$ can only take integer values). As $N$ and $N_0$ get larger, this equation better describes the situation.
Here, $\tau = 1/\lambda$ is in fact the mean lifetime, and is related to the half life,$\tau_{1/2}$ via
$$\tau = \frac{\tau_{1/2}}{\ln 2}$$
(from http://hyperphysics.phy-astr.gsu.edu/hbase/Nuclear/meanlif.html)
Best Answer
An example that might help:
Start with a big pile of coins. Flip them. Remove the heads. About half remain.
Take the remainder and flip them. Remove the heads. About half remain.
Take the remainder and flip them. Remove the heads. About half remain.
The analogy: An atom has a 50% chance of decaying in some particular interval $T_{1/2}$. After each of those intervals, half are left.