Answer
0.1.
Work Step by Step
Use equations 30–4 and 30–6 to find the fraction left after 2.5 years, or 30 months.
$$N=N_0e^{-\lambda t}$$
$$\frac{N}{N_0}= e^{-\lambda t}=e^{-(ln\;2) t/T_{1/2}}$$
$$= e^{-(ln\;2) (30\;months)/(9\;months)}=0.0992\approx 0.1$$
$$\frac{N}{N_0}= 0.1$$
Only 1 significant figure is kept because the problem stated "about 9 months".