PDG [@pdg]. Integrating the area below these structures we obtain $$\begin{aligned}
R &=& \frac{\Gamma(D^0 \to \bar{K}^0 a_0(980), a_0(980) \to \pi^0
\eta)}{\Gamma(D^0 \to \bar{K}^0 f_0(980), f_0(980) \to \pi^+\pi^-)}
\nonumber \\
& = & 6.7 \pm 1.3 , \label{ratioth}\end{aligned}$$ where we have added a $20\%$ theoretical error due to uncertainties in the extraction of the background.

Experimentally we find from the PDG and the Refs. [@Muramatsu:2002jp; @Rubin:2004cq], $$\begin{aligned}
&& \Gamma(D^0 \to \bar{K}^0 a_0(980), a_0(980) \to \pi^0 \eta)
\nonumber \\
&&= (6.5 \pm 2.0) \times 10^{-3}, \\
&& \Gamma(D^0 \to \bar{K}^0 f_0(980), f_0(980) \to
\pi^+\pi^-)\nonumber \\
&&= (1.22^{+0.40}_{-0.24}) \times 10^{-3} .\end{aligned}$$

The ratio that one obtains from there is $$\begin{aligned}
R = 5.33^{+2.4}_{-1.9} . \label{ratioex}\end{aligned}$$

The agreement found between Eq. (\[ratioth\]) and Eq. (\[ratioex\]) is good, within errors. This is, hence, a prediction that we can do parameter free.

As we mentioned, the explicit consideration of the $\eta \eta$ channel in the meson meson interaction, required to use a cut off $q_{\rm max} = 600$ MeV [@weihong] to agree with experimental amplitudes, smaller than in Ref. [@npa] where this channel was omitted. We use the same cut off here. Yet, we want to show explicitly that the ratio obtained does not get spoiled 