$\frac{\mathcal{B}(B^{+}_{c} \to K^+ K_S^0)}{\mathcal{B}(B^+\to K_S^0 \pi^+)} \times \frac{\it{f}_c}{\it{f}_u}$

ExperimentMeasurement [10-2]$\Delta\chi^2$ReferenceComments
Average$< 5.8$
LHCb$< 5.8$Phys.Lett.B 726,646 (2013)
Quoted upper limits are at 90% confidence level (CL), unless mentioned otherwise.