FIT INFORMATION show precise values?

 
An overall fit to and 16 branching ratios uses 28 measurements to determine 9 parameters. The overall fit has a $\chi {}^{2}$ = 32.6 for 19 degrees of freedom.
 
The following off-diagonal array elements are the correlation coefficients <$\mathit \delta $p$_{i}\delta $p$_{j}$> $/$ ($\mathit \delta $p$_{i}\cdot{}\delta $p$_{j}$), in percent, from the fit to parameters ${{\mathit p}_{{{i}}}}$, including the branching fractions, $\mathit x_{i}$ = $\Gamma _{i}$ $/$ $\Gamma _{total}$.
 
 x16 100
 x18 17 100
 x20 81 13 100
 x69 0 0 0 100
 x72 0 0 0 46 100
 x75 0 0 0 39 83 100
 x76 0 0 0 41 19 16 100
 x79 0 0 0 29 14 11 52 100
 x129 0 0 0 15 7 6 6 4 100
   x16  x18  x20  x69  x72  x75  x76  x79  x129
 
    Mode Fraction (Γi / Γ)Scale factor
Γ16  ${{\mathit B}_{{{s}}}^{0}}$ $\rightarrow$ ${{\mathit D}_{{{s}}}^{-}}{{\mathit \pi}^{+}}$ ($2.98$ $\pm0.13$) $ \times 10^{-3}$ 
Γ18  ${{\mathit B}_{{{s}}}^{0}}$ $\rightarrow$ ${{\mathit D}_{{{s}}}^{-}}{{\mathit \pi}^{+}}{{\mathit \pi}^{+}}{{\mathit \pi}^{-}}$ ($6.1$ $\pm1.0$) $ \times 10^{-3}$ 
Γ20  ${{\mathit B}_{{{s}}}^{0}}$ $\rightarrow$ ${{\mathit D}_{{{s}}}^{\mp}}{{\mathit K}^{\pm}}$ ($2.25$ $\pm0.12$) $ \times 10^{-4}$ 
Γ69  ${{\mathit B}_{{{s}}}^{0}}$ $\rightarrow$ ${{\mathit J / \psi}{(1S)}}{{\mathit \phi}}$ ($1.01$ $\pm0.04$) $ \times 10^{-3}$ 
Γ72  ${{\mathit B}_{{{s}}}^{0}}$ $\rightarrow$ ${{\mathit J / \psi}{(1S)}}{{\mathit \eta}}$ ($4.45$ $\pm0.25$) $ \times 10^{-4}$ 1.1
Γ75  ${{\mathit B}_{{{s}}}^{0}}$ $\rightarrow$ ${{\mathit J / \psi}{(1S)}}{{\mathit \eta}^{\,'}}$ ($3.53$ $\pm0.22$) $ \times 10^{-4}$ 1.1
Γ76  ${{\mathit B}_{{{s}}}^{0}}$ $\rightarrow$ ${{\mathit J / \psi}{(1S)}}{{\mathit \pi}^{+}}{{\mathit \pi}^{-}}$ ($2.00$ $\pm0.17$) $ \times 10^{-4}$ 1.7
Γ79  ${{\mathit B}_{{{s}}}^{0}}$ $\rightarrow$ ${{\mathit J / \psi}{(1S)}}{{\mathit f}_{{{0}}}{(980)}}$ , ${{\mathit f}_{{{0}}}}$ $\rightarrow$ ${{\mathit \pi}^{+}}{{\mathit \pi}^{-}}$ ($1.23$ $\pm0.15$) $ \times 10^{-4}$ 2.1
Γ129  ${{\mathit B}_{{{s}}}^{0}}$ $\rightarrow$ ${{\mathit \phi}}{{\mathit \phi}}$ ($1.83$ $\pm0.14$) $ \times 10^{-5}$