Let the acceleration of the solid sphere rolling down the incline without slipping be $a_{roll}$. The formula for acceleration of a body rolling down an inclined plane is given by: $a_{roll} = \frac{g \sin\theta}{1 + \frac{I}{mR^2}}$ For a solid sphere, the moment of inertia about its central axis is $I = \frac{2}{5}mR^2$. Substituting this into the formula: $a_{roll} = \frac{g \sin\theta}{1 + \frac{\frac{2}{5}mR^2}{mR^2}} = \frac{g \sin\theta}{1 + \frac{2}{5}} = \frac{g \sin\theta}{\frac{7}{5}} = \frac{5}{7}g \sin\theta$

When the solid sphere slips down the incline without rolling, it implies there is no friction to provide the torque for rolling. In this case, it behaves like a particle sliding down a smooth inclined plane. Let its acceleration be $a_{slip}$. $a_{slip} = g \sin\theta$

The ratio of their accelerations is: $\frac{a_{roll}}{a_{slip}} = \frac{\frac{5}{7}g \sin\theta}{g \sin\theta} = \frac{5}{7} = 5:7$