On a temperature scale 'X', the boiling point of water is 65$$^\circ$$X and the freezing point is $$-15^\circ$$X. Assuming that the X scale is linear. The equivalent temperature corresponding to $$-95^\circ$$X on the Fahrenheit scale would be

We need to convert $$-95°$$X to the Fahrenheit scale, using Celsius as an intermediate.

On the X scale, the freezing point is $$-15°$$X and the boiling point is $$65°$$X, giving a range of $$65 - (-15) = 80$$ divisions. The conversion formula from X to Celsius is:

$$\frac{T_X - (-15)}{80} = \frac{T_C - 0}{100}$$

Substituting $$T_X = -95$$:

$$\frac{-95 + 15}{80} = \frac{T_C}{100}$$

$$\frac{-80}{80} = \frac{T_C}{100}$$

$$T_C = -100°\text{C}$$

Now, converting to Fahrenheit:

$$T_F = \frac{9}{5}T_C + 32 = \frac{9}{5}(-100) + 32 = -180 + 32 = -148°\text{F}$$

So, the answer is $$-148°$$F.