We are given two temperature scales: the X-scale and the Celsius scale. The relationship between two linear temperature scales can be expressed as follows:

$$\frac{X - X_{\text{freeze}}}{X_{\text{boil}} - X_{\text{freeze}}} = \frac{C - C_{\text{freeze}}}{C_{\text{boil}} - C_{\text{freeze}}}$$

Here, $$X_{\text{freeze}}$$ and $$X_{\text{boil}}$$ are the freezing and boiling points of water on the X-scale, while $$C_{\text{freeze}}$$ and $$C_{\text{boil}}$$ are the freezing and boiling points of water on the Celsius scale.

We are given the following values:

X-scale:

$$X_{\text{boil}} = 65^{\circ} \mathrm{X}$$

$$X_{\text{freeze}} = -15^{\circ} \mathrm{X}$$

Celsius scale:

$$C_{\text{boil}} = 100^{\circ} \mathrm{C}$$

$$C_{\text{freeze}} = 0^{\circ} \mathrm{C}$$

Our goal is to find the equivalent temperature of $$-95^{\circ} \mathrm{X}$$ on the Fahrenheit scale.

Step 1: Convert $$-95^{\circ} \mathrm{X}$$ to Celsius:

Use the relationship between X-scale and Celsius scale:

$$\frac{-95 - (-15)}{65 - (-15)} = \frac{C - 0}{100 - 0}$$

Simplify and solve for C:

$$\frac{-80}{80} = \frac{C}{100}$$ $$C = -100^{\circ} \mathrm{C}$$

Step 2: Convert $$-100^{\circ} \mathrm{C}$$ to Fahrenheit:

Use the conversion formula between Celsius and Fahrenheit:

$$T_F = \frac{9}{5}T_C + 32$$

Substitute the Celsius temperature:

$$T_F = \frac{9}{5} \times (-100) + 32$$

$$T_F = -180 + 32$$

$$T_F = -148^{\circ} \mathrm{F}$$

So, the equivalent temperature corresponding to $$-95^{\circ} \mathrm{X}$$ on the Fahrenheit scale is $$-148^{\circ} \mathrm{F}$$.