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

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: <br/><br/> $$\frac{X - X_{\text{freeze}}}{X_{\text{boil}} - X_{\text{freeze}}} = \frac{C - C_{\text{freeze}}}{C_{\text{boil}} - C_{\text{freeze}}}$$ <br/><br/> 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. <br/><br/> We are given the following values: <br/><br/> X-scale:<br/><br/> $X_{\text{boil}} = 65^{\circ} \mathrm{X}$<br/><br/> $X_{\text{freeze}} = -15^{\circ} \mathrm{X}$ <br/><br/> Celsius scale:<br/><br/> $C_{\text{boil}} = 100^{\circ} \mathrm{C}$<br/><br/> $C_{\text{freeze}} = 0^{\circ} \mathrm{C}$ <br/><br/> Our goal is to find the equivalent temperature of $-95^{\circ} \mathrm{X}$ on the Fahrenheit scale. <br/><br/> Step 1: Convert $-95^{\circ} \mathrm{X}$ to Celsius: <br/><br/> Use the relationship between X-scale and Celsius scale: <br/><br/> $\frac{-95 - (-15)}{65 - (-15)} = \frac{C - 0}{100 - 0}$ <br/><br/> Simplify and solve for C: <br/><br/> $\frac{-80}{80} = \frac{C}{100}$ $C = -100^{\circ} \mathrm{C}$ <br/><br/> Step 2: Convert $-100^{\circ} \mathrm{C}$ to Fahrenheit: <br/><br/> Use the conversion formula between Celsius and Fahrenheit: <br/><br/> $T_F = \frac{9}{5}T_C + 32$ <br/><br/> Substitute the Celsius temperature: <br/><br/> $T_F = \frac{9}{5} \times (-100) + 32$<br/><br/> $T_F = -180 + 32$<br/><br/> $T_F = -148^{\circ} \mathrm{F}$ <br/><br/> So, the equivalent temperature corresponding to $-95^{\circ} \mathrm{X}$ on the Fahrenheit scale is $-148^{\circ} \mathrm{F}$.