On a temperature scale X,the boiling point of | vedclass

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(D) The relationship between any linear temperature scale and the Fahrenheit scale is given by the formula:$\frac{X - X_{	ext{freezing}}}{X_{	ext{bo

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$-148$

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(D) The relationship between any linear temperature scale and the Fahrenheit scale is given by the formula:$\frac{X - X_{	ext{freezing}}}{X_{	ext{boiling}} - X_{	ext{freezing}}} = \frac{F - 32}{212 - 32}$Given:$X_{	ext{boiling}} = 65^{\circ} X$$X_{	ext{freezing}} = -15^{\circ} X$$X = -95^{\circ} X$Substituting the values into the formula:$\frac{-95 - (-15)}{65 - (-15)} = \frac{F - 32}{180}$$\frac{-95 + 15}{65 + 15} = \frac{F - 32}{180}$$\frac{-80}{80} = \frac{F - 32}{180}$$-1 = \frac{F - 32}{180}$$-180 = F - 32$$F = -180 + 32 = -148^{\circ} F$

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

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