PhysicsHard251×since 2002Q7524On a temperature scale 'X\mathrm{X}X', the boiling point of water is 65∘X65^{\circ} \mathrm{X}65∘X and the freezing point is −15∘X-15^{\circ} \mathrm{X}−15∘X. Assume that the X\mathrm{X}X scale is linear. The equivalent temperature corresponding to −95∘X-95^{\circ} \mathrm{X}−95∘X on the Farenheit scale would be:A−148∘F-148^{\circ} \mathrm{F}−148∘FB−48∘F-48^{\circ} \mathrm{F}−48∘FC−63∘F-63^{\circ} \mathrm{F}−63∘FD−112∘F-112^{\circ} \mathrm{F}−112∘FCheck answerSkip