$\mathbf{Gauge}\mathbf{}\mathbf{pressure}\mathbf{}\mathbf{=}\mathbf{Total}\mathbf{}\mathbf{pressure}\mathbf{-}\mathbf{atmospheric}\mathbf{}\mathbf{pressure}$

$\mathbf{Gauge}\mathbf{}\mathbf{pressure}\mathbf{}\mathbf{=}\mathbf{3}\mathbf{.}\mathbf{96}\mathbf{}\mathbf{}\overline{)\mathbf{atm}}\mathbf{\times}\frac{\mathbf{14}\mathbf{.}\mathbf{7}\mathbf{}\mathbf{psi}}{\mathbf{1}\mathbf{}\overline{)\mathbf{atm}}}\mathbf{-}\mathbf{14}\mathbf{.}\mathbf{7}\mathbf{}\mathbf{psi}\mathbf{=}$**43.5 psi**

An automobile tire has a maximum rating of 38.0 psi (gauge pressure). Note: The gauge pressure is the difference between the total pressure and atmospheric pressure. In this case, assume that atmospheric pressure is 14.7 psi. The total pressure (gauge pressure + atmospheric pressure) in the tire after warming on a hot day is 3.96 atm.

Does the pressure in the tire exceed its maximum rating?

a) No, since 3.45 atm would result in a gauge reading of 36.0 psi.

b) No, since 3.96 atm would result in a gauge reading of 22.2 psi.

c) Yes, since 3.96 atm would result in a gauge reading of 58.2 psi.

d) Yes, since 3.96 atm would result in a gauge reading of 43.5 psi.

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