Answer
$ \text{it will not offer protection to temperatures as low as}:$
$$-15^{\circ} \mathrm{F}=-26.1^{\circ} \mathrm{C}$$
Work Step by Step
$\text{First We convert the Fahrenheit temperature to Celsius.}$ $$^{\circ} \mathrm{C}=\left(^{\circ} \mathrm{F}-32^{\circ} \mathrm{F}\right) \frac{\mathrm{5}^{\circ} \mathrm{C}}{\mathrm{9}^{\circ} \mathrm{F}}=\left(-15^{\circ} \mathrm{F}-32^{\circ} \mathrm{F}\right) \frac{\mathrm{5}^{\circ} \mathrm{C}}{\mathrm{9}^{\circ} \mathrm{F}}=-26^{\circ} \mathrm{C} .$$
We konw that automobile engine coolant has antifreeze which protects to $-22^{\circ} \mathrm{C}$ only.
$\therefore \text{it will not offer protection to temperatures as low as}:$ $$-15^{\circ} \mathrm{F}=-26.1^{\circ} \mathrm{C}$$
