Answer: -3.212 \#__Y You take your cup of coffee outside on a cold day. The temperature, in ${ }^{\circ} \mathrm{C}$, of the cup of coffee is modeled by the function $c(t)=-0.01 t^{3}+75$, where $t$ is measured in minutes. How long does it take for the temperature of the coffee to decrease from $70^{\circ} \mathrm{C}$ to $50^{\circ} \mathrm{C}$ ?

Step 1 ：Solve for \(t\) when \(c(t) = 70\): \(-0.01t^3 + 75 = 70\) which simplifies to \(-0.01t^3 = -5\). Solving for \(t\) gives \(t = (500)^{1/3}\) which is approximately \(7.94\) minutes.

Step 2 ：Solve for \(t\) when \(c(t) = 50\): \(-0.01t^3 + 75 = 50\) which simplifies to \(-0.01t^3 = -25\). Solving for \(t\) gives \(t = (2500)^{1/3}\) which is approximately \(13.57\) minutes.

Step 3 ：Subtract the time it takes for the coffee to cool to 70°C from the time it takes to cool to 50°C: \(13.57 - 7.94 = 5.63\) minutes.

Step 4 ：\(\boxed{5.63}\) minutes is the time it takes for the temperature of the coffee to decrease from 70°C to 50°C.