Ryan places a bottle of water inside a cooler. As the water cools, its temperature \( C(t) \) in degrees Celsius is given by the following function, where \( t \) is the number of minutes since the bottle was placed in the cooler. \[ C(t)=4+20 e^{-0.045 t} \] Ryan wants to drink the water when it reaches a temperature of 17 degrees Celsius. How many minutes should he leave it in the coolen? Round your answer to the nearest tenth, and do not round any interimediate computations.

To find out how many minutes \( t \) Ryan should leave the bottle in the cooler for it to reach 17 degrees Celsius, we need to set the equation \( C(t) = 17 \) and solve for \( t \) in the equation: \[ 17 = 4 + 20 e^{-0.045 t} \] First, we can simplify the equation: \[ 17 - 4 = 20 e^{-0.045 t} \implies 13 = 20 e^{-0.045 t} \] Now, divide both sides by 20: \[ e^{-0.045 t} = \frac{13}{20} \] Next, take the natural logarithm of both sides: \[ -0.045 t = \ln\left(\frac{13}{20}\right) \] Solving for \( t \): \[ t = \frac{\ln\left(\frac{13}{20}\right)}{-0.045} \] Calculating \( \ln\left(\frac{13}{20}\right) \): \[ \ln\left(\frac{13}{20}\right) \approx -0.1744 \] Now substituting back into the equation for \( t \): \[ t \approx \frac{-0.1744}{-0.045} \approx 3.87 \] Rounding to the nearest tenth: \[ t \approx 3.9 \] Ryan should leave the bottle in the cooler for approximately 3.9 minutes.