The function \( P(x)=\frac{120}{1+372 e^{-0.133 x}} \) models the percentage, \( P(x) \), of Americans who are \( x \) years old and have some degree of heart disease. What is the percentage, to the nearest tenth, of 39 -year olds with some degree of heart disease?

To find the percentage of 39-year-olds with some degree of heart disease using the function \( P(x) = \frac{120}{1 + 372 e^{-0.133 x}} \), we need to substitute \( x = 39 \) into the function. First, we calculate \( e^{-0.133 \cdot 39} \): \[ -0.133 \cdot 39 = -5.197 \] Now, we calculate \( e^{-5.197} \): \[ e^{-5.197} \approx 0.0055 \quad (\text{using a calculator}) \] Next, we substitute this value back into the function: \[ P(39) = \frac{120}{1 + 372 \cdot 0.0055} \] Calculating \( 372 \cdot 0.0055 \): \[ 372 \cdot 0.0055 \approx 2.046 \] Now, we can substitute this back into the equation: \[ P(39) = \frac{120}{1 + 2.046} = \frac{120}{3.046} \] Calculating \( \frac{120}{3.046} \): \[ P(39) \approx 39.3 \] Thus, the percentage of 39-year-olds with some degree of heart disease, rounded to the nearest tenth, is: \[ \boxed{39.3} \]