Problem 44 - Algebra and Trigonometry

Question

The logistic growth function $$ P(x)=\frac{90}{1+271 e^{-0.122 x}} $$ models the percentage, \(P(x),\) of Americans who are \(x\) years old with some coronary heart disease. Use the function to solve Exercises \(43-46\) What percentage of 80 -year-olds have some coronary heart disease?

Step-by-Step Solution

Answer

Approximately 52.94% of 80-year-olds in the United States have coronary heart disease according to the given logistic growth model.

Step 1: Understand the Problem

We need to find out the percentage of 80-year-olds who have coronary heart disease. This can be determined by substituting \(x = 80\) into the given logistic growth function \(P(x)=\frac{90}{1+271 e^{-0.122 x}}\).

Step 2: Substitute the Value

Let's substitute the value of \(x\) into the given function: \(P(80)=\frac{90}{1+271 e^{-0.122 \cdot 80}}\).

Step 3: Simplify the Expression

Now, simplify the expression inside the exponential function: \(P(80)=\frac{90}{1+271 e^{-9.76}}\) and further to: \(P(80)=\frac{90}{1+271 e^{-0.0000058}}\).

Step 4: Calculate the Final Result

Now, perform the calculations \(P(80)=\frac{90}{1+271 \cdot (0.0000058)} = 52.94\% \) approximately.