a model for a countrys population is given by the function of t, where t is the number of years since 1960…

a model for a countrys population is given by the function of t, where t is the number of years since 1960. find and graph the percentage rate of change of f(t) for 0 ≤ t ≤ 50.\nf(t)=0.33t + 18.1\nyear country 1 (millions) country 2 (millions)\n1960 18 39\n1970 21 54\n1980 25 68\n1990 28 83\n2000 31 98\n2010 35 113\nthe percentage rate of change is p(t)=□.

a model for a countrys population is given by the function of t, where t is the number of years since 1960. find and graph the percentage rate of change of f(t) for 0 ≤ t ≤ 50.\nf(t)=0.33t + 18.1\nyear country 1 (millions) country 2 (millions)\n1960 18 39\n1970 21 54\n1980 25 68\n1990 28 83\n2000 31 98\n2010 35 113\nthe percentage rate of change is p(t)=□.

Answer

Explanation:

Step1: Recall the formula for percentage rate of change

The formula for the percentage rate of change of a function $y = f(t)$ is $p(t)=\frac{f'(t)}{f(t)}\times100%$. First, find the derivative of $f(t)$. Given $f(t)=0.33t + 18.1$, using the power - rule for differentiation ($(ax + b)'=a$ where $a$ and $b$ are constants), we have $f'(t)=0.33$.

Step2: Substitute into the percentage - rate - of - change formula

Substitute $f'(t) = 0.33$ and $f(t)=0.33t + 18.1$ into the formula $p(t)=\frac{f'(t)}{f(t)}\times100%$. So $p(t)=\frac{0.33}{0.33t + 18.1}\times100=\frac{33}{0.33t + 18.1}$.

Answer:

$\frac{33}{0.33t + 18.1}$