the function c(t)=3000(1.195)^t where t is the number of weeks since april 19, 2020 and c(t) is the number…

Explanation:

Step1: Recall average - rate - of - change formula

The average rate of change of a function $y = f(x)$ from $x = a$ to $x = b$ is $\frac{f(b)-f(a)}{b - a}$.

Step2: Calculate $C(1)$ and $C(10)$ for the first part

For $t = 1$, $C(1)=3000(1.195)^{1}=3000\times1.195 = 3585$. For $t = 10$, $C(10)=3000(1.195)^{10}$. Using a calculator, $(1.195)^{10}\approx6.1177$, so $C(10)=3000\times6.1177 = 18353.1$. The average rate of change from $t = 1$ to $t = 10$ is $\frac{C(10)-C(1)}{10 - 1}=\frac{18353.1 - 3585}{9}=\frac{14768.1}{9}\approx1640.9$.

Step3: Calculate $C(19)$ and $C(20)$ for the second part

For $t = 19$, $C(19)=3000(1.195)^{19}$. Using a calculator, $(1.195)^{19}\approx43.777$, so $C(19)=3000\times43.777 = 131331$. For $t = 20$, $C(20)=3000(1.195)^{20}$. Using a calculator, $(1.195)^{20}\approx52.313$, so $C(20)=3000\times52.313 = 156939$. The average rate of change from $t = 19$ to $t = 20$ is $\frac{C(20)-C(19)}{20 - 19}=\frac{156939 - 131331}{1}=25608$.

Answer:

1640.9 25608