#### Answer

See below for answers and explanations.

#### Work Step by Step

a. Using the given graph, we have $\lim_{t\to t_1}p(t)=20$ and $p(t_1)=20$
b. We can conclude that $p$ is continuous at $t_1$ because $\lim_{t\to t_1^-}p(t)=\lim_{t\to t_1^+}p(t)=p(t_1)$. That is, $p$ is defined at $t_1$ and the limit exists and is equal to the function value at $t_1$.
c. Using the given graph, we have $\lim_{t\to t_2}p(t)=40$ and $p(t_2)=10$
d. We can conclude that $p$ is not continuous at $t_2$ because $\lim_{t\to t_2}p(t)\ne p(t_2)$
e. We have $\lim_{t\to t_3^-}p(t)=50$, $\lim_{t\to t_3^+}p(t)=70$, $p(t_3)=70$; thus $\lim_{t\to t_3}p(t)$ does not exist.
f. We can conclude that $p$ is not continuous at $t_3$ because $\lim_{t\to t_3}p(t)$ does not exist.
g. We have $\lim_{t\to t_4^-}p(t)=100$, $p(t_4)=100$,
h. $t_4$ means the end of the course, and $100$ means the maximum percentage that can be achieved. The limit in (g) means that as time approaches the end of the course, the student learned close to $100\%$ of the content. The function value in (g) means that at the end of the course, the student learned $100\%$ of the content.