Question

let e be a colleges enrollment (in thousands of students) at t years since the college began. some pairs of values of t and e are shown in the table.

t (years) | e (thousands of students)
--- | ---
0 | 4
1 | 6
2 | 8
3 | 10
4 | 12

choose the correct linear model.
a. graph
b. graph
c. graph
d. graph

Explanation:

Step1: Analyze the table data

From the table, when \( t = 0 \), \( E = 4 \); \( t = 1 \), \( E = 6 \); \( t = 2 \), \( E = 8 \); \( t = 3 \), \( E = 10 \); \( t = 4 \), \( E = 12 \). The slope \( m=\frac{\Delta E}{\Delta t}=\frac{6 - 4}{1 - 0}=2 \). So the linear equation is \( E=2t + 4 \).

Step2: Check the graphs

• For \( t = 0 \), \( E = 4 \), so the graph should pass through \( (0,4) \).
• For \( t = 1 \), \( E = 6 \); \( t = 2 \), \( E = 8 \), etc. The slope is 2, so the line should have a steepness corresponding to slope 2.
• Looking at the options, we need to find the graph that has \( y \)-intercept 4 (since \( E \) is like \( y \)) and slope 2. The graph that matches this (passes through (0,4), (1,6), (2,8) etc. with slope 2) is the correct one. From the visual, we check the intercept and the rate of increase. The graph that starts at (0,4) and increases by 2 for each \( t \) increase of 1 is the correct linear model. (Assuming the correct graph is the one with intercept 4 and slope 2, likely one of the options with those points. Since we analyze the table, the linear model has \( E = 2t+4 \), so the graph should have \( y \)-intercept 4 and slope 2. So we identify the graph that matches this. )

Answer:

(Assuming the correct graph is the one with \( y \)-intercept 4 and slope 2, for example, if the options are as per the visual, the correct graph would be the one that passes through (0,4), (1,6) etc. So depending on the options, say if option D is the one with that, then D. But since we need to check the graphs, the key is the intercept and slope. So the correct linear model graph is the one with \( E = 2t + 4 \), so the graph that has \( t = 0 \) at \( E = 4 \) and increases by 2 each year. So the answer is the graph that matches this, e.g., if the fourth graph (D) is that, then D. )

(Note: Since the exact graph labels are not fully clear, but based on the table, the linear equation is \( E = 2t + 4 \), so the graph should have a \( y \)-intercept of 4 and a slope of 2. So the correct graph is the one that starts at (0,4) and goes up 2 for each 1 in \( t \).)