Question

let v be the value (in dollars) of a companys stock at t years since 2000. some pairs of values of t and v are shown in the following table. complete parts a to d.

t (years)	v (dollars)
2	12
4	8
6	4
7	2

a. create a scattergram of the data. then draw model.
choose the correct scattergram.

a.
b.
c.
d.
(graphs of scatterplots with lines are shown for options a, b, c, d.)

Explanation:

Step1: Analyze the data trend

From the table, as \( t \) (years) increases, \( v \) (stock value) decreases: \( t = 1, v = 14 \); \( t = 2, v = 12 \); \( t = 4, v = 8 \); \( t = 6, v = 4 \); \( t = 7, v = 2 \). So the scatter points should show a negative correlation (decreasing as \( t \) increases).

Step2: Evaluate the scattergrams

- Option A: Points show an increasing trend (positive correlation) – incorrect.
- Option B: Let's check the values. For \( t = 1 \), \( v \) should be 14, but the initial point here seems too high? Wait, no, let's check the axes. Wait, the key is the trend. Wait, no, let's re - check. Wait, the data has \( t \) from 1 - 7, and \( v \) decreasing. Option B: The points seem to have a negative slope, but let's check the coordinates. Wait, the third option (C) and B. Wait, no, let's look at the values. For \( t = 1 \), \( v = 14 \); \( t = 2 \), \( v = 12 \); \( t = 4 \), \( v = 8 \); \( t = 6 \), \( v = 4 \); \( t = 7 \), \( v = 2 \). So when \( t \) increases by 1 (from 1 to 2), \( v \) decreases by 2; from 2 to 4 (increase by 2), \( v \) decreases by 4 (from 12 to 8); from 4 to 6 (increase by 2), \( v \) decreases by 4 (from 8 to 4); from 6 to 7 (increase by 1), \( v \) decreases by 2 (from 4 to 2). So the rate of decrease is consistent (slope of - 2). Now, looking at the scattergrams, Option B: Let's check the first point. If \( t = 1 \), \( v \) should be 14. In Option B, the first point is around \( t = 0 \) with high \( v \), which is incorrect. Option C: Wait, maybe the axes are labeled differently? Wait, no, the correct scattergram should have points at \( (1,14) \), \( (2,12) \), \( (4,8) \), \( (6,4) \), \( (7,2) \). Let's check the slope between \( (1,14) \) and \( (2,12) \): \( \frac{12 - 14}{2 - 1}=-2 \). Between \( (2,12) \) and \( (4,8) \): \( \frac{8 - 12}{4 - 2}=\frac{-4}{2}=-2 \). Between \( (4,8) \) and \( (6,4) \): \( \frac{4 - 8}{6 - 4}=\frac{-4}{2}=-2 \). Between \( (6,4) \) and \( (7,2) \): \( \frac{2 - 4}{7 - 6}=-2 \). So the line has a slope of - 2. Now, looking at the options, Option B: The points seem to fit? Wait, no, maybe I made a mistake. Wait, the correct scattergram should have the points plotted as \( (1,14) \), \( (2,12) \), \( (4,8) \), \( (6,4) \), \( (7,2) \). Let's check the x - axis (t) and y - axis (v) scales. In Option B, when \( t = 1 \), \( v \) is around 14? Wait, no, maybe the correct one is Option B? Wait, no, let's check Option C. Wait, the key is the trend. The data is decreasing, so we can eliminate A and D (which have increasing trends). Now between B and C. Let's check the values. For \( t = 1 \), \( v = 14 \); \( t = 2 \), \( v = 12 \); \( t = 4 \), \( v = 8 \); \( t = 6 \), \( v = 4 \); \( t = 7 \), \( v = 2 \). So the difference in \( t \) from 1 to 7 is 6, and \( v \) goes from 14 to 2, a decrease of 12, so slope \( \frac{2 - 14}{7 - 1}=\frac{-12}{6}=-2 \). Now, looking at Option B: The line has a negative slope, and the points seem to align with the data. Wait, maybe the correct answer is B? Wait, no, let's re - examine. Wait, the x - axis in Option B: when \( t = 0 \), \( v \) is around 18? But our data starts at \( t = 1 \), \( v = 14 \). Option C: The x - axis starts at 2? No, the original problem's scattergrams: Let's assume the correct one is B? Wait, no, maybe I made a mistake. Wait, the data points are \( (1,14) \), \( (2,12) \), \( (4,8) \), \( (6,4) \), \( (7,2) \). So when we plot these, as \( t \) increases, \( v \) decreases. So the scattergram with a negative slope (decreasing) is either B or C. Now, let's check the spacing. From \( t = 1 \) to \( t = 2 \) (difference 1), \( v \) decreases by 2; \( t = 2 \) to \( t = 4 \) (difference 2), \( v \) decreases by 4; \( t = 4 \) to \( t = 6 \) (difference 2), \( v \) decreases by 4; \( t = 6 \) to \( t = 7 \) (difference 1), \( v \) decreases by 2. So the rate is consistent (slope - 2). Now, looking at Option B: The points seem to have a slope of - 2. So the correct scattergram is B? Wait, no, maybe the answer is B. Wait, I think the correct answer is B.

Answer:

B.