Question

the scatter plot below shows the cost (y), in dollars, of pear trees based on their ages (x), in years.
cost of pear trees
based on the scatter plot, which equation represents
the line of best fit for the cost of the pear trees?
a. \\( y = 11.8x \\)
b. \\( y = 11.8x + 29.2 \\)
c. \\( y = 15.7x \\)
d. \\( y = 15.7x + 40.0 \\)

1. the scatter plot below shows the relationship between the time, in minutes, and the distance, in miles, that janet walked on several occasions.

walking distance
based on the line of best fit, which is most likely the number of miles janet would walk in 105 minutes?
a. 4
b. 5
c. 6
d. 7

1. a scatter plot is shown below. which equation in slope - intercept form describes the line of best fit for the scatter plot?

a. \\( y = 0.37x - 4 \\)
b. \\( y = 0.37x \\)
c. \\( y = 2.7x - 4 \\)
d. \\( y = 2.7x \\)

Explanation:

Problem 1 (Pear Tree Cost Scatter Plot)

Step1: Analyze the y-intercept

The line of best fit for the pear tree cost should have a positive y - intercept (since when \(x = 0\) (age 0), the cost should be a positive value, not zero). So equations with \(y=mx\) (A and C) are less likely.

Step2: Analyze the slope

Now compare the slopes. The slope of the line of best fit from the scatter plot (visually) should be around 11.8 (not 15.7 which is steeper). So the equation \(y = 11.8x+29.2\) (option B) is the best fit.

Step1: Find the slope of the line of best fit

From the scatter plot, we can take two points on the line of best fit. For example, when \(x = 0\), \(y\approx0.5\) (approximate y - intercept), and when \(x = 90\), \(y = 6\). The slope \(m=\frac{y_2 - y_1}{x_2 - x_1}=\frac{6 - 0.5}{90-0}=\frac{5.5}{90}\approx0.061\) (wait, no, better to use the trend). Wait, actually, the line of best fit has a slope such that the rate of walking is about \(\frac{1}{15}\) miles per minute? Wait, no, let's look at the options. Wait, the question is about 105 minutes. From the scatter plot, the line of best fit passes near (0,0.5) and (90,6). The equation of the line of best fit: let's calculate the slope \(m=\frac{6 - 0.5}{90-0}=\frac{5.5}{90}\approx0.061\), and the equation \(y=mx + b\), \(b\approx0.5\). But looking at the options, when \(x = 105\), let's check the trend. The line of best fit at \(x = 90\) is at \(y = 6\), so at \(x=105\) (15 minutes more), since the slope is constant, the increase in \(y\) is about \(\frac{6 - 0.5}{90}\times15\approx0.92\), so \(y\approx6 + 0.92\approx6.92\)? Wait, no, maybe I made a mistake. Wait, the options are A.4, B.5, C.6, D.7? Wait, the original problem's options are not fully visible, but from the marked answer (the circle), it's B.5? Wait, no, let's re - evaluate. The line of best fit: when \(x = 0\), \(y\) is around 0.5, and the slope is \(\frac{5 - 0.5}{75-0}=\frac{4.5}{75}=0.06\). So for \(x = 105\), \(y=0.06\times105 + 0.5=6.3 + 0.5 = 6.8\), but maybe the intended way is to see the trend. The line of best fit at \(x = 90\) is at \(y = 6\), so at \(x = 105\) (15 minutes later), the distance should be a bit more than 6? Wait, no, maybe the options are A.4, B.5, C.6, D.7. Wait, the marked answer in the image is B (the circle is on B). So the most likely number of miles Janet would walk in 105 minutes is 5? Wait, maybe my initial analysis is wrong. Let's look at the scatter plot again. The line of best fit passes through points like (15,1), (30,2), (45,3), (60,4), (75,5), (90,6). Oh! Wait, that's a better way. So the line of best fit has a slope of \(\frac{1}{15}\) (since for every 15 minutes, distance increases by 1 mile). So the equation is \(y=\frac{1}{15}x\) (since when \(x = 15\), \(y = 1\); \(x = 30\), \(y = 2\), etc.). So when \(x = 105\), \(y=\frac{105}{15}=7\)? Wait, no, (15,1): \(1=\frac{1}{15}\times15\), (30,2): \(2=\frac{1}{15}\times30\), so the equation is \(y=\frac{1}{15}x\). Then at \(x = 105\), \(y=\frac{105}{15}=7\)? But the marked answer in the image is B (5). Wait, maybe the scatter plot's line of best fit is not passing through (15,1) exactly. Wait, the points on the scatter plot: the line of best fit is a bit below that. Let's take two points on the line of best fit: (0,0) and (90,6), slope \(m=\frac{6}{90}=\frac{1}{15}\), so \(y=\frac{1}{15}x\). At \(x = 105\), \(y = 7\). But the marked answer in the image is B (5). Wait, maybe I misread the options. The options are A.4, B.5, C.6, D.7. Wait, the original image shows that the marked option is B (5). Maybe the line of best fit is a bit less steep. Let's calculate the slope between (15,1) and (75,5): slope \(m=\frac{5 - 1}{75 - 15}=\frac{4}{60}=\frac{1}{15}\approx0.0667\). At \(x = 105\), \(y=1+\frac{1}{15}\times(105 - 15)=1+\frac{1}{15}\times90=1 + 6=7\). But the marked answer is B. Maybe the question has a typo or my analysis is wrong. Wait, the problem says "most likely the number of miles Janet would walk in 105 minutes". From the scatter plot, the line of best fit at \(x = 90\) is at \(y = 5\) (maybe), so at \(x = 105\), it's 5? No,…

Step1: Analyze the y-intercept

The scatter plot has points that when \(x = 0\), \(y\) is near 0, but wait, no, looking at the scatter plot, when \(x = 0\), \(y\) is near 0, but the line of best fit: let's calculate the slope. Take two points on the line of best fit. For example, (0, - 4) and (10,23) (no, better to take two points from the scatter plot. Let's take (2,5) and (3,9). Wait, no, the options have slopes 0.37 and 2.7. Let's calculate the slope between two points. Let's take (0, - 4) (approximate y - intercept) and (10,23) no, wait, the scatter plot: when \(x = 2\), \(y\approx5\); when \(x = 3\), \(y\approx9\). The slope \(m=\frac{9 - 5}{3 - 2}=4\)? No, that's not right. Wait, the options: A: \(y = 0.37x-4\), B: \(y = 0.37x\), C: \(y = 2.7x-4\), D: \(y = 2.7x\). Let's calculate the slope using two points on the line of best fit. Let's take (0, - 4) (if the line of best fit crosses the y - axis at - 4) and (4,7) (approximate point on the line). Then slope \(m=\frac{7-(-4)}{4 - 0}=\frac{11}{4}=2.75\approx2.7\). So the equation is \(y = 2.7x-4\) (option C), since the y - intercept is - 4 (from the line of best fit crossing the y - axis near - 4) and slope is 2.7.

Step2: Verify the slope and y-intercept

The slope of 2.7 is steeper than 0.37, and the y - intercept of - 4 matches the visual of the line of best fit (when \(x = 0\), \(y\) is near - 4). So the equation \(y = 2.7x-4\) (option C) is correct.

Answer:

B. \(y = 11.8x + 29.2\)

Problem 2 (Janet's Walking Distance)