Question

1. margie heated a beaker of water in science class. the scatter plot below shows the temperature (y), in degrees celsius (°c), of water based on the number of minutes (x) she heated the water. temperatures of water in beaker which equation describes the line of best fit for the temperature of the water based on the number of minutes margie heated the water? a. $y = 5.3x + 12$ b. $y = 5.3x + 23$ c. $y = -5.3x + 23$ d. $y = -5.3x + 50$ 11. malia went on a bicycle trip. the scatter plot below shows the distance she had traveled for various lengths of time during the bicycle trip. which equation describes the line of best fit for the scatter plot? malia’s bicycle trip a. $d = \frac{11}{50}m + \frac{13}{20}$ b. $d = \frac{50}{11}m + \frac{13}{20}$ c. $d = \frac{11}{50}m + 50$ d. $d = \frac{50}{11}m + 50$ 12. henry counted the number of scratches on cars of different ages. henry then created the scatter plot below showing the relationship between the age of the car (x) and the number of scratches found on the car (y). car scratches based on the line of best fit, what is most likely the number of scratches that could be found on a car that is 30 years old? a. 30 b. 74 c. 87 d. 100

Explanation:

First Sub - Question (Margie's Water Heating)

Step1: Analyze the slope

The scatter plot shows a positive correlation (as \(x\) increases, \(y\) increases), so the slope should be positive. Eliminate options C and D (which have negative slopes).

Step2: Analyze the y - intercept

When \(x = 0\) (time = 0 minutes), the temperature (y - intercept) is around 20 - 23. Option A has a y - intercept of 12, which is too low. Option B has a y - intercept of 23, which is more consistent with the graph.

Step1: Analyze the slope

The scatter plot shows a positive correlation. To estimate the slope, we can use two points. Looking at the graph, when \(m = 0\), \(d\) is around 0 - 1 (close to \(\frac{13}{20}=0.65\)). When \(m = 200\), \(d\approx45\). The slope \(m=\frac{\Delta d}{\Delta m}\). Let's check the options. Option B has a slope of \(\frac{50}{11}\approx4.55\), which is too steep. Option A has a slope of \(\frac{11}{50} = 0.22\), too flat. Wait, maybe a better way: when \(m = 0\), the y - intercept (d - intercept) is around \(\frac{13}{20}\) (0.65) or 0 - 1. Option C and D have y - intercepts of 50, which is way too high. Now, for the slope, if we take two points, say (0, \(\frac{13}{20}\)) and (200, 45). The slope \(=\frac{45-\frac{13}{20}}{200 - 0}\approx\frac{45}{200}=\frac{9}{40}=0.225\), but \(\frac{50}{11}\approx4.5\) is wrong. Wait, maybe I made a mistake. Wait, the options: A: \(d=\frac{11}{50}m+\frac{13}{20}\), B: \(d = \frac{50}{11}m+\frac{13}{20}\), C: \(d=\frac{11}{50}m + 50\), D: \(d=\frac{50}{11}m+50\). The y - intercept when \(m = 0\) should be small (around 0 - 1), so C and D are out. Now, the slope: if we look at the graph, as \(m\) (time) increases, \(d\) (distance) increases. Let's take two points: (0, \(\frac{13}{20}\)) and (40, 10). The slope \(=\frac{10-\frac{13}{20}}{40-0}=\frac{\frac{200 - 13}{20}}{40}=\frac{187}{20\times40}=\frac{187}{800}\approx0.23\), which is close to \(\frac{11}{50}=0.22\). Wait, but \(\frac{50}{11}\) is about 4.5, which would mean at \(m = 11\), \(d=\frac{50}{11}\times11+\frac{13}{20}=50 + 0.65 = 50.65\), which is too high. So the correct answer should be A? Wait, no, maybe I messed up. Wait, the graph: when \(m = 0\), distance is around 5 - 10? Wait, the y - axis is distance (miles), x - axis is time (minutes). At \(m = 0\), the first point is around 5 miles? Wait, the first point is at (0, 5) maybe? Wait, the graph shows at \(m = 0\), \(d\) is around 5? Wait, the user's graph: "Distance (miles)" on y - axis, "Time (minutes)" on x - axis. The first point is at (0, 5) maybe? Wait, the options: A: \(d=\frac{11}{50}m+\frac{13}{20}\) (\(\frac{13}{20}=0.65\)), B: \(d=\frac{50}{11}m+\frac{13}{20}\), C: \(d=\frac{11}{50}m + 50\), D: \(d=\frac{50}{11}m+50\). If at \(m = 0\), \(d\) is around 5, then C and D have \(d = 50\) at \(m = 0\), which is wrong. A has \(d=\frac{13}{20}\approx0.65\) at \(m = 0\), wrong. B has \(d=\frac{13}{20}\approx0.65\) at \(m = 0\), wrong. Wait, maybe the graph is misread. Alternatively, maybe the correct answer is B. Wait, I think I made a mistake in the first analysis. Let's try again. The slope of the line of best fit: if we take two points, say (0, \(\frac{13}{20}\)) and (200, 45). The slope is \(\frac{45-\frac{13}{20}}{200}=\frac{\frac{900 - 13}{20}}{200}=\frac{887}{4000}\approx0.22\), which is \(\frac{11}{50}=0.22\). But the y - intercept: when \(m = 0\), the distance is around 0 - 1, so \(\frac{13}{20}\) is 0.65, which is reasonable. But when \(m = 200\), \(d=\frac{11}{50}\times200+\frac{13}{20}=44 + 0.65 = 44.65\), which is close to 45. So option A. But wait, the other option B: \(d=\frac{50}{11}m+\frac{13}{20}\), when \(m = 11\), \(d = 50+\frac{13}{20}=50.65\), which is too high. So the correct answer is A?

Step1: Analyze the line of best fit

The line of best fit has a positive slope. We need to find the number of scratches for a 30 - year - old car. Looking at the graph, the line of best fit at \(x = 30\) (age = 30 years) is around 87. Option A (30) is too low, B (74) is low, D (100) is too high.

Answer:

B. \(y = 5.3x+23\)

Second Sub - Question (Malia's Bicycle Trip)