QUESTION IMAGE
Question
the scatter plot shows the relationship between the time, in minutes, and the distance, in miles, that julie walked on several occasions.
chart: walking distances, x-axis: time (minutes) from 0 to 105, y-axis: distance (miles) from 0 to 6, with a line of best fit and data points
based on the line of best fit, which is most likely the number of miles julie would walk in 105 minutes?
options: 5, 6, 7 (partial view of options)
Step1: Analyze the line of best fit
The line of best fit in the graph shows a linear relationship between time (x - axis, minutes) and distance (y - axis, miles). We can observe the slope or the pattern of the line. For example, at \(x = 15\) minutes, the line is around \(y = 1\) mile; at \(x = 45\) minutes, \(y = 3\) miles. The rate of change (slope) can be calculated as \(\frac{\Delta y}{\Delta x}=\frac{3 - 1}{45 - 15}=\frac{2}{30}=\frac{1}{15}\) miles per minute.
Step2: Predict for 105 minutes
Using the slope, the distance \(y\) at \(x = 105\) minutes can be found by \(y - y_1=m(x - x_1)\). Taking a point on the line, say \((15,1)\), \(m=\frac{1}{15}\). So \(y - 1=\frac{1}{15}(105 - 15)\). Calculate \(105 - 15 = 90\), then \(\frac{1}{15}\times90 = 6\), so \(y=1 + 6=7\)? Wait, no, maybe better to observe the pattern. Wait, when \(x = 15\), \(y = 1\); \(x = 30\), \(y = 2\); \(x = 45\), \(y = 3\); \(x = 60\), \(y = 4\); \(x = 75\), \(y = 5\); \(x = 90\), \(y = 6\); \(x = 105\), \(y = 7\)? Wait, no, the options have 5,6,7. Wait, maybe my initial slope calculation was wrong. Wait, looking at the graph, the line at \(x = 75\) is at \(y = 5\), \(x = 90\) would be \(y = 6\), \(x = 105\) would be \(y = 7\)? Wait, but let's check the options. Wait, maybe the line at \(x = 105\) is around \(y = 7\)? Wait, no, maybe I misread. Wait, the options are 5,6,7. Wait, let's see the grid. The x - axis is 0,15,30,45,60,75,90,105. The y - axis is 0,1,2,3,4,5,6. Wait, at \(x = 75\), the line is at \(y = 5\), \(x = 90\) would be \(y = 6\), \(x = 105\) would be \(y = 7\)? But wait, the options: one of them is 7? Wait, the original problem's options (from the image) seem to have 5,6,7. Wait, maybe my first analysis was wrong. Wait, let's re - check. The line of best fit: when \(x = 15\), \(y = 1\); \(x = 30\), \(y = 2\); \(x = 45\), \(y = 3\); \(x = 60\), \(y = 4\); \(x = 75\), \(y = 5\); \(x = 90\), \(y = 6\); \(x = 105\), \(y = 7\). So the distance at 105 minutes is 7? Wait, but the options: if the options are 5,6,7, then 7 is an option. Wait, maybe I made a mistake earlier. Wait, let's count the intervals. From 15 to 105 is 90 minutes, which is 6 intervals of 15 minutes. Each 15 minutes, distance increases by 1 mile. So from 15 minutes (1 mile), 6 more intervals: 1+6 = 7 miles.
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7 (assuming the option with 7 is the correct one, like option D if labeled as D. 7)