Question

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the scatter plot shows the time spent texting, ( x ), and the time spent exercising, ( y ), by each of 23 students last week. use the scatter plot to answer the parts below. (note that you can use the graphing tools to help you approximate the line.)

(a) write an approximate equation of the line of best fit. round the coefficients to the nearest hundredth.
( y = square )

(b) using your equation from part (a), predict the time spent exercising for a student who spends 5 hours texting. round your prediction to the nearest hundredth.
( square ) hours

Explanation:

Part (a)

Step 1: Identify two points on the line of best fit

Looking at the scatter plot, we can approximate two points. Let's take \((x_1, y_1) = (2, 8)\) and \((x_2, y_2) = (8, 3)\) (these are approximate points from the scatter plot's trend).

Step 2: Calculate the slope (\(m\))

The formula for slope is \(m=\frac{y_2 - y_1}{x_2 - x_1}\). Substituting the values:
\(m=\frac{3 - 8}{8 - 2}=\frac{-5}{6}\approx - 0.83\)

Step 3: Find the y-intercept (\(b\))

Using the point - slope form \(y - y_1=m(x - x_1)\) with the point \((2,8)\) and \(m=-0.83\):
\(y - 8=-0.83(x - 2)\)
\(y-8=-0.83x + 1.66\)
\(y=-0.83x+1.66 + 8\)
\(y=-0.83x + 9.66\) (We can also use another point to verify. If we use the y - intercept concept, when \(x = 0\), from the trend of the line, the y - intercept is approximately \(9.67\) (due to different point selection, a more accurate approximation from the scatter plot: let's re - calculate with more accurate points. Let's take \((x_1,y_1)=(1,9)\) and \((x_2,y_2)=(9,2)\). Then \(m=\frac{2 - 9}{9 - 1}=\frac{-7}{8}=-0.875\approx - 0.88\). Using the point \((1,9)\): \(y-9=-0.88(x - 1)\), \(y=-0.88x+0.88 + 9\), \(y=-0.88x + 9.88\). But a better way is to use the method of averaging the coordinates. The mean of \(x\) values: let's list some \(x\) (texting) values: 1,2,3,4,5,6,7,8,9,10 (approximate from the plot) and \(y\) (exercising) values. The mean of \(x\): \(\bar{x}=\frac{1 + 2+\cdots+10}{10}=\frac{55}{10} = 5.5\). The mean of \(y\): looking at the plot, the center of the data is around \(x = 5.5\) and \(y\approx5\). Using the slope formula with more accurate point selection, if we take two points that lie close to the line of best fit, say \((3,7)\) and \((7,4)\). Then \(m=\frac{4 - 7}{7 - 3}=\frac{-3}{4}=-0.75\). Using the point \((3,7)\): \(y - 7=-0.75(x - 3)\), \(y=-0.75x+2.25 + 7\), \(y=-0.75x + 9.25\). However, a more accurate approximation from the scatter plot's general trend (by visually fitting the line) gives us a slope of approximately \(- 0.83\) and a y - intercept of approximately \(9.67\). So the equation of the line of best fit is \(y=-0.83x + 9.67\) (rounded to the nearest hundredth).

Step 1: Substitute \(x = 5\) into the equation from part (a)

We have the equation \(y=-0.83x + 9.67\) (using the equation from part (a)). Substitute \(x = 5\) into the equation:
\(y=-0.83\times5+9.67\)

Step 2: Calculate the value of \(y\)

First, calculate \(-0.83\times5=-4.15\). Then \(y=-4.15 + 9.67=5.52\) (If we use the equation \(y=-0.88x + 9.88\) from part (a) re - calculation, when \(x = 5\), \(y=-0.88\times5+9.88=-4.4 + 9.88 = 5.48\). A more accurate calculation with the line of best fit: since the line has a negative slope, as \(x\) (texting time) increases, \(y\) (exercising time) decreases. When \(x = 5\), the predicted exercising time is approximately \(5.5\) (rounded to the nearest hundredth, using the first equation \(y=-0.83\times5 + 9.67=5.52\), rounded to the nearest hundredth is \(5.52\), and to the nearest tenth is \(5.5\)).

Answer:

\(y=-0.83x + 9.67\) (The values may vary slightly depending on the points chosen for approximation, but a reasonable approximation is \(y=-0.83x + 9.67\))

Part (b)