Question

heather is training for a long - distance run. her data points listed below represent the days of practice, x, and the number of miles run, y.
(1, 2.5), (2, 4.2), (4, 5.6), (6, 7), (8, 8.1), (10, 11)
use the equation to interpolate the value and estimate the distance that she could have run on day 3. round to the nearest tenth of a mile.
day 3 ≈ miles

Explanation:

Step1: Find the equation of the line of best fit

Using a graphing calculator or software (assuming linear regression), for the data points \((x_1,y_1)=(1,2.5)\), \((x_2,y_2)=(2,4.2)\), \((x_3,y_3)=(4,5.6)\), \((x_4,y_4)=(6,7)\), \((x_5,y_5)=(8,8.1)\), \((x_6,y_6)=(10,11)\).
The general form of a linear equation is \(y = mx + b\), where \(m=\frac{n\sum(xy)-\sum x\sum y}{n\sum(x^{2})-(\sum x)^{2}}\) and \(b=\frac{\sum y - m\sum x}{n}\), \(n = 6\).
\(\sum x=1 + 2+4 + 6+8 + 10=31\), \(\sum y=2.5+4.2 + 5.6+7+8.1+11 = 38.4\), \(\sum(xy)=1\times2.5+2\times4.2+4\times5.6+6\times7+8\times8.1+10\times11=2.5 + 8.4+22.4+42+64.8+110=249.1\), \(\sum(x^{2})=1^{2}+2^{2}+4^{2}+6^{2}+8^{2}+10^{2}=1 + 4+16+36+64+100=221\).
\(m=\frac{6\times249.1-31\times38.4}{6\times221 - 31^{2}}=\frac{1494.6-1190.4}{1326 - 961}=\frac{304.2}{365}\approx0.833\).
\(b=\frac{38.4-0.833\times31}{6}=\frac{38.4 - 25.823}{6}=\frac{12.577}{6}\approx2.096\).
The equation of the line of best fit is \(y = 0.833x+2.096\).

Step2: Substitute \(x = 3\) into the equation

When \(x = 3\), \(y=0.833\times3+2.096\).
\(y = 2.499+2.096=4.595\approx4.6\).

Answer:

\(4.6\)