Question

week 1 2 3 4 5 6 7 8 laps 24 27 31 38 39 42 47 48 write the equation for the best - fit line for the data. round to the nearest hundredth, if necessary. y = x + part b incorrect 2 tries left please try again how many laps should keon expect to swim in week 12? about laps

Explanation:

Step1: Calculate the slope \(m\)

The formula for the slope \(m\) of the best - fit line is \(m=\frac{n\sum_{i = 1}^{n}x_iy_i-\sum_{i = 1}^{n}x_i\sum_{i = 1}^{n}y_i}{n\sum_{i = 1}^{n}x_i^{2}-(\sum_{i = 1}^{n}x_i)^{2}}\), where \(n = 8\) (number of data points), \(x_i\) are the week numbers (\(x_1 = 1,x_2=2,\cdots,x_8 = 8\)) and \(y_i\) are the lap numbers (\(y_1 = 24,y_2=27,\cdots,y_8 = 48\)).

\(\sum_{i=1}^{8}x_i=1 + 2+3+4+5+6+7+8=\frac{8\times(8 + 1)}{2}=36\)

\(\sum_{i=1}^{8}y_i=24 + 27+31+38+39+42+47+48=296\)

\(\sum_{i=1}^{8}x_iy_i=1\times24+2\times27+3\times31+4\times38+5\times39+6\times42+7\times47+8\times48\)
\(=24+54+93+152+195+252+329+384=1483\)

\(\sum_{i=1}^{8}x_i^{2}=1^{2}+2^{2}+3^{2}+4^{2}+5^{2}+6^{2}+7^{2}+8^{2}\)
\(=1 + 4+9+16+25+36+49+64=204\)

\(m=\frac{8\times1483-36\times296}{8\times204 - 36^{2}}=\frac{11864-10656}{1632 - 1296}=\frac{1208}{336}\approx3.59\)

Step2: Calculate the y - intercept \(b\)

The formula for the y - intercept \(b\) is \(b=\overline{y}-m\overline{x}\), where \(\overline{x}=\frac{\sum_{i = 1}^{n}x_i}{n}=\frac{36}{8}=4.5\) and \(\overline{y}=\frac{\sum_{i = 1}^{n}y_i}{n}=\frac{296}{8}=37\)

\(b = 37-3.59\times4.5=37 - 16.155=20.845\approx20.85\)

The equation of the best - fit line is \(y = 3.59x+20.85\)

Step3: Predict for \(x = 12\)

Substitute \(x = 12\) into the equation \(y=3.59x + 20.85\)

\(y=3.59\times12+20.85=43.08+20.85=63.93\approx64\)

Answer:

The equation of the best - fit line is \(y = 3.59x+20.85\). Keon should expect to swim about \(64\) laps in week \(12\).