Question

selena took a pizza out of the oven, and it started to cool to room temperature (68°f). she will serve the pizza when it reaches 130°f. she took the pizza out at 5:00 p.m. use regression to find when she can serve it. selena can serve the pizza after about minutes. (round up to the nearest whole number as needed.) time (min) 5 8 10 15 20 25 temperature (°f) 306 269 236 201 190 173

Explanation:

Step1: Assume a linear - regression model

Let the time be \(x\) (in minutes) and the temperature be \(y\) (in °F). The general form of a linear - regression equation is \(y = mx + b\), where \(m\) is the slope and \(b\) is the y - intercept. We can use the least - squares method to find \(m\) and \(b\). First, calculate the following sums from the data points \((x_i,y_i)\):
\(n = 6\) (number of data points)
\(\sum_{i = 1}^{n}x_i=5 + 8+10 + 15+20+25=83\)
\(\sum_{i = 1}^{n}y_i=306 + 269+236+201+190+173 = 1375\)
\(\sum_{i = 1}^{n}x_i^2=5^2 + 8^2+10^2 + 15^2+20^2+25^2=25 + 64+100 + 225+400+625 = 1449\)
\(\sum_{i = 1}^{n}x_iy_i=5\times306+8\times269 + 10\times236+15\times201+20\times190+25\times173=1530+2152+2360+3015+3800+4325 = 17182\)
The slope \(m=\frac{n\sum_{i = 1}^{n}x_iy_i-\sum_{i = 1}^{n}x_i\sum_{i = 1}^{n}y_i}{n\sum_{i = 1}^{n}x_i^2-(\sum_{i = 1}^{n}x_i)^2}\)
\(m=\frac{6\times17182-83\times1375}{6\times1449 - 83^2}=\frac{103092-114125}{8694 - 6889}=\frac{- 11033}{1805}\approx - 6.11\)
The y - intercept \(b=\frac{\sum_{i = 1}^{n}y_i-m\sum_{i = 1}^{n}x_i}{n}\)
\(b=\frac{1375-(-6.11)\times83}{6}=\frac{1375 + 507.13}{6}=\frac{1882.13}{6}\approx313.69\)
So the regression equation is \(y=-6.11x + 313.69\)

Step2: Solve for \(x\) when \(y = 130\)

Set \(y = 130\) in the equation \(y=-6.11x + 313.69\)
\(130=-6.11x + 313.69\)
\(6.11x=313.69 - 130\)
\(6.11x = 183.69\)
\(x=\frac{183.69}{6.11}\approx30.06\)

Answer:

30