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 linear - regression equation be $T = at + b$, where $T$ is the temperature and $t$ is the time. We have the data points $(t_1,T_1)=(5,306),(t_2,T_2)=(8,269),(t_3,T_3)=(10,236),(t_4,T_4)=(15,201),(t_5,T_5)=(20,190),(t_6,T_6)=(25,173)$.

Step2: Calculate the means of $t$ and $T$

$\bar{t}=\frac{5 + 8+10+15+20+25}{6}=\frac{83}{6}\approx13.83$
$\bar{T}=\frac{306 + 269+236+201+190+173}{6}=\frac{1375}{6}\approx229.17$

Step3: Calculate the slope $a$

$a=\frac{\sum_{i = 1}^{6}(t_i-\bar{t})(T_i - \bar{T})}{\sum_{i = 1}^{6}(t_i-\bar{t})^2}$
$\sum_{i = 1}^{6}(t_i-\bar{t})(T_i - \bar{T})=(5 - 13.83)(306-229.17)+(8 - 13.83)(269 - 229.17)+(10 - 13.83)(236-229.17)+(15 - 13.83)(201-229.17)+(20 - 13.83)(190-229.17)+(25 - 13.83)(173-229.17)$
$=-8.83\times76.83-5.83\times39.83-3.83\times6.83 + 1.17\times(-28.17)+6.17\times(-39.17)+11.17\times(-56.17)$
$=-688.41 - 232.21-26.16-32.96 - 241.68-627.42$
$=-1848.84$
$\sum_{i = 1}^{6}(t_i-\bar{t})^2=(5 - 13.83)^2+(8 - 13.83)^2+(10 - 13.83)^2+(15 - 13.83)^2+(20 - 13.83)^2+(25 - 13.83)^2$
$=(-8.83)^2+(-5.83)^2+(-3.83)^2+(1.17)^2+(6.17)^2+(11.17)^2$
$=77.97+34.09+14.66+1.37+38.07+124.77$
$=280.93$
$a=\frac{-1848.84}{280.93}\approx - 6.58$

Step4: Calculate the intercept $b$

$b=\bar{T}-a\bar{t}=229.17-(-6.58)\times13.83=229.17 + 91.96=321.13$
So the regression equation is $T=-6.58t + 321.13$

Step5: Find the time when $T = 130$

$130=-6.58t+321.13$
$6.58t=321.13 - 130$
$6.58t=191.13$
$t=\frac{191.13}{6.58}\approx29$

Answer:

29