10. the table below shows the recorded temperature on a certain day starting at 6:00 a.m.\n| time | 6:00 |…

Explanation:

Step1: Let (x) be the number of hours since 6:00 a.m. and (y) be the temperature. Calculate the necessary sums.

Let (n = 8) (number of data - points). (\sum_{i = 1}^{n}x_{i}=0 + 0.75+1.5 + 2.5+3+3.75+4.5+5=21) (\sum_{i = 1}^{n}y_{i}=65 + 67+68+72+74+75+77+80 = 578) (\sum_{i = 1}^{n}x_{i}^{2}=0^{2}+0.75^{2}+1.5^{2}+2.5^{2}+3^{2}+3.75^{2}+4.5^{2}+5^{2}) (=0 + 0.5625+2.25+6.25+9+14.0625+20.25+25 = 77.375) (\sum_{i = 1}^{n}x_{i}y_{i}=0\times65+0.75\times67 + 1.5\times68+2.5\times72+3\times74+3.75\times75+4.5\times77+5\times80) (=0 + 50.25+102+180+222+281.25+346.5+400 = 1582)

Step2: Calculate the slope (m) of the regression line.

The formula for the slope (m) is (m=\frac{n\sum_{i = 1}^{n}x_{i}y_{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}}) Substitute (n = 8), (\sum_{i = 1}^{n}x_{i}=21), (\sum_{i = 1}^{n}y_{i}=578), (\sum_{i = 1}^{n}x_{i}^{2}=77.375), and (\sum_{i = 1}^{n}x_{i}y_{i}=1582) into the formula: [ \begin{align*} m&=\frac{8\times1582-21\times578}{8\times77.375 - 21^{2}}\ &=\frac{12656-12138}{619 - 441}\ &=\frac{518}{178}\ & = 2.91 \end{align*} ]

Step3: Calculate the y - intercept (b) of the regression line.

The formula for the y - intercept (b) is (b=\frac{\sum_{i = 1}^{n}y_{i}-m\sum_{i = 1}^{n}x_{i}}{n}) Substitute (n = 8), (m = 2.91), (\sum_{i = 1}^{n}x_{i}=21), and (\sum_{i = 1}^{n}y_{i}=578) into the formula: [ \begin{align*} b&=\frac{578-2.91\times21}{8}\ &=\frac{578 - 61.11}{8}\ &=\frac{516.89}{8}\ &=64.61 \end{align*} ]

The regression equation is (y = 2.91x+64.61)

Step4: Predict the temperature at 2:00 p.m.

Since 2:00 p.m. is 8 hours after 6:00 a.m., substitute (x = 8) into the regression equation (y = 2.91x+64.61) (y=2.91\times8 + 64.61=23.28+64.61 = 87.89\approx87.9^{\circ})

Answer:

a) (y = 2.91x+64.61) b) (87.9^{\circ})