the average monthly sales from 2006 to 2014 for a store are shown in this table.\n| year (x) | average…

Explanation:

Step1: Let (x = 0) represent 2006.

So the data - points are ((0,3700),(1,3500),(2,3900),(3,4300),(4,4000),(5,4200),(6,4300),(7,4600),(8,4600)).

Step2: Calculate the means of (x) and (y).

(\bar{x}=\frac{0 + 1+2+3+4+5+6+7+8}{9}=\frac{36}{9}=4) (\bar{y}=\frac{3700 + 3500+3900+4300+4000+4200+4300+4600+4600}{9}=\frac{37100}{9}\approx4122.22)

Step3: Calculate the slope (m) of the best - fit line.

[ \begin{align*} m&=\frac{\sum_{i = 1}^{n}(x_i-\bar{x})(y_i - \bar{y})}{\sum_{i = 1}^{n}(x_i-\bar{x})^2}\ \sum_{i = 1}^{9}(x_i - 4)(y_i-4122.22)&=(0 - 4)(3700 - 4122.22)+(1 - 4)(3500 - 4122.22)+(2 - 4)(3900 - 4122.22)+(3 - 4)(4300 - 4122.22)+(4 - 4)(4000 - 4122.22)+(5 - 4)(4200 - 4122.22)+(6 - 4)(4300 - 4122.22)+(7 - 4)(4600 - 4122.22)+(8 - 4)(4600 - 4122.22)\ &=(- 4)(-422.22)+(-3)(-622.22)+(-2)(-222.22)+(-1)(177.78)+0+(1)(77.78)+(2)(177.78)+(3)(477.78)+(4)(477.78)\ &=1688.88 + 1866.66+444.44-177.78+0 + 77.78+355.56+1433.34+1911.12\ &=7600 \end{align*} ] [ \begin{align*} \sum_{i = 1}^{9}(x_i - 4)^2&=(0 - 4)^2+(1 - 4)^2+(2 - 4)^2+(3 - 4)^2+(4 - 4)^2+(5 - 4)^2+(6 - 4)^2+(7 - 4)^2+(8 - 4)^2\ &=16 + 9+4 + 1+0+1+4+9+16\ &=60 \end{align*} ] (m=\frac{7600}{60}\approx126.67)

Step4: Calculate the (y) - intercept (b) of the best - fit line.

Using (y=mx + b) and ((\bar{x},\bar{y})), we have (4122.22=126.67\times4 + b) (b=4122.22-506.68 = 3615.54) So the best - fit line is (y = 126.67x+3615.54)

Step5: For 2018, (x = 12) (since (x = 0) is 2006).

Substitute (x = 12) into the equation: (y=126.67\times12+3615.54) (y = 1520.04+3615.54=5135.58\approx5100)

Answer:

C. $5,100