Use the following information for questions
$\begin{array}{c}
\begin{array}{l}\
\underline{\text { Month }} \
\text { Jan } \
\text { Feb } \
\text { Mar } \
\text { Apr } \
\text { May } \
\text { Jun } \
\text { Jul } \
\text { Aug }
\end{array}
\begin{array}{ll}
\text { \# Machine } & \text { Maintenance } \
\underline{\text { Hours (X)}} & \underline{ \text { Costs(Y) }}\
3,000 & \$ 440 \
4,500 & \$ 690 \
8,000 & \$ 510 \
7,000 & \$ 600 \
6,000 & \$ 550 \
9,000 & \$ 980 \
3,500 & \$ 840 \
5,500 & \$ 600
\end{array}
\end{array}$

-What is the equation of the line using the high-low method and this data?
A)$980 = 170 + (0.09X)
B)Y = $170 + (0.09 x 9,000)
C)Y = 170+ ($0.09 X)
D)X = 170+ ($0.09 Y)

Question

Use the following information for questions 
 $\begin{array}{c}
\begin{array}{l}\
\underline{\text { Month }} \
\text { Jan } \
\text { Feb } \
\text { Mar } \
\text { Apr } \
\text { May } \
\text { Jun } \
\text { Jul } \
\text { Aug }
\end{array}
\begin{array}{ll}
\text { \# Machine } & \text { Maintenance } \
 \underline{\text { Hours (X)}} & \underline{ \text { Costs(Y) }}\
3,000 & \$ 440 \
4,500 & \$ 690 \
8,000 & \$ 510 \ 
7,000 & \$ 600 \
6,000 & \$ 550 \
9,000 & \$ 980 \
3,500 & \$ 840 \
5,500 & \$ 600 
\end{array}
\end{array}$

-What is the equation of the line using the high-low method and this data?
A)$980 = 170 + (0.09X)
B)Y = $170 + (0.09 x 9,000)
C)Y = 170+ ($0.09 X)
D)X = 170+ ($0.09 Y)

Quizplus · Accepted Answer

Based on the given information, high-low method is used to find the equation of the line. The equation calculated will be in the form of Y = a + bX, where a is the fixed cost (intercept) and b is the variable cost per unit (slope). In this case, we are given the following information:
- Highest point (H): 11eb11f1_5402_c484_ad38_9f8520f61223_TB1239_00 (assume this is Y)
- Lowest point (L): TB1239_00 (assume this is X)
- Difference in cost (D): $980 - $170 = $810
- Difference in quantity (Q): 11eb11f1_5402_c484_ad38_9f8520f61223_TB1239_00 - TB1239_00 = 9,000 units
To calculate the variable cost per unit (b), we divide the difference in cost (D) by the difference in quantity (Q):
b = D/Q = $810/9,000 = $0.09

To calculate the fixed cost (a), we can use either the highest or lowest point and substitute the known values for Y, X, and b:
a = Y - bX = $170 - ($0.09 x TB1239_00) = $170
Therefore, the equation of the line using the high-low method and this data is:
Y = $170 + ($0.09 X)

Use the Following Information for Questions -What Is the Equation of the Line Using the High-Low