Q. 119 - Precalculus Enhanced with Graphing Utilities

Question

Commonwealth Edison Company supplies electricity to residential customers for a monthly customer charge of $11.47 plus 11 cents per kilowatt-hour for up to 600 kilowatt-hours.

(a) Write a linear equation that relates the monthly charge C, in dollars, to the number x of kilowatt-hours used in a month, $0\le x\le 600$

(b) Graph this equation.

(c) What is the monthly charge for using 200 kilowatthours?

(d) What is the monthly charge for using 500 kilowatthours?

(e) Interpret the slope of the line.

Step-by-Step Solution

Answer

Part (a)The linear equation that relates the monthly charge C, in dollars, to the number x of kilowatt hours used in a month is: $C=0.11x+11.47 , 0\le x\le 600$

Part (b) Graph of the equation.

Part (c) Monthly charge for using 200 kilowatthours is: $\$33.47$

Part (d) Monthly charge for using 500 kilowatthours is: $\$66.47$

Part (e) Each additional kW-hr used adds $\$0.11$ to the bill.

Part (a) Step 1. Given information

Commonwealth Edison Company supplies electricity to residential customers for a monthly customer charge of $11.47 plus 11 cents per kilowatt-hour for up to 600 kilowatt-hours.

Part (a) Step 2. Finding the linear equation

The required equation is: $C=0.11x+11.47$, $0\le x\le 600$

Part (b) Step 1. Graphing the equation

Graph of the equation: $C=0.11x+11.47$, $0\le x\le 600$

Part (c) Step 1. Finding the monthly charge for using 200 kilowatthours

Monthly charge for using 200 kilowatthours:

Put $x=200$ in the above linear equation

$C=0.11\times 200+11.47$

$C=22+11.47$

$C=33.47$

Part (d) Step 1. Finding the monthly charge for using 500 kilowatthours

Monthly charge for using 500 kilowatthours:

Put $x=500$ in the above linear equation

$$\begin{gathered} C=0.11\times 500+11.47 \\ C=55+11.47 \\ C=66.47 \end{gathered}$$

Part (e) Step 1. Interpreting slope of the line

Each additional kW-hr used adds $\$0.11$ to the bill.