Chapter 1.3, Problem 119AYU

Electricity Rates in Florida Florida Power & Light Company supplies electricity to residential customers for a monthly customer charge of $\$8.01$ plus $8.89$ cents per kilowatt hour (kWh) for up to $1000$ kilowatt hours.

Write a linear equation that relates the monthly charge $C$, in dollars, to the number $x$ of kilowatt hours used in a month, $0\le x\le 1000.$

Graph this equation.

What is the monthly charge for using $200$ kilowatt hours?

What is the monthly charge for using $500$ kilowatt hours?

Interpret the slope of the line.

To determine

A linear equation representing relation between the monthly charge $C$ in dollars, and the number $x$ of kilowatt hours used in a month $0\le x\le 1000$, ifelectricity is supplied for a monthly customer charge of $\$8.01$ plus $8.89¢$ per kilowatt hour $\left(\text{kWh}\right)$ for up to $1000$ kilowatt hours.

Answer to Problem 119AYU

Solution:

A linear equation that relates the monthly charge $C$ in dollars, to the number $x$ of kilowatt hours used in a month $0\le x\le 1000$ is $C\left(x\right)=0.0889x+8.01$

Explanation of Solution

Given Information:

Florida Power & Light Company supplies electricity to residential customers for a monthly customer charge of $\$8.01$ plus $8.89¢$ per kilowatt hour $\left(\text{kWh}\right)$ for up to $1000$ kilowatt hours.

Explanation:

First, convert $8.89¢$ in dollar.

$1\text{\$}=100\text{¢}$.

Thus, divide $8.89$ by $100$.

$\Rightarrow \frac{8.89}{100}=0.0889$

Therefore, $8.89¢$ is equal to $0.0889\$$.

Number of kilowatt hours used in a month is $x$.

Thus, the cost for $x$ kilowatt per houris $0.0889\cdot x$.

So the linear equation that relates the monthly charge $C$ in dollars, to the number $x$ of kilowatt hours used in a month $0\le x\le 1000$ is,

$C\left(x\right)=0.0889x+8.01$

To determine

To graph: The linear equation that relates the monthly charge $C$ in dollars, to the number $x$ of kilowatt hours used in a month $0\le x\le 1000$, ifelectricity is supplied for a monthly customer charge of $\$8.01$ plus $8.89¢$ per kilowatt hour $\left(\text{kWh}\right)$ for up to $1000$ kilowatt hours.

Explanation of Solution

Given Information:

Florida Power & Light Company supplies electricity to residential customers for a monthly customer charge of $\$8.01$ plus $8.89$ cents per kilowatt hour $\left(\text{kWh}\right)$ for up to $1000$ kilowatt hours.

Graph:

From part (a), the linear equation that relates the monthly charge $C$ in dollars, to the number $x$ of kilowatt hours used in a month $0\le x\le 1000$. $C\left(x\right)=0.0889x+8.01$.

The graph is as follows;

Interpretation:

The graph shows the graph of the linear equation $C\left(x\right)=0.0889x+8.01$.

To determine

The monthly charge for using $200$ kilowatt hours, ifelectricity is supplied for a monthly customer charge of $\$8.01$ plus $8.89¢$ per kilowatt hour $\left(\text{kWh}\right)$ for up to $1000$ kilowatt hours.

Answer to Problem 119AYU

Solution:

The monthly charge for using $200$ kilowatt hours is $25.79$.

Explanation of Solution

Given Information:

Florida Power & Light Company supplies electricity to residential customers for a monthly customer charge of $\$8.01$ plus $8.89$ cents per kilowatt hour $\left(\text{kWh}\right)$ for up to $1000$ kilowatt hours.

Explanation:

From part (a), the linear equation that represents the relation between the monthly charge $C$ in dollars, and the number $x$ of kilowatt hours used in a month $0\le x\le 1000$ is $C\left(x\right)=0.0889x+8.01$.

To find the monthly charge for using $200$ kilowatt hours, plug $x=200$ in linear equation $C\left(x\right)=0.0889x+8.01$.

Thus, $C\left(x\right)=0.0889\left(200\right)+8.01$

$\Rightarrow C\left(x\right)=17.78+8.01$

$\Rightarrow C\left(x\right)=25.79$

Thus, the monthly charge for using $200$ kilowatt hours is $25.79$.

To determine

The monthly charge for using $500$ kilowatt hours. If electricity is supplied for a monthly customer charge of $\$8.01$ plus $8.89¢$ per kilowatt hour $\left(\text{kWh}\right)$ for up to $1000$ kilowatt hours.

Answer to Problem 119AYU

Solution:

The monthly charge for using $500$ kilowatt hours is $52.46$.

Explanation of Solution

Given Information:

Florida Power & Light Company supplies electricity to residential customers for a monthly customer charge of $\$8.01$ plus $8.89$ cents per kilowatt hour $\left(\text{kWh}\right)$ for up to $1000$ kilowatt hours.

Explanation:

From part (a), the linear equation that represents the relation between the monthly charge $C$ in dollars and the number $x$ of kilowatt hours used in a month $0\le x\le 1000$ is $C\left(x\right)=0.0889x+8.01$.

To find the monthly charge for using $500$ kilowatt hours, plug $x=500$ in linear equation $C\left(x\right)=0.0889x+8.01$.

Thus, $C\left(x\right)=0.0889\left(500\right)+8.01$

$\Rightarrow C\left(x\right)=44.45+8.01$

$\Rightarrow C\left(x\right)=52.46$

Thus, the monthly charge for using $500$ kilowatt hours is $52.46$.

To determine

The slope of the line, ifelectricity is supplied for a monthly customer charge of $\$8.01$ plus $8.89¢$ per kilowatt hour $\left(\text{kWh}\right)$ for up to $1000$ kilowatt hours.

Answer to Problem 119AYU

Solution:

The slope of the line is $C\left(x\right)=0.0889x+8.01$ is $0.0889$.

Explanation of Solution

Given Information:

Florida Power & Light Company supplies electricity to residential customers for a monthly customer charge of $\$8.01$ plus $8.89$ cents per kilowatt hour $\left(\text{kWh}\right)$ for up to $1000$ kilowatt hours.

Explanation:

From part (a), the linear equation that represents the relation between the monthly charge $C$ in dollars and the number $x$ of kilowatt hours used in a month $0\le x\le 1000$. $C\left(x\right)=0.0889x+8.01$.

The slope point form of the equation of the line is $y=mx+c$ where $m$ is a slope and $c$ is the $y$ intercept.

The equation $C\left(x\right)=0.0889x+8.01$ is of the form $y=mx+c$.

Comparing the equation $y=0.0889x+8.01$ with $y=mx+c$,

$\Rightarrow m=0.0889$

Therefore, the slope of the line is $C\left(x\right)=0.0889x+8.01$ is $0.0889$.