Electricity Rates in Florida. Florida Power & Light Company supplies electricity to residential customers for a monthly customer charge of $5.69 plus 8.48 cents per kilowatt-hour for up to 1000 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 ≤ x ≤ 1000. (b) Graph this equation. (c) What is the monthly charge for using 200 kilowatt-hours? (d) What is the monthly charge for using 500 kilowatt-hours?(e) Interpret the slope of the line.

a. C = 8.48x + 5.69

b. (X,C),(0,5.69),(50,429.69),Add more
points for graphing.

c. C = 8.48*200 + 5.69 = $1701.69

d. Same procedure as c.

e. Slope = 8.48

(a) To write a linear equation that relates the monthly charge C, in dollars, to the number x of kilowatt-hours used in a month, we can break it down into two parts: the fixed cost and the variable cost. The fixed cost is the customer charge of $5.69, which does not change based on the number of kilowatt-hours used. The variable cost is 8.48 cents per kilowatt-hour for up to 1000 kilowatt-hours. However, it is important to note that the variable cost only applies if the usage is up to 1000 kilowatt-hours. For usage beyond 1000 kilowatt-hours, a different rate will be charged.

Let's break down the equation into the two cases:

For 0 ≤ x ≤ 1000 kilowatt-hours:
C = $5.69 + (8.48 cents/kWh) * x

For x > 1000 kilowatt-hours:
A different equation or rate would be used; however, the details for that rate are not provided in the given information.

(b) To graph the equation, plot the number of kilowatt-hours used, x, on the x-axis and the monthly charge, C, on the y-axis. The equation from part (a) can be graphed as a straight line for the range of kilowatt-hours from 0 to 1000.

(c) To find the monthly charge for using 200 kilowatt-hours, substitute x = 200 into the equation from part (a):
C = $5.69 + (8.48 cents/kWh) * 200

(d) To find the monthly charge for using 500 kilowatt-hours, substitute x = 500 into the equation from part (a):
C = $5.69 + (8.48 cents/kWh) * 500

(e) The slope of the line represents the rate of increase in the monthly charge per kilowatt-hour used. In this case, the slope is 8.48 cents/kWh, which means that for every additional kilowatt-hour used, the monthly charge increases by 8.48 cents.

(a) The linear equation that relates the monthly charge C, in dollars, to the number x of kilowatt-hours used in a month is given by:

C = 5.69 + 0.0848x

(b) To graph this equation, we can plot the points (x, C) for different values of x within the given range, 0 ≤ x ≤ 1000. The slope-intercept form of the equation is y = mx + b, where m is the slope and b is the y-intercept. In this case, the slope is 0.0848 and the y-intercept is 5.69. The graph is a straight line with positive slope.

(c) To find the monthly charge for using 200 kilowatt-hours, we substitute x = 200 into the equation:
C = 5.69 + 0.0848(200)
C = 5.69 + 16.96
C = $22.65

The monthly charge for using 200 kilowatt-hours is $22.65.

(d) To find the monthly charge for using 500 kilowatt-hours, we substitute x = 500 into the equation:
C = 5.69 + 0.0848(500)
C = 5.69 + 42.40
C = $48.09

The monthly charge for using 500 kilowatt-hours is $48.09.

(e) The slope of the line (0.0848) in this context represents the additional cost per kilowatt-hour used. It shows how much the monthly charge increases for each additional kilowatt-hour used. In this case, for every additional kilowatt-hour used, the monthly charge increases by 8.48 cents.