An electric utility company determines the monthly bill by charging 85.4 cents p

An electric utility company determines the monthly bill by charging 85.4 cents per​ kilowatt-hour (kWh) used plus a base charge of ​$17.55 per month.​ Thus, the monthly charge is given by the function $f\left(W\right)=0.854W+17.55$ ​dollars, where W is the number of​ kilowatt-hours. a. Find f (6000​) and explain what it means. b. What is the monthly charge if 9500 kWh are​ used? a. Select the correct choice below and fill in the answer box to complete your choice. ​(Simplify your answers. Type integers or​ decimals.) A. $f\left(6000\right)=$ nothing. The charge for 6000 kWh is$nothing.
B. $f\left(6000\right)=$ nothing.
The charge for nothing kWh is ​$6000. b. The monthly charge is ​$nothing if 9500 kWh are used.
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Step 1
Given that
An electric utility company determines the monthly bill by charging 85.4 cents per​ kilowatt-hour (kWh) used plus a base charge of ​$17.55 per month.​ Thus, the monthly charge is given by the function $f\left(W\right)=0.854W+17.55$ ​dollars, where W is the number of​ kilowatt-hours. Step 2 To find a. f(6000). b. What is the monthly charge if 9500 kWh is used. a. To find: f(6000). The monthly charge is given by function, $f\left(W\right)=0.854W+17.55$ So to find f(6000) substitute $W=6000\in f\left(W\right)=0.854W+17.55$, $⇒f\left(6000\right)=0.854\left(6000\right)+17.55$ $⇒f\left(6000\right)=5141.55$ Therefore there is monthly charge of$5141.55 if 6000 kWh is used.
Step 3
b. To find What is the monthly charge if 9500 kWh is used.
Substitute $W=9500\in f\left(W\right)=0.854W+17.55$,
$f\left(9500\right)=0.854\left(9500\right)+17.55$
$⇒f\left(9500\right)=8130.55$
Therefore the monthly charge if 9500 kWh is used is \$8130.55.