Answer:
102.78$
Step-by-step explanation:
First, you need to multiply the consumed potence by the hours of operation per day:
![247W*19(h)/(day) =4693(W*h)/(day)](https://img.qammunity.org/2020/formulas/engineering/college/6fvfs8hwzo8hnwaltzdsry1cflif4ohlb1.png)
Now, a kilowatt is equivalent to 1,000 watts, so let's convert from watts to kilowatts:
![4693(W*h)/(day)*(1kW)/(1000W) =4.693(kWh)/(day)](https://img.qammunity.org/2020/formulas/engineering/college/275izzfcj5jio2am0ukqe2ykfnm94h0hre.png)
We have determined how many kWh the refrigerator consumes per day, now let's multiply the previous result by 365 in order to find how many kWh it consumes per year:
![4.693(kWh)/(day)*(365 days)/(1 year)=1712.945(kWh)/(year)](https://img.qammunity.org/2020/formulas/engineering/college/x5ogoxtdacrn2xrln4ik0fpk9ciz97tz41.png)
Finally let's multiply the amount of kilowatts per year by the cost of electricity:
![1712.945(kWh)/(year)*(0.06\$)/(kWh) =102.7767(\$)/(year)\approx102.78(\$)/(year)](https://img.qammunity.org/2020/formulas/engineering/college/ruukh8fwrcqgsl6lihkgpkgzorzarnjuv9.png)