I need part 2 of this energy question

A small air compressor operates on a 2.3 hp (horsepower) electric motor for 6.8 hours a day.
How much energy is consumed by the motor daily? 1 hp equals about 750 watts.
Answer is: 42228000J

This is the part I need help with!!!

If electricity costs 22 cents per kilowatt-hour, how much does it cost to run the compressor each day? Answer in units of cent

Thank you!!

Just checking your answer:

2.3 * 750 = 1725 watts / second

multiply by 6.8 hours

(6.8 * 3600) * 1725 = 42228000 J

Excellent. On to part 2:

So, your motor runs at 1725 W (or 1.725 kW) for 6.8 hours

Look at the units. We have kilo Watts. We want kiloWatt hours. Let's multiply!

1.725 * 6.8 = 11.73 kWh

at $0.22 / kWh this comes to:

11.73 * .22 = $2.58

Checking our answer by calculating a different way:

Remember that 1 Watt is 1 Jule / 1 second. So to convert (like we did above):

1 kilo Watt hour multiplied by 3600 seconds is 3600000 J - that is it costs 22 cents for 3600000 Jules of energy.

We used 42228000 Jules in the entire day so

42228000 /3600000 * .22 = $2.58

E = 2.3 hp*750 watts *60seconds*60minutes*6.8hours = 42225800J

Hope this helps :)