Answer:
Approximately [tex]\$ 3.02[/tex].
Explanation:
Note that the electric rate in this question is in the unit dollar-per-[tex]{\rm kWh}[/tex], where [tex]1\; {\rm kWh}[/tex] is the energy to run an appliance of power [tex]1\; {\rm kW}[/tex] for an hour.
Number of minutes for which the air conditioner is running in that day: [tex]15 \times 24 = 360\; \text{minute}[/tex]. Apply unit conversion and ensure that this time is measured in hours (same as the unit of the electric rate.)
[tex]\begin{aligned} \text{time} &= 360\; \text{minute} \times \frac{1\; \text{hour}}{60\; \text{minute}} = 6\; \text{hour} \end{aligned}[/tex].
The power of this air conditioner is:
[tex]\begin{aligned} \text{power} &= \text{voltage} \times \text{current} \\ &= 240\; {\rm V} \times 21\; {\rm A} \\ &= 5040\; {\rm W} \\ &= 5.04\; {\rm kW} \end{aligned}[/tex].
Thus, the energy that this air conditioner would consume would be:
[tex]\begin{aligned}\text{energy} &= \text{power} \times \text{time} \\ &= 5.04\; {\rm kW} \times 6\; \text{hour} \\ &= 30.24\; {\rm kWh} \end{aligned}[/tex].
At a rate of [tex]0.1[/tex] dollar-per-[tex]{\rm kWh}[/tex], the cost of that much energy would be approximately [tex]3.02[/tex] dollars (rounded to the nearest cent.)
