In 2010, the average annual electricity consumption for a U. S. residential utility customer was 11,500 kWh, an average of nearly 960 kWh per month. If Plant Scherer were to supply 50% of its electricity to residential customers, how many residences could the plant supply with electricity if it were running at 60% average capacity for an entire year

If Plant Scherer supplies 50% of its electricity to residential customers, it could supply 849,962 residential customers per year.

Step-by-step explanation:

Plant Scherer capacity = 930 megawatts * 4 plants = 3,720 MW

3,720 MW * 1,000 = 3,720,000 kW

At 60% production, it will produce 2,232,000 kW  (3,720,000 * 60%)

50% for residential utility customers = 1,116,000 kW  (2,232,000 * 50%)

Average annual consumption per residence = 11,500 kWh

Per hour consumption = 11,500/8,760 = 1.313 kW per residence

Therefore, the number of residential customers it could supply  in a year

= 1,116,000/1.313  

= 849,962 residential customers .

choice b is correct

step-by-step explanation:

we have been given the following system of inequalities;

we are required to graph the solution to the system of inequalities. using a graphing tool, the graph of the system is as shown in the attachment. the continuous line represents the graph of while the dotted line represents .