Ms. Lee is shopping for a new electric water heater. Brand A costs $575 and uses 437 kilowatt-hours per month. Brand B costs $633 and uses 394 kilowatt-hours per month. If electricity costs $0.18 per kilowatt-hour, how much would Ms. Lee save on electricity

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andyzhuangzixian andyzhuangzixian · Answer 1 · 2024-01-19T03:59:21+01:00

Answer: $7.74

Step-by-step explanation:

For Brand A:

Monthly electricity cost = Electricity rate (per kilowatt-hour) × Kilowatt-hours used per month

Monthly electricity cost for Brand A = $0.18/kWh × 437 kWh

For Brand B:

Monthly electricity cost for Brand B = $0.18/kWh × 394 kWh

Now, let's calculate the monthly electricity costs for each brand:

Monthly electricity cost for Brand A = $0.18/kWh × 437 kWh = $78.66

Monthly electricity cost for Brand B = $0.18/kWh × 394 kWh = $70.92

To find the savings, we subtract the monthly electricity cost of Brand B from Brand A:

Savings = Monthly electricity cost for Brand A - Monthly electricity cost for Brand B

Savings = $78.66 - $70.92

Therefore, Ms. Lee would save $7.74 per month on electricity by choosing Brand B over Brand A.