Answer:
a. 3942 kWh
b. $473.04
c. 1314 kWh
d. $157.68
Step-by-step explanation:
<h3>a. </h3>
There are 1000 W in 1 kW, so 450 W = 0.450 kW. The energy used per day is ...
(0.45 kW)(24 h) = 10.8 kWh . . . . energy per day
Then in a 365-day year, the energy used is
(365 da/yr)(10.8 kWh/da) = 3942 kWh/yr
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<h3>b.</h3>
At the rate of $0.12/kWh, the cost of running the pump is ...
($0.12/kWh)(3942 kWh/yr) = $473.04/yr
__
<h3>c.</h3>
Switching the pump off for 1/3 of the time will save 1/3 of the energy found in part (a):
1/3(3942 kWh) = 1314 kWh . . . . energy saved by switching off the pump
__
<h3>d.</h3>
The savings will be 1/3 of the cost of running the pump full time:
1/3($473.04/yr) = $157.68/yr
Answer:
90%
can u plz make me brainliest?
Step-by-step explanation:
Answer:
Wait,
Step-by-step explanation:
<h2>Where are Part A and Part B? It doesn't show the questions, just the graph.</h2>
A=3b+2
a-2=3b
(a-2)/3=b
therefore b=(a-2)/3
X 3 0 4
Y 0 4 -4
Hope it helps
And don’t forget to mark as brainliest
