what is the meanig of cepalco
1. what is the meanig of cepalco
Cagayan Electric Power and Light Company, Inc. (CEPALCO) is among the prime movers in the Philippine electric utility industry having been operating in the country for six decades, since 1952.
2. how much electricity was consume by a household if the rate is php10.00 and they have paid php 750.00? A. kWh B. 25 kWh C. 50 kWh D. 75 kWh
Answer:
Letter D. 75KwhStep-by-step explanation:
Divide 750php to 10 phpThe answer is 75 Kwhwhy. because 10php for 1 Kwh
If you divide 750php to 10 means.
the answer is 75 Kwh
3. the cost of electric consumption is computed by multiplying the consumption (in kwh) by the per kwh rate. if a household has consumed 8 kwh while the rate is P12.00, how much does the cost a.96.00 b.P106.00 c.P 116.00 d. P 256.00
Answer:
the answer is A.96
00
Step-by-step explanation:
JUST LIKE MATH TAKE THE 12 THEN × IT TO THE 8KWH
4. 2. Mr. Rickman recorded their electric consumption for four months as follows: April - 0345 kwh; May - 0412 kwh; June - 0478 kwh; July - 0560 kwh. The initial reading 0235 kwh.What was the average monthly electric consumption of Mr. Rickman?*A. 81.25 kwhB. 71.25 kwhC. 61.25 kwhD. 51.25 kwh
Answer:
61.25kwh
Step-by-step explanation:
pa brainleast
5. 21. Anne consumed 250 kwh of electricity in March. If the rate is P 6.50 per kwh. How much will be the electric bill for the month of March? A. 1625 B. 1630 C. 1630.50 D.P 1640
Answer: A. 1 625
Step-by-step explanation: 250 × 6.50 = 1625
6. June 13, Use figure above in answering number 1-5. D.1548 kwh C.1538 kwh Date Signed A 1) What is the reading shown on figure A? 1458 kwh B. 1448 kwh 2)) What is the reading shown on figure B? B. 6602 kwh A. 5602 kwh C.6502 kwh D.6702 kwh 3)) What is the reading shown on figure C? A. 7050 kwh B. 7150 kwh C.7250 kwh D.7251 kwh 4.) What is the reading shown on figure D? A 8812 kwh B. 8712 kwh C.7912 kwh D.7812 kwh 5.) What is the reading shown on figure E? A. 6535 kwh B. 6553 kwh C.6435 kwh D.6453 kwh 6. What is the electric consumption if the Present Reading = 4 185 and the Previous Reading = 4100? A. 85 kwh B. 95 kwh C. 105 kwh D. 185 kwh 7. Compute the electric consumption if the Present Reading = 6 437 and the Previous Reading = 6137? A. 300 kwh B.310 kwh C. 320 kwh D.330 kwh 8) Solve the electric consumption if the Present Reading = 5 987and the Previous Reading = 5842? A. 145 kwh B. 155 kwh C. 165 kwh D. 175 kwh A. 20 kwh 9) A DVD player has a power rating of 30 watts. If the DVD player is used for six hours, what is the Electric consumption Number of watts x Tim electric power consumption? 1000 10) Find the cost of electricity if the electrical consumption is 30 kilowatt-hours and the cost per ilowatt-hour is Php. 8.50. B..18 kwh C..16 kwh D..15 kwh A PhP 215 B. PhP 225 C. PhP 325 D. PhP 335 11) It used to show the relation of the parts to the whole and to each other. A. Circle Graph B. Line graph C. Pictograph D. Bar Graph 12The circle represents the entire quantity, the whole unit or A. 80% B. 90% C. 100% D. 200% 13) A one complete revolution to a circle B. 200 degrees is equal to A. 100 degrees C. 360 degrees D. 480 degrees
Answer:
1.C.1538
2.A.5602
3.D.7251
4.A.8812
5.B.6553
7. 2.) Mrs. Reyes consumed 155 kwh of electricity in May. If the rate is P5.50 per kwh, how much will be the electric bill for the month of May?
Answer:
just multiply 155 in P5.50 and the answer will be 852.5 just analize the question po
8. 19. An electric fan has a power rating of 65 W. If it is used for3 h, what is the electric consumption?A.0.195 kWhC. 1.5 kWhb. 1.2 kWhd. 1.915 kWh
Answer:
c 1.5kwh
Step-by-step explanation:
sana maka tulong
9. For numbers 16-20 Fill in the chart with your answers. month January- Present Reading 5706 kwh Previous Reading 5438 kwh kwh Used 16._______month febuary- Present 5850 kWh previous 17._______ kwh used 144 kWhmonth March- present 18._______ previous 5850 kWh, kwh used 201 kwhmonth April- present 6298 kWh previous 19._______kwh used 247 kWh month May- present 6494 kWh, previous 6298 kWh, kwh used 20.______
ANSWER:
16. 268 kWh
17. 5,706 kWh
18. 6,051 kWh
19. 6,051 kWh
20. 196 kWh
#CarryonLearning10. How long would it take an appliance rated at 3kW draw 72 kWh of electrical energy?
Answer:
multiply the power rating (watts) of the appliance by the amount of time (hrs) you use the appliance and divide by 1000.
11. Bornok’s electric meter reading last may 16 was 5632 kWh. The meter reading today (june 16) is 5674 kWh. The rate of electricity is 1.15pesos per kWh. How much will he be paying for the month?
Answer:
Given:
x=5632
y=5674
r=1.15
Solution:
(y-x)r
(5674-5632)1.15
(42)1.15
=P48.3 will Bornok's paying for the month.
Step-by-step explanation:
Hope it helps
12. 2. Marites wanted to know how much electricity her family consumes in a day. The meter reads 2 756 kWh. After 24 hours, it reads 2 765 kWh. At this rate, what will be their electric consumption in 30 days?
Answer:
To find out their daily electric consumption, we need to subtract the initial reading from the final reading and then divide the result by the number of days.
Daily electric consumption = (2 765 kWh - 2 756 kWh) ÷ 1 day = 9 kWh/day
Then, to calculate their electric consumption for 30 days, we can simply multiply their daily consumption by the number of days:
Electric consumption for 30 days = 9 kWh/day x 30 days = 270 kWh
Therefore, Marites' family is expected to consume 270 kWh of electricity in 30 days, assuming their usage remains consistent.
13. An electric fan has a power rating of 60 W. If it is used for 3 h, what isthe electric consumption? O 0. 195kWhO 1.2 kWhO 1.5 kWhO 1. 915kWh
Answer:
0.120
Explaination:
60 ÷ 1000 = 0.06
0.06 × 3 = 0.12
14. the new rate per kilowatt hour of electricity is 0.125. What is the resulting rate if the original rate per kWh was is 7.95? a.80.75 b.807.50 c.8075 d.8.075
Answer: B. 807.50Step-by-step explanation:#CarryOnLearning
15. what is the electric consumption for the month if the previous reading is 03487 kwh and the present reading is 03826 kwhincomplete, nonsense answer and no solution=reportcorrect, complete answer and have solution= auto brainliest heart five star rate
Answer:
03826-03487=339
Step-by-step explanation:
yan po, iminus mo ang present sa previous. thanks
16. The present reading is 8743 kWh while the previous reading is 8691 kWh. What is the electric consumption? a. 55 kwh b. 51 kwh c. 56 kwh d. 52 kwh
Answer:
d.
52 kwh
Step-by-step explanation:
17. How long would it take an appliance rated at 2kW draw 48 kWh of electrical energy?
Answer:
thanks sa points
Explanation:
18. mrs. ocampo's family consumed 450 kwh of electricity in april. if the rate is ph.5.50 per kwh, how much will be the electric bill for the month of april?
Answer:
PhP 2,475
Step-by-step explanation:
electric bill = consumption x rate = 450 kwh * 5.50 pesos / kwh = 2,475
19. 180 kWh 60 kWh Score: Choose from the choices below the correct answer of the following: 664 kWh 250 kwh 260 kWh 179 kWh 662 kwh Look at the electric bill below, use the information to answer the questions. JSG COMPANY DATE METER READING January 980 kwh February 1040 kwh March 1230 kwh April 1410 kwh May 1642 kwhpasagot Naman po kailangan na ngayon
HENCE, THE ANSWER IS
(1). 60 KWH
(2). 250 KWH
(3). 180 KWH
(4). 662 KWH
Step-by-step explanation:
THE SOLUTION IS IN THE PICTUREhope it helps
20. the cost of electric consu mption is computed by multiplying the consumption (in kwh)by the per kwh rate.if a household has consumed 8 kwh while the rate is P12.00,how much does the cost A.P96.00 B.P106.00 C.P116.00 D.P256.00
Answer:
A. 96
Step-by-step explanation:
the consumed electric (8) multiply the rate per electric (12) equal to 96
21. Analyze and solve. (Please show your solution) 1.Last month, my electric meter reading was 43 411 kWh. This month, the reading is 43 619 kWh How many kWh did I consume? 2 On March 28, 2021. an electric meter reads 1029 kWh and on April 28, 2021, the meter reads 1265 kWh. How much is the amount of electricity consumed if the rate is P9.42?
Answer:
25kwh
34kwh
I HOPE HELP THIS TO YOU
22. 10. Pagsunod-sunurin ang bilang ng mga kilowat na nagamit A. 135 kwh,130 kwh, 140 kwh, 120 kwh, 110 kwh, 100 kwh B. 140 kwh 135 kwh 130 kwh 120 kwh, 110 kwh, 100 kwh C. 100 kwh, 110 kwh, 120 kwh 130 kwh 135 kwh, 140 kwh mula sa pinakamataas hanggang sa pinakakaunting nagamit9. ano ang datos sa pahabang asks?wag sasagot pag hindi alam.
Answer:
D
Explanation:
10. Pagsunod-sunurin ang bilang ng mga kilowat na nagamit A. 135 kwh,130 kwh, 140 kwh, 120 kwh, 110 kwh, 100 kwh B. 140 kwh 135 kwh 130 kwh 120 kwh, 110 kwh, 100 kwh C. 100 kwh, 110 kwh, 120 kwh 130 kwh 135 kwh, 140 kwh mula sa pinakamataas hanggang sa pinakakaunting nagamit
23. What is the power rating of a space heater that uses 1.2 kWh of energy in 45 min?
Answer: 0.026 kWh
Explanation: The quantity that has to do with the rate at which a certain amount of work is done is known as power. Power is the rate at which work is done. It is the work/time ratio. Mathematically, it is computed using the following equation. Power = Work / time or P = W / t
24. A 10 hp motor runs at rated load for 5 hours. how many kwh is consumed
A 10 hp motor runs at rated load for 5 hours. how many kwh is consumed
25. A flat iron operated 0.75 hr and used 3 kWh of electrical energy. What is the power rating of the flat iron?
Answer:
5 kWh of electrical energy iron
Explanation:
hope it's help
26. SITUATION: The Rodriguez residence's electric meter reading last month was 4445 kWh. This month, their meter is 4697 kWh. How much kWh of electricity were consumed and how much will they pay if the energy rate per kWh is Php 8.50? What are the present reading, previous reading, kWh used, and the Total Amount due? With solution po.
Answer:
PRESENT READING - 4697 kWh
PREVIOUS READING - 4445 kWh
kWh USED - 252 kWh
TOTAL AMOUNT DUE - Php 2,142.00
To get the kWh use we must subtract the previous reading to present reading.4697kWh - 4445kWh = 252kWhWe get the kWh used which is 252kWhTo find the total amount due multiply the kWh used to the energy rate.252 × 8.50 = 2,142We get Php 2,142THEREFORE, THE kWh OF ELECTRICITY CONSUMED IS 252 kWh AND THE TOTAL AMOUNT RODRIGUEZ RESIDENCE'S WILL PAY IS Php 2,142
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LETS STUDY
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27. How many electricity was consumed by a household if the rate per kilowatt is Php 10.00 and they have paid Php 750.00? A. 5 KWh B. 25 KWh C. 50 KWh D. 75 KWh
Answer:
75
Step-by-step explanation:
750/10 = 75
Since they paid 750.00 and the rate per kilowatt is 10.00 you have to divide 750.00 by 10 to get how much they consumed.
28. A radio with a power rating of 100W uses 0.4 kWh in one day. For how many hours was the radio on during that day?
Answer:
4h
Explanation:
[tex]kWh=\frac{Wh}{1000} \\kWh=0.4\\W=100\\h=?\\0.4=\frac{100h}{1000} \\\frac{10}{1}(0.4=\frac{h}{10}) \frac{10}{1} \\\4=h[/tex]
29. A flat iron operated for 0.75 hour and used 3.0 kWh of electrical energy. What is the power rating of the flat iron?
Answer:
1 kWh is equal to 0.25
Explanation:
Because 75 divided by 3 is 25.
So you consume 1 kWh per 0.25 hour
30. B. solve for the electric consumption previous reading present reading electric power consumption 1. 3798 kwh 2954 kwh 2. 6128 kwh 1974 kwh 3. 4165 kwh 3869 kwh 4. 2130 kwh 1240 kwh 5. 5030 kwh 4259 kwh pa answer po
Answer:
1. 844 kwh 2.4154 kwh 3.296 kwh 4.890 kwh 5.771 kwh
hope it helps
Step-by-step explanation: