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Leona [35]
3 years ago
7

A 200 Watt incandescent bulb is on for eight hours a day during the month of April. How much will it cost to operate that bulb f

or the month if AEP's current rate is $0.10/kWh?
Advanced Placement (AP)
2 answers:
sammy [17]3 years ago
7 0

Answer: ¢4. 8

Explanation:

Power consumed = 200 watt

= 200/1000 = 0.2kw

The bulb lights for 8 hrs a day for the month of april,

April has 30 days, therefore total hrs = 8 x 30 = 240 hrs

Energy usage = 0.2 x 240 = 48 kWh

At the rate of ¢0.10/kWh, total cost will be;

0.1 x 48 = ¢4.8

cupoosta [38]3 years ago
4 0

Answer: $4.8

A 1000W bulb switched on for 1 hr will consume 1KWh (or 1 unit of electricity)

A 200W bulb switched on for 1 hr will consume 200/1000 units of electricity.

A 200W bulb switched on for 8hrs will consume 8 × 200 ÷ 1000 = 1.6units

There are 30days in a April,

A 200W bulb switched on for 8hrs/day for 30 days will consume

1.6 × 30 = 48 units of electricity or 48KWh of electricity.

Cost of operating bulb if AEP's current rate is $0.10/kWh

48KWh × $0.10/kWh

= $4.8 monthly.

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*The correct answer is IJ = JK


•explanation:

Given:
AB is the perpendicular bisector of
IK. ⇒ AB divides the line segment IK in two equal parts i.e. IJ = JK and the angle formed at the point of intersection J is 90° ⇒ ∠AJI = 90°. In ΔAIJ, By angle sum property of a triangle ∠AJI + ∠AIJ + ∠IAJ = 180° ( But ∠AJI = 90° ) ∠AIJ + ∠IAJ = 90° ⇒ ∠IAJ < 90° So, ∠IAJ is not a right angle. Its not given IK is a perpendicular bisector so AJ = BJ need NOT be true. As A does not lie on the line IK so A can not be the mid point of IK.

•Hence, we conclude the correct statement is IJ = JK

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