According to the US Census, there are about 117,000,000 households in the US. If every household replaced 1 60W incandescent bulb that is used for 35 hours per week with a 10W LED bulb, how many kWh of electricity would this save the country per year?

The country is saved 1.0647×10^10 kWh of electricity per year.

Step-by-step explanation:

Power saved per household = 60 W - 10 W = 50 W = 50/1000 = 0.05 kW

Power saved in 117,000,000 households = 117,000,000 × 0.05 kW = 5,850,000 kW

Number of hours the bulb is used per week = 35 hours

Number of hours the bulb is used per year (52 weeks) = 52 × 35 = 1,820 hours

Quantity of electricity saved per year = 5,850,000 kW × 1,820 hours = 1.0647×10^10 kWh

10,647,000,000 kWh per year of electricity is saved.

Step-by-step explanation:

Given:

Time, t = 35 hours/week

In 1 year, assume 52 weeks in 1 year:

35 hours/week * 52 weeks/1year

= 1820 hours/year

Bulb used by households = 60 W

Power of replaced bulbs = 10 W

Power saved per household = 60 - 10

= 50 W

Power saved in 117,000,000 = 117,000,000 × 50

= 5,850,000,000 W

= 5,850,000 kW

Energy saved = power x time

= 5,850,000 × 1820

= 10,647,000,000 kWh per year.

6mm^3 or .6cm^3

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  36     6

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6     6   6

factor trees are nice.

answer: on graphs, the open and closed circles, or vertical asymptotes drawn as dashed lines us identify discontinuities.

as before, graphs and tables allow us to estimate at best.

when working with formulas, getting zero in the denominator indicates a point of discontinuity.

when working with piecewise-defined functions, check for discontinuities at the transition points where one piece ends and the next begins.

step-by-step explanation: