The use of electricity in the US in 1902 was 6.03×10

Question

9
kilowatt-hours.

The use of electricity in the US in 1950 was 4.3×10
11
kilowatt-hours.

What was the total kilowatt-hours of electricity used during these two years?

(1 point)

○ ($6.03 \times 10^9$) + ($4.3 \times 10^{11}$) = 0.063×10
11
+ ($4.3 \times 10^{11}$) = 4.3603×10
11


○ ($6.03 \times 10^9$) \times ($4.3 \times 10^{11}$) = 25.929×10
2
=2.6×10
1


○ ($6.03 \times 10^9$) + ($4.3 \times 10^{11}$) = 10.33×10
20
=1.033×10
19


○ ($6.03 \times 10^9$) \times ($4.3 \times 10^{11}$) = 25.929×10
20
=2.6×10
19

Answer 1

To calculate the total electricity used during the years 1902 and 1950, we'll add the two values:

1. Electricity used in 1902: \(6.03 \times 10^9\) kilowatt-hours
2. Electricity used in 1950: \(4.3 \times 10^{11}\) kilowatt-hours

To add these numbers effectively, we need to express both numbers in a similar format. Here, we'll convert the first number to the same exponent as the second:

1. Convert \(6.03 \times 10^9\) to match the exponent of \(10^{11}\): \[ 6.03 \times 10^9 = 0.0603 \times 10^{11} \]

Now we can sum: \[ 0.0603 \times 10^{11} + 4.3 \times 10^{11} = (0.0603 + 4.3) \times 10^{11} \] \[ = 4.3603 \times 10^{11} \]

So, the correct answer is: \[ ($6.03 \times 10^9$) + ($4.3 \times 10^{11}$) = 4.3603 \times 10^{11} \]

Thus, the correct selection from your given options is:

○ ($6.03 \times 10^9$) + ($4.3 \times 10^{11}$) = 0.063×10^{11} + ($4.3 \times 10^{11}$) = 4.3603×10^{11}