The use of electricity in the US in 1902 was 6.03 x 10^9 kilowatt-hours.

Question

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

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

(1 point)
Responses

(6.03×109)+(4.3×1011)=0.063×1011+(4.3×1011)=4.3603×1011
open paren 6 point 0 3 times 10 to the 9th power close paren plus open paren 4 point 3 times 10 to the 11th power close paren is equal to 0 point 0 6 3 times 10 to the 11th power plus open paren 4 point 3 times 10 to the 11th power close paren is equal to 4 point 3 6 0 3 times 10 to the 11th power

(6.03×109)×(4.3×1011)=25.929×1020=2.6×1019
open paren 6 point 0 3 times 10 to the 9th power close paren times open paren 4 point 3 times 10 to the 11th power close paren is equal to 25 point 9 2 9 times 10 to the 20th power is equal to 2 point 6 times 10 to the 19th power

(6.03×109)+(4.3×1011)=10.33×1020=1.033×1019
open paren 6 point 0 3 times 10 to the 9th power close paren plus open paren 4 point 3 times 10 to the 11th power close paren is equal to 10 point 3 3 times 10 to the 20th power is equal to 1 point 0 3 3 times 10 to the 19th power

(6.03×109)×(4.3×1011)=25.929×102=2.6×101

Answer 1

To find the total kilowatt-hours of electricity used during the years 1902 and 1950, you'll need to add the two values together.

Given:

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

First, convert \( 6.03 \times 10^9 \) so that it has the same exponent as \( 4.3 \times 10^{11} \):

\[ 6.03 \times 10^9 = 0.0603 \times 10^{11} \]

Now, add the two amounts:

\[ (0.0603 \times 10^{11}) + (4.3 \times 10^{11}) = (0.0603 + 4.3) \times 10^{11} = 4.3603 \times 10^{11} \]

Thus, the correct total kilowatt-hours of electricity used during these two years is:

\[ 4.3603 \times 10^{11} \]

So, the correct response is: \[ (6.03 \times 10^9) + (4.3 \times 10^{11}) = 0.063 \times 10^{11} + (4.3 \times 10^{11}) = 4.3603 \times 10^{11} \]