The use of electricity in the US in 1902 was 6.03 x 10^9 kilowatt-hours.
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?
(2 points)
Responses
(6.3×109)×(4.3×1011)=25.929×1020=2.6×1019
open paren 6 point 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
(6.3×109)+(4.3×1011)=0.063×1011+(4.3×1011)=4.363×1011
open paren 6 point 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 3 times 10 to the 11th power
(6.3×109)+(4.3×1011)=10.33×1020=1.033×1019
open paren 6 point 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
(6.3×109)×(4.3×1011)=25.929×102=2.6×101
The correct response is:
(6.03 x 10^9) + (4.3 x 10^11) = 4.30603 x 10^11
Therefore, the total kilowatt-hours of electricity used during these two years was 4.30603 x 10^11.