Question

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?

(1 point)
Responses

(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

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 squared is equal to 2 point 6 times 10 to the 1th power

(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

Answer 1

Convert to the same power of 10: 6.03×10^9 = 0.0603×10^11.
Then 0.0603×10^11 + 4.3×10^11 = 4.3603×10^11 kWh.

So the total is 4.3603 × 10^11 kilowatt-hours (≈ 436,030,000,000 kWh).