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)=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)=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)=25.929×102=2.6×101 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 squared is equal to 2 point 6 times 10 to the 1th power (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

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? (2 points) Responses (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)=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)=25.929×102=2.6×101 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 squared is equal to 2 point 6 times 10 to the 1th power (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

Answer 1

The correct response is (6.3×10^9) + (4.3×10^11) = 4.363×10^11.

Therefore, the total kilowatt-hours of electricity used during these two years is 4.363×10^11 kilowatt-hours.

Answer 2

so it would be the answer: (6.3 * 10^9) + (4.3 * 10^11) = 0.063 * 10^11 + (4.3 * 10^11) = 4.363 * 10^11?

Answer 3

Yes, that is correct. The total kilowatt-hours of electricity used during the two years is 4.363 * 10^11 kilowatt-hours.

Answer 4

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. how many times more was the electricity used in 1950 than 1902? express your answer in scientific notation.

Answer 5

To find out how many times more electricity was used in 1950 than in 1902, we divide the electricity used in 1950 by the electricity used in 1902:

(4.3 * 10^11) / (6.03 * 10^9)

Dividing the values, we get:

(4.3 / 6.03) * 10^(11-9)
= 0.712871287 * 10^2
= 7.1287 * 10^1
= 71.287

Therefore, the electricity used in 1950 was 71.287 times more than in 1902, expressed in scientific notation as 7.1287 * 10^1.