Add 7.3×10^7+1.8×10^7 . Which is the correct sum?(1 point)

Responses

9.1×10^7
9.1 times 10 superscript 7 baseline

9.1+10^7
9.1 plus 10 superscript 7 baseline

9.1×10^14
9.1 times 10 superscript 14 baseline

8.1×10^7

The correct sum is 9.1×10^7.

The population of a city is given as 3.45×10^6 . A nearby city has a population of 1.3×10^5 . What is the sum of the populations of the two cities in scientific notation?(1 point)

Responses

4.75×10^5
4.75 times 10 superscript 5 baseline

4.75×10^11
4.75 times 10 superscript 11 baseline

35.8+10^5
35.8 plus 10 superscript 5 baseline

3.58×10^6
3.58 times 10 superscript 6 baseline

The sum of the populations of the two cities in scientific notation is 3.58×10^6.

An empty dump truck weighs about 2.8×10^5 pounds. How much does it weigh if 15,300 pounds of trash is added? The answer should be in scientific notation.(1 point)

Responses

29,530 pounds
29,530 pounds

4.33×10^5 pounds
4.33 times 10 superscript 5 baseline pounds

2.953×10^5 pounds
2.953 times 10 superscript 5 baseline pounds

15,302.8 pounds

The weight of the dump truck with 15,300 pounds of trash added is 4.43×10^5 pounds.

The half-life of uranium-235 is 700,000,000 years. The half-life of uranium-238 is 4.463×10^11 years longer. What is the half-life of uranium-238? The answer should be in decimal form.(1 point)

Responses

447,000,000,000
447,000,000,000

516,300,000,000
516,300,000,000

11,463,000,000
11,463,000,000

4.4637×10^11

The half-life of uranium-238 is 4.463×10^11 years.

A mobile phone holds 1.28×10^11 bytes of storage. You can also pay more money to get an additional 384,000,000,000 bytes. How much storage would your phone have if you buy the additional bytes? The answer should be in scientific notation.(1 point)

Responses

385.28×10^11
385.28 times 10 superscript 11 baseline

385,280,000,000
385,280,000,000

512×10^9
512 times 10 superscript 9 baseline

5.12×10^11
5.12 times 10 superscript 11 baseline

If you buy the additional 384,000,000,000 bytes, your phone would have a total storage of 5.12×10^11 bytes.

Is he right