What percentage of a radioactive substance remains after 7.00 half-lives have elapsed?

Explanation:

We want to know the percentage of the radio active element after 7half life

Let assume the original mass or quantity of the radioactive element is No

After first half life it remains No/2

Second half life No/4

Third half life No/8

Fourth half life No/16

Fifth half life No/32

Sixth half life No/64

Seventh half life No/128

So after the seventh half life the radioactive element has decay to No/128

Then, the that remain is

The remaining amount / Original amount × 100

%remain = (No/128 / No) × 100

No cancel No

%remain = 100/128

%remain = 0.78%

t = 1/400 h   ⇔ 0.0025 h ⇔ 9s ⇔ 0.15 m

any of the answers are correct.