Question #5a784

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Question #5a784

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Answer 1 · 2018-02-01T20:20:15

$12.5 %$ $12.5%$

Explanation:

$17, 190 = 3 \cdot 5, 730$ $17, 190 = 3 * 5,730$

after $17, 190$ $17, 190$ years, $3$ $3$ half-lives will have passed.

after $1$ $1$ half-life, ${(\frac{1}{2})}^{1}$ $(1/2)^1$ of the carbon-14 remains.

${(\frac{1}{2})}^{1} = \frac{1}{2} = 50 %$ $(1/2)^1 = 1/2 = 50%$

after $3$ $3$ half-lives, ${(\frac{1}{2})}^{3}$ $(1/2)^3$ of the carbon-14 remains.

${(\frac{1}{2})}^{3} = \frac{1}{8} = 12.5 %$ $(1/2)^3 = 1/8 = 12.5%$

after $17, 190$ $17, 190$ years, $12.5 %$ $12.5%$ of the carbon-14 remains.

Answer 2 · 2018-02-01T20:54:59

12.5%

Explanation:

If the half life is 5730 years and you want to know how much is left after 17190 years, the first thing you need to do is find out how many 'half life' cycles the sample has gone through.

In this case 3 ( $\frac{17190}{5730}$ $17190/5730$ = 3).

Therefore the sample has been halved three times.

1st half life period= $\frac{100%}{2}$ $"100%"/"2"$ = 50%

2nd half life period = $\frac{50%}{2}$ $"50%"/"2"$ = 25%

3rd half life period = $\frac{25%}{2}$ $"25%"/"2"$ = 12.5%

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Question #5a784

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