Given that

Initially the radioactive carbon-14 present in the burial cloth of the egyptian mummy is 560g.

The half life of carbon-14 is 5730 years.

The total amount of carbon-14 decayed after t years is given by

\(\displaystyle{A}={A}_{{0}}{\left({0.5}\right)}^{{{\frac{{{1}}}{{{h}}}}}}\) (1)

Where, \(\displaystyle{A}_{{0}}\) initial amount of carbon-14

t-time in years,

h-Half life of carbon-14

a) To complete table using the following calculation:

Now,

1) t=0 years, h=5730 years,\(\displaystyle{A}_{{0}}={560}\)g

\(\displaystyle{A}={A}_{{0}}{\left({0.5}\right)}^{{{\frac{{{1}}}{{{h}}}}}}\)

\(\displaystyle{A}={560}{\left({0.5}\right)}^{{{\frac{{{0}}}{{{5730}}}}}}\)

\(\displaystyle{A}={560}{\left({0.5}\right)}^{{0}}={560}{\left({1}\right)}\)

\(\displaystyle{A}={560}\)g

2) t=5730 years, h=5730 years,\(\displaystyle{A}_{{0}}={560}\)g

\(\displaystyle{A}={560}{\left({0.5}\right)}^{{{\frac{{{5730}}}{{{5730}}}}}}\)

\(\displaystyle{A}={560}{\left({0.5}\right)}^{{1}}={560}{\left({0.5}\right)}\)

\(\displaystyle{A}={280}\)g

3) t=11460 years, h=5730 years,\(\displaystyle{A}_{{0}}={560}\)g

\(\displaystyle{A}={560}{\left({0.5}\right)}^{{{\frac{{{11460}}}{{{5730}}}}}}\)

\(\displaystyle{560}{\left({0.5}\right)}^{{2}}={560}{\left({0.25}\right)}\)

\(\displaystyle{A}={140}\)g

3) t=17190 years, h=5730 years,\(\displaystyle{A}_{{0}}={560}\)g

\(\displaystyle{A}={560}{\left({0.5}\right)}^{{{\frac{{{17190}}}{{{5730}}}}}}\)

\(\displaystyle{A}={560}{\left({0.5}\right)}^{{3}}={560}{\left({0.125}\right)}\)

\(\displaystyle{A}={70}\)g

The table becomes

\(\displaystyle{b}{e}{g}\in{\left\lbrace{a}{r}{r}{a}{y}\right\rbrace}{\left\lbrace{\left|{l}\right|}{l}{\left|{l}\right|}\right\rbrace}{h}{l}\in{e}{t}{\left(\text{in years}\right)}&{m}{\left(\text{amoun of radioactive material}\right)}\backslash{h}{l}\in{e}{0}&{560}\backslash{h}{l}\in{e}{5730}&{280}\backslash{h}{l}\in{e}{11460}&{140}\backslash{h}{l}\in{e}{17190}&{70}\backslash{h}{l}\in{e}{e}{n}{d}{\left\lbrace{a}{r}{r}{a}{y}\right\rbrace}\)

b)To find the exponential function that models the amount of carbon-14 in the cloth, y, after t years:

\(\displaystyle{A}={A}_{{0}}{e}^{{-{k}{t}}}\)

Here, At t=0, y=560 g,

k- rate of decay,

t- time in years.

The exponential function is

\(\displaystyle{y}={560}{e}^{{-{k}{t}}}\)

c) To find t, for k=49.5% :

49.5% of original amount of carbon-14 is

\(\displaystyle{49.5}\%\times{560}={\frac{{{49.5}}}{{{100}}}}\times{560}\)

\(\displaystyle={49.5}\times{5.6}\)

\(\displaystyle{49.5}\%\times{560}={277.2}\)g

Now, A0 =560, A= 277.2 g, h=5730 years, from (1)

\(\displaystyle{A}={A}_{{0}}{\left({0.5}\right)}^{{{\frac{{{t}}}{{{h}}}}}}\)

\(\displaystyle{277.2}={560}{\left({0.5}\right)}^{{{\frac{{{t}}}{{{5730}}}}}}\)

\(\displaystyle{\left({0.5}\right)}^{{{\frac{{{t}}}{{{5730}}}}}}={\frac{{{277.2}}}{{{560}}}}\)

\(\displaystyle{\left({0.5}\right)}^{{{\frac{{{t}}}{{{5730}}}}}}={0.495}\)

\(\displaystyle{\frac{{{t}}}{{{5730}}}}{\ln{{0.5}}}={\ln{{0.495}}}\)

\(\displaystyle{t}={5730}{\left({\frac{{{\ln{{0.495}}}}}{{{\ln{{0.5}}}}}}\right)}\)

\(\displaystyle{t}={5730}{\left({\frac{{-{0.7032}}}{{-{0.693}}}}\right)}\)

\(\displaystyle{t}={5730}{\left({1.015}\right)}\)

\(\displaystyle{t}={5814.33}\) years

Thus, the mummy was burried before 5814.33 years.

Initially the radioactive carbon-14 present in the burial cloth of the egyptian mummy is 560g.

The half life of carbon-14 is 5730 years.

The total amount of carbon-14 decayed after t years is given by

\(\displaystyle{A}={A}_{{0}}{\left({0.5}\right)}^{{{\frac{{{1}}}{{{h}}}}}}\) (1)

Where, \(\displaystyle{A}_{{0}}\) initial amount of carbon-14

t-time in years,

h-Half life of carbon-14

a) To complete table using the following calculation:

Now,

1) t=0 years, h=5730 years,\(\displaystyle{A}_{{0}}={560}\)g

\(\displaystyle{A}={A}_{{0}}{\left({0.5}\right)}^{{{\frac{{{1}}}{{{h}}}}}}\)

\(\displaystyle{A}={560}{\left({0.5}\right)}^{{{\frac{{{0}}}{{{5730}}}}}}\)

\(\displaystyle{A}={560}{\left({0.5}\right)}^{{0}}={560}{\left({1}\right)}\)

\(\displaystyle{A}={560}\)g

2) t=5730 years, h=5730 years,\(\displaystyle{A}_{{0}}={560}\)g

\(\displaystyle{A}={560}{\left({0.5}\right)}^{{{\frac{{{5730}}}{{{5730}}}}}}\)

\(\displaystyle{A}={560}{\left({0.5}\right)}^{{1}}={560}{\left({0.5}\right)}\)

\(\displaystyle{A}={280}\)g

3) t=11460 years, h=5730 years,\(\displaystyle{A}_{{0}}={560}\)g

\(\displaystyle{A}={560}{\left({0.5}\right)}^{{{\frac{{{11460}}}{{{5730}}}}}}\)

\(\displaystyle{560}{\left({0.5}\right)}^{{2}}={560}{\left({0.25}\right)}\)

\(\displaystyle{A}={140}\)g

3) t=17190 years, h=5730 years,\(\displaystyle{A}_{{0}}={560}\)g

\(\displaystyle{A}={560}{\left({0.5}\right)}^{{{\frac{{{17190}}}{{{5730}}}}}}\)

\(\displaystyle{A}={560}{\left({0.5}\right)}^{{3}}={560}{\left({0.125}\right)}\)

\(\displaystyle{A}={70}\)g

The table becomes

\(\displaystyle{b}{e}{g}\in{\left\lbrace{a}{r}{r}{a}{y}\right\rbrace}{\left\lbrace{\left|{l}\right|}{l}{\left|{l}\right|}\right\rbrace}{h}{l}\in{e}{t}{\left(\text{in years}\right)}&{m}{\left(\text{amoun of radioactive material}\right)}\backslash{h}{l}\in{e}{0}&{560}\backslash{h}{l}\in{e}{5730}&{280}\backslash{h}{l}\in{e}{11460}&{140}\backslash{h}{l}\in{e}{17190}&{70}\backslash{h}{l}\in{e}{e}{n}{d}{\left\lbrace{a}{r}{r}{a}{y}\right\rbrace}\)

b)To find the exponential function that models the amount of carbon-14 in the cloth, y, after t years:

\(\displaystyle{A}={A}_{{0}}{e}^{{-{k}{t}}}\)

Here, At t=0, y=560 g,

k- rate of decay,

t- time in years.

The exponential function is

\(\displaystyle{y}={560}{e}^{{-{k}{t}}}\)

c) To find t, for k=49.5% :

49.5% of original amount of carbon-14 is

\(\displaystyle{49.5}\%\times{560}={\frac{{{49.5}}}{{{100}}}}\times{560}\)

\(\displaystyle={49.5}\times{5.6}\)

\(\displaystyle{49.5}\%\times{560}={277.2}\)g

Now, A0 =560, A= 277.2 g, h=5730 years, from (1)

\(\displaystyle{A}={A}_{{0}}{\left({0.5}\right)}^{{{\frac{{{t}}}{{{h}}}}}}\)

\(\displaystyle{277.2}={560}{\left({0.5}\right)}^{{{\frac{{{t}}}{{{5730}}}}}}\)

\(\displaystyle{\left({0.5}\right)}^{{{\frac{{{t}}}{{{5730}}}}}}={\frac{{{277.2}}}{{{560}}}}\)

\(\displaystyle{\left({0.5}\right)}^{{{\frac{{{t}}}{{{5730}}}}}}={0.495}\)

\(\displaystyle{\frac{{{t}}}{{{5730}}}}{\ln{{0.5}}}={\ln{{0.495}}}\)

\(\displaystyle{t}={5730}{\left({\frac{{{\ln{{0.495}}}}}{{{\ln{{0.5}}}}}}\right)}\)

\(\displaystyle{t}={5730}{\left({\frac{{-{0.7032}}}{{-{0.693}}}}\right)}\)

\(\displaystyle{t}={5730}{\left({1.015}\right)}\)

\(\displaystyle{t}={5814.33}\) years

Thus, the mummy was burried before 5814.33 years.