Let us first evaluate left side of the statements \(\displaystyle\frac{{{3.56}\cdot{10}^{{2}}}}{{{1.09}\cdot{10}^{{4}}}}:\)

\(\displaystyle\frac{{{3.56}{x}{10}^{{2}}}}{{{1.09}{x}{10}^{{4}}}}=\frac{{356}}{{10900}}\sim{0.0326}={3.26}{x}{10}^{{-{{2}}}}\)

Let us next evaluate the right side of the statements \(\displaystyle{\left({4.08}{x}{10}^{{2}}\right)}{\left({1.95}{x}{10}^{{-{{6}}}}\right)}:\)

\(\displaystyle{\left({4.08}{x}{10}^{{2}}\right\rbrace}{\left({1.95}{x}{10}^{{-{{6}}}}\right)}={\left({408}\right)}{\left({0.00000195}\right)}\sim{0.0007956}={7.956}{x}{10}^{{-{{4}}}}\)

The large the power of 10, the large the number is. This then implies that \(\displaystyle{3.56}{x}{10}^{{-{{2}}}}\) is larger than \(\displaystyle{7.956}{x}{10}^{{-{{4}}}}\).

\(\displaystyle\frac{{{3.56}{x}{10}^{{2}}}}{{{1.09}{x}{10}^{{4}}}}={3},{26}{x}{10}^{{-{{2}}}}{>}{7.956}{x}{10}^{{-{{4}}}}={\left({4.08}{x}{10}^{{2}}\right)}{\left({1.95}{x}{10}^{{-{{6}}}}\right)}\)

Thus the third statement \(\displaystyle\frac{{{3.56}\cdot{10}^{{2}}}}{{{1.09}\cdot{10}^{{4}}}}{>}{\left({4.08}{x}{10}^{{2}}\right)}{\left({1.95}{x}{10}^{{-{{6}}}}\right)}\) is then correct.

\(\displaystyle\frac{{{3.56}{x}{10}^{{2}}}}{{{1.09}{x}{10}^{{4}}}}=\frac{{356}}{{10900}}\sim{0.0326}={3.26}{x}{10}^{{-{{2}}}}\)

Let us next evaluate the right side of the statements \(\displaystyle{\left({4.08}{x}{10}^{{2}}\right)}{\left({1.95}{x}{10}^{{-{{6}}}}\right)}:\)

\(\displaystyle{\left({4.08}{x}{10}^{{2}}\right\rbrace}{\left({1.95}{x}{10}^{{-{{6}}}}\right)}={\left({408}\right)}{\left({0.00000195}\right)}\sim{0.0007956}={7.956}{x}{10}^{{-{{4}}}}\)

The large the power of 10, the large the number is. This then implies that \(\displaystyle{3.56}{x}{10}^{{-{{2}}}}\) is larger than \(\displaystyle{7.956}{x}{10}^{{-{{4}}}}\).

\(\displaystyle\frac{{{3.56}{x}{10}^{{2}}}}{{{1.09}{x}{10}^{{4}}}}={3},{26}{x}{10}^{{-{{2}}}}{>}{7.956}{x}{10}^{{-{{4}}}}={\left({4.08}{x}{10}^{{2}}\right)}{\left({1.95}{x}{10}^{{-{{6}}}}\right)}\)

Thus the third statement \(\displaystyle\frac{{{3.56}\cdot{10}^{{2}}}}{{{1.09}\cdot{10}^{{4}}}}{>}{\left({4.08}{x}{10}^{{2}}\right)}{\left({1.95}{x}{10}^{{-{{6}}}}\right)}\) is then correct.