Von insgesamt $55$ Teilen des Streifens sind $22$ Teile grau gefärbt.

Bruch: ${\displaystyle \frac{2}{5}}\backslash dfrac\{2\}\{5\}$

Prozentsatz: ${\displaystyle \frac{2}{5}}={\displaystyle \frac{40}{100}}=40\mathrm{\%}\backslash dfrac\{2\}\{5\}=\backslash dfrac\{40\}\{100\}=40\backslash \%$

Umwandlung in einen Bruch

$75\mathrm{\%}={\displaystyle \frac{75}{100}}={\displaystyle \frac{3}{4}}75\backslash \%=\backslash dfrac\{75\}\{100\}=\backslash dfrac\{3\}\{4\}$

${\displaystyle \frac{3}{4}}\backslash dfrac\{3\}\{4\}$ des Quadrats müssen grau eingefärbt werden.

$100\mathrm{\%}\widehat{=}16{\text{ cm}}^2\phantom{x}\phantom{10}1\mathrm{\%}\widehat{=}{\displaystyle \frac{16}{100}}=\mathrm{0,16}{\text{ cm}}^2\phantom{x}\phantom{1}75\mathrm{\%}\widehat{=}\mathrm{0,16}\cdot 75=12{\text{ cm}}^{2}100\backslash \%\; \backslash \; \backslash hat\{=\}\; \backslash \; 16\backslash cm^2\; \backslash \backslash \; \backslash phantom\{x\}\backslash \backslash \; \backslash hphantom\{10\}1\backslash \%\; \backslash \; \backslash hat\{=\}\; \backslash \; \backslash dfrac\{16\}\{100\}=0\{,\}16\backslash cm^2\; \backslash \backslash \; \backslash phantom\{x\}\backslash \backslash \; \backslash hphantom\{1\}75\backslash \%\; \backslash \; \backslash hat\{=\}\; \backslash \; 0\{,\}16\backslash cdot\; 75=12\backslash cm^2$