olá, questão nao tem nada de financeira.... resolução: X => número de pesquisados 20% de X => nunca experimentaram drogas, logo 80% de X experimentaram drogas, ok? desses 80%, 65% consumiram alcool => 65% de 80% de X e 1400 restantes de 80% consomem outros tipos.... X - (20%*X) - (80%*65%*X) = 1400 ...
![\sqrt[2]{4}=2
\sqrt[]{8+\sqrt[]{14+\sqrt[3]{6+2}}}
\sqrt[3]{6+2}=\sqrt[3]({2}^{3})}=2
\sqrt[]{8+\sqrt[]{14+2}}
\sqrt[]{14+2}}=4
\sqrt[]{8+4}= \sqrt[]{{2}^{2}\times3}}=2 \sqrt[]{3}](/latexrender/pictures/61b554498972d0ba22ef1d8c29cc82da.png "\sqrt[2]{4}=2
\sqrt[]{8+\sqrt[]{14+\sqrt[3]{6+2}}}
\sqrt[3]{6+2}=\sqrt[3]({2}^{3})}=2
\sqrt[]{8+\sqrt[]{14+2}}
\sqrt[]{14+2}}=4
\sqrt[]{8+4}= \sqrt[]{{2}^{2}\times3}}=2 \sqrt[]{3}")
![\sqrt[]{8+\sqrt[]{14+\sqrt[3]{6+\sqrt[]{4}}}} = 2\sqrt[]{3}](/latexrender/pictures/4ee9d8d31f4a1f780aa25e08c25b7629.png "\sqrt[]{8+\sqrt[]{14+\sqrt[3]{6+\sqrt[]{4}}}} = 2\sqrt[]{3}")
![R=\frac{P}{[\frac{{(1+i)}^{n}-1}{{(1+i)}^{n}\times i}]}}](/latexrender/pictures/a76f58ced90a2f5867941d2d37ff6fae.png "R=\frac{P}{[\frac{{(1+i)}^{n}-1}{{(1+i)}^{n}\times i}]}}")