Question

Va rog, ajutati-ma la 6 si 8​

Answer 1

$6)A=\left\{0,2,4,6,8\right\} = > cardA=5$

Numărul tuturor submulțimilor lui A este:

${2}^{cardA} = {2}^{5} = 32$

=>număr cazuri posibile=32

Numărul tuturor submulțimilor lui A cu 3 elemente este :

$C_{5}^{3}=\frac{5!}{3!(5-3)!}=\frac{5!}{3!\times2!} = \frac{1 \times 2 \times 3 \times 4 \times 5}{1 \times 2 \times 3 \times 1 \times 2} = \frac{4 \times 5}{2} = \frac{20}{2} = 10$

=>număr cazuri favorabile=10

$P = \frac{nr. \: cazuri \: favorabile}{nr. \: cazuri \: posibile} = \frac{10}{32} = \frac{10}{ {2}^{5} }$

$=>D. \: \frac{10}{ {2}^{5} }$

$8) \alpha \: \in \: (\pi, \frac{3\pi}{2} ) = > cadranul \: III$

$cos \alpha = - \frac{3}{5}$

${sin}^{2} \alpha + {cos}^{2} \alpha = 1$

${sin}^{2} \alpha + {( - \frac{3}{5}) }^{2} = 1$

${sin}^{2} \alpha + \frac{9}{25} = 1$

${sin}^{2} \alpha = 1 - \frac{9}{25}$

${sin}^{2} \alpha = \frac{16}{25} = > sin \alpha = \pm \sqrt{ \frac{16}{25} } = \pm \frac{4}{5}$

$\alpha \: \in \: (\pi, \frac{3\pi}{2} ) = > sin \alpha < 0 = > sin \alpha = - \frac{4}{5}$

$sin2 \alpha = 2sin \alpha cos \alpha = 2 \times ( - \frac{4}{5} ) \times ( - \frac{3}{5} ) = \frac{24}{25}$

$= > B. \: \frac{24}{25}$