- A$1224$
- B$1042$
- C$540$
- ✓$539$
$=\left[\begin{array}{lll}9^{2}+12^{2}-15^{2} & -10^{2}+13^{2}+16^{2} & 11^{2}-14^{2}+17^{2}\end{array}\right]$
$\left[\begin{array}{l}1 \\ 1 \\ 1\end{array}\right]$
$=\left[9^{2}+12^{2}-15^{2}-10^{2}+13^{2}+16^{2}+11^{2}-14^{2}+17^{2}\right]$
$=[539]$
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$x+3 y+k^{2} z=0$
$3 x+y+3 z=0$
has a non-zero solution $(x, y, z)$ for some $k \in R ,$ then $x +\left(\frac{ y }{ z }\right)$ is equal to
$(1)$ Probability that the selected bag is $B _3$ and the chosen ball is green equals $\frac{3}{10}$
$(2)$ Probability that the chosen ball is green equals $\frac{39}{80}$
$(3)$ Probability that the chosen ball is green, given that the selected bag is $B_3$, equals $\frac{3}{8}$
$(4)$ Probability that the selected bag is $B_3$, given that the chosen balls is green, equals $\frac{5}{13}$