If $\frac{x^2}{\mathrm{a}}+\frac{2 x y}{\mathrm{~h}}+\frac{y^2}{\mathrm{~b}}=0$ represents a pair of straight lines and slope of one of the lines is twice that of the other, then $a b: h^2$ is
If $\overline{\mathrm{a}}, \overline{\mathrm{b}}$ and $\overline{\mathrm{c}}$ are unit coplanar vectors, then the scalar triple product $\left[\begin{array}{lll}2 \overline{\mathrm{a}}-\overline{\mathrm{b}} & 2 \overline{\mathrm{~b}}-\overline{\mathrm{c}} & 2 \overline{\mathrm{c}}-\overline{\mathrm{a}}\end{array}\right]$ has the value
If a random variable X has the following probability distribution values
| $\mathrm{X}$ | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 |
|---|---|---|---|---|---|---|---|---|
| $\mathrm{P(X):}$ | 0 | $\mathrm{k}$ | $\mathrm{2k}$ | $\mathrm{2k}$ | $\mathrm{3k}$ | $\mathrm{k^2}$ | $\mathrm{2k^2}$ | $\mathrm{7k^2+k}$ |
Then $P(X \geq 6)$ has the value
Let the vectors $\overline{\mathrm{a}}, \overline{\mathrm{b}}, \overline{\mathrm{c}}$ be such that $|\bar{a}|=2,|\bar{b}|=4$ and $|\bar{c}|=4$. If the projection of $\bar{b}$ on $\bar{a}$ is equal to the projection of $\bar{c}$ on $\bar{a}$ and $\bar{b}$ is perpendicular to $\bar{c}$, then the value of $|\vec{a}+\bar{b}-\bar{c}|$ is