In a triangle $A B C$, with usual notations, the sides $\mathrm{a}, \mathrm{b}, \mathrm{c}$ are such that they are roots of the equation $x^3-11 x^2+38 x-40=0$ then $\frac{\cos \mathrm{A}}{\mathrm{a}}+\frac{\cos \mathrm{B}}{\mathrm{b}}+\frac{\cos \mathrm{C}}{\mathrm{c}}=$

The value of $\tan \left[2 \tan ^{-1} \frac{1}{5}-\frac{\pi}{4}\right]$ is

If $\mathrm{X} \sim \mathrm{B}(35, \mathrm{p})$ such that $7 \mathrm{P}(\mathrm{X}=0)=\mathrm{P}(\mathrm{X}=1)$ then the value of $\frac{\mathrm{P}(\mathrm{X}=15)}{\mathrm{P}(\mathrm{X}=20)}$ is equal to

A student studies for X number of hours during a randomly selected school day. The probability that X can take the values, has the following form, where k is some constant.

$$ \mathrm{P}(\mathrm{X}=x)= \begin{cases}0.2, & \text { if } x=0 \\ \mathrm{k} x, & \text { if } x=1 \text { or } 2 \\ \mathrm{k}(6-x), & \text { if } x=3 \text { or } 4 \\ 0, & \text { otherwise }\end{cases} $$

The probability that the student studies for at most two hours is