$X$ is the random variable that can take any one of the values, $0,1,7,11$ and 12 . The probability mass function for $X$ is

$$ \begin{aligned} & \mathrm{P}(X=0)=0.4 ; \mathrm{P}(X=1)=0.3 ; \mathrm{P}(X=7)=0.1 ; \\ & \mathrm{P}(X=11)=0.1 ; \mathrm{P}(X=12)=0.1 \end{aligned} $$

Then, the variance of $X$ is

$$ \text { The value of }\mathop {\lim }\limits_{x \to \infty } \left(x-\sqrt{x^2+x}\right) \text { is equal to }$$

Let $y$ be the solution of the initial value problem $y^{\prime}+0.8 y+0.16 y=0$ where $y(0)=3$ and $y^{\prime}(0)=4.5$. Then, $y(1)$ is equal to__________ (rounded off to 1 decimal place).