Find mean of the following frequency table:

Step 1: Find class marks (mid-points)
$0$–$6\Rightarrow x_1=\dfrac{0+6}{2}=3$; $6$–$12\Rightarrow x_2=9$; $12$–$18\Rightarrow x_3=15$; $18$–$24\Rightarrow x_4=21$; $24$–$30\Rightarrow x_5=27$.
Step 2: Use $\bar{x}=\dfrac{\sum f_ix_i}{\sum f_i}$
$\sum f_i=5+9+10+12+4=40$.
$\sum f_ix_i=5(3)+9(9)+10(15)+12(21)+4(27)$
$\hspace{2.8cm}=15+81+150+252+108=606$.
Step 3: Compute the mean
\[ \bar{x}=\frac{\sum f_ix_i}{\sum f_i}=\frac{606}{40}=15.15. \] \[ \boxed{\text{Mean }=15.15} \]
The product of $\sqrt{2}$ and $(2-\sqrt{2})$ will be:
If a tangent $PQ$ at a point $P$ of a circle of radius $5 \,\text{cm}$ meets a line through the centre $O$ at a point $Q$ so that $OQ = 12 \,\text{cm}$, then length of $PQ$ will be:
In the figure $DE \parallel BC$. If $AD = 3\,\text{cm}$, $DE = 4\,\text{cm}$ and $DB = 1.5\,\text{cm}$, then the measure of $BC$ will be: