Find the mode from the following table: 
Step 1: Identify the modal class.
The class with the highest frequency is \(30–40\), so it is the modal class.
Step 2: Write the given data.
\[ L = 30, \quad f_1 = 23, \quad f_0 = 21, \quad f_2 = 14, \quad h = 10 \] Step 3: Apply the formula for mode.
\[ \text{Mode} = L + \left(\dfrac{f_1 - f_0}{2f_1 - f_0 - f_2}\right) \times h \] Step 4: Substitute the values.
\[ \text{Mode} = 30 + \left(\dfrac{23 - 21}{2(23) - 21 - 14}\right) \times 10 \] \[ = 30 + \left(\dfrac{2}{46 - 35}\right) \times 10 = 30 + \dfrac{20}{11} = 31.82 \] Step 5: Conclusion.
Hence, the mode = 31.82 (approx.).
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:
The mean of the following table will be:
| Class-interval | 0-2 | 2-4 | 4-6 | 6-8 | 8-10 |
|---|---|---|---|---|---|
| Frequency (f) | 3 | 1 | 5 | 4 | 7 |
From the following data, the modal class of the table will be:
\[ \begin{array}{|c|c|c|c|c|c|} \hline \text{Class-interval} & 0-10 & 10-20 & 20-30 & 30-40 & 40-50 \\ \hline \text{Frequency (f)} & 11 & 21 & 23 & 14 & 5 \\ \hline \end{array} \]
Find the Mean from the Following Table
Given data:
\[ \begin{array}{|c|c|} \hline \text{Class-interval} & \text{Frequency (f)} \\ \hline 0-10 & 3 \\ 10-20 & 10 \\ 20-30 & 11 \\ 30-40 & 9 \\ 40-50 & 7 \\ \hline \end{array} \]