Question

please find the answer for this its emergency​

Answer 1

Don't be worried on what we're given to find. The given thing which has to be found just means the value of m+ n!

How?!  Because,

$\dfrac{m^3+3mn(m+n)+n^3}{m^2+2mn+n^2}\ =\ \dfrac{(m+n)^3}{(m+n)^2}\ =\ (m+n)^{3-2}\ =\ m+n$

So we have just only to find the value of m + n.

Let me write the given data,

$\begin{tabular}{|c|c|}\cline{1-2}Class&Frequency\\ \cline{1-2} 0-6 & 10 \\ \cline{1-2} 6-12 & m \\ \cline{1-2} 12-18 & 4 \\ \cline{1-2} 18-24 & 7 \\ \cline{1-2} 24-30 & n \\ \cline{1-2} 30-36 & 4 \\ \cline{1-2} 36-42 & 1 \\ \cline{1-2}\end{tabular}$

Consider the total frequency 40.

$\begin{aligned}&10+m+4+7+n+4+1=40\\ \\ \Longrightarrow\ \ &26+m+n=40\\ \\ \Longrightarrow\ \ &\mathbf{m+n=14}\end{aligned}$

Hence 14 is the answer.

Answer 2

Question :-

If the mean of the following data is 14.7. Then find the value of

$\sf \dfrac{ {m}^{3} + 3mn(m + n) + {n}^{3} }{ {m}^{2} + 2mn + {n}^{2} }$

$\begin{tabular}{|c|c|c|c|c|c|c|c|c|}\cline{1-9}Class: & 0-6 & 6-12 & 12-18 & 18-24 & 24-30 & 30-36& 36-42 & Total \\ Frequency: & 10 & m & 4 & 7 & n & 9 & 1 & 40\\\cline{1-9}\end{tabular}$

Answer :-

14

Solution :-

$\begin{tabular}{|c|c|}\cline{1-2}Class&Frequency\\ \cline{1-2} 0-6 & 10 \\ \cline{1-2} 6-12 & m \\ \cline{1-2} 12-18 & 4 \\ \cline{1-2} 18-24 & 7 \\ \cline{1-2} 24-30 & n \\ \cline{1-2} 30-36 & 4 \\ \cline{1-2} 36-42 & 1 \\ \cline{1-2}Total & 40 \\ \cline{1-2}\end{tabular}$

From table

Find $\sf \sum fi$

$\sf \sum fi = 10 + m + 4 + 7 + n + 4 +1 = 26 + m + n$

Given

Total = 40

$\sf \implies \sum fi = 40$

$\sf \implies 26 + m + n = 40$

$\sf \implies m + n = 40 - 26$

$\sf \implies m + n = 14$

i) m + n = 14

Cubing on both sides

(m + n)³ = (14)³

m³ + 3mn(m + n) + n³ = 14³ ----(1)

[Since (x + y)³ = x³ + 3xy(x + y) + y³ ]

ii) m + n = 14

Squaring on both sides

(m + n)² = (14)²

m² + 2mn + n² = 14² -----(2)

[Since (x + y)² = x² + 2xy + y²]

Dividing eq (1) and (2)

$\sf \dfrac{ {m}^{3} + 3mn(m + n) + {n}^{3} }{ {m}^{2} + 2mn + {n}^{2} } = \dfrac{14^{3} }{ {14}^{2} }$

$\sf \dfrac{ {m}^{3} + 3mn(m + n) + {n}^{3} }{ {m}^{2} + 2mn + {n}^{2} } = 14^{3 - 2}$

$\bf \because \dfrac{ {a}^{m} }{ {a}^{n} } = {a}^{m - n}$

$\sf \dfrac{ {m}^{3} + 3mn(m + n) + {n}^{3} }{ {m}^{2} + 2mn + {n}^{2} } = 14^{1}$

$\bf \dfrac{ {m}^{3} + 3mn(m + n) + {n}^{3} }{ {m}^{2} + 2mn + {n}^{2} } = 14$