Question

Simplify√48x⁵y²
A. 4x³y√3
B. 4x²y√3
C. 4xy√3x
D. 4x²y√3x

Answer 1

Step-by-step explanation:

Here, we have to find the mean, median & mode of the following data :

$\begin{gathered}\boxed{\begin{array}{c|c|c|c|c|c}\bf Weight\: (in \: kg)&\sf 48&\sf 50&\sf 52 &\sf 54&\sf 58\\\frac{\qquad \quad \qquad}{}&\frac{\quad \qquad \qquad}{}&\frac{\qquad \quad\qquad}{}&\frac{\qquad \quad \qquad}{}&\frac{\quad \qquad \qquad}{}&\frac{\qquad \quad\qquad}{}\\\bf No. \: of \: players &\sf 4&\sf 3&\sf 2&\sf 2 &\sf 1\end{array}}\end{gathered}$

Weight(inkg)No.ofplayers484503522542581

❏ Firstly, let's calculate for mean.

$\begin{gathered}\boxed{\begin{array}{cccc}\bf X_i \: &\bf F_i \: &\bf F_i \times X_i \: \\\frac{\qquad \qquad \qquad \qquad}{}&\frac{\qquad \qquad \qquad \qquad}{}&\frac{\qquad \qquad \qquad \qquad\qquad}{}\\\sf 48&\sf 4&\sf 192 \\\\\sf 50 &\sf 3&\sf 150 \\\\\sf 52 &\sf 2&\sf 104 \\\\\sf 54&\sf 2&\sf 108\\\\\sf 58 &\sf 1 &\sf 58\\\frac{\qquad \qquad \qquad \qquad}{}&\frac{\qquad \qquad \qquad \qquad}{}&\frac{\qquad \qquad \qquad \qquad\qquad}{}\\ &\sf 12& \sf 612\end{array}}\end{gathered}$

Xi4850525458Fi4322112Fi×Xi19215010410858612

~ Calculating the mean weight.

Formula to be used for finding Mean :-

\begin{gathered}\star\;{\boxed{\sf{\pink{Mean = \dfrac{ \sum \: (f_i \times x_i)}{ \sum \: f_i} }}}}\\ \\\end{gathered}⋆Mean=∑fi∑(fi×xi)

\begin{gathered}\dag\;{\underline{\frak{Putting\:the\:values\:,}}}\\ \\\end{gathered}†Puttingthevalues,

\begin{gathered} \bf{\longrightarrow \: Mean \: = \cancel\dfrac{612}{12} } \\ \\ \ \ \longrightarrow \underline{\boxed{\bf{51 \: kg}}} \: \purple{\bigstar}\end{gathered}⟶Mean=12612  ⟶51kg★

\therefore\:{\underline{\sf{Mean\:of\:given\:data\:is\: {\textsf{\textbf{51\:kg}}}.}}}∴Meanofgivendatais51kg.

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❏ Now, calculating for median.

$\begin{gathered}\boxed{\begin{array}{cccc}\bf X_i \: &\bf F_i \: &\bf Cumulative \: Frequency \: \\\frac{\qquad \qquad \qquad \qquad}{}&\frac{\qquad \qquad \qquad \qquad}{}&\frac{\qquad \qquad \qquad \qquad\qquad}{}\\\sf 48&\sf 4&\sf 7 \\\\\sf 50 &\sf 3&\sf 4 \\\\\sf 52 &\sf 2&\sf 9 \\\\\sf 54&\sf 2&\sf 11\\\\\sf 58 &\sf 1 &\sf 12\\\frac{\qquad \qquad \qquad \qquad}{}&\frac{\qquad \qquad \qquad \qquad}{}&\frac{\qquad \qquad \qquad \qquad\qquad}{}\\ &\sf 12& \end{array}}\end{gathered}$Xi4850525458Fi

Answer 2

Answer:

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