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Given the point P(2,3).
We have to find the distance from x-axis i.e perpendicular distance from x-axis.
In the graph , we see in order to determine the perpendicular distance the point of x-axis to be taken is (2,0)
Therefore, the distance is 3-0=3.
Hence, the distance of point P(2,3) from x-axis is 3 units.
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Yogita Ingle 6 years, 2 months ago
3, 4, 6, 7, 8, 14
Mean is given by: Mean={tex}\frac{Sum of observations}{total number ofobservations}{/tex}
{tex}Mean=\frac{3+4+6+7+8+14}{6}=\frac{42}{6}=7{/tex}
Thus, {tex}{\overline{x}}=7.{/tex}
Now,{tex} {\Delta}x_{1}={\overline{x}-x_{1}}=7-3=4{/tex}
{tex}|{\Delta}x_{2}|=|{\overline{x}-x_{2}}|=|7-4|=3{/tex}
{tex}|{\Delta}x_{3}|=|{\overline{x}-x_{3}}|=|7-6|=1{/tex}
{tex}|{\Delta}x_{4}|=|{\overline{x}-x_{4}}|=|7-7|=0{/tex}
{tex}|{\Delta}x_{5}|=|{\overline{x}-x_{5}}|=|7-8|=1{/tex}
{tex}{\Delta}x_{6}|=|{\overline{x}-x_{6}}|=|7-14|=7{/tex}
Sum of deviations of the variate values={tex}|{\Delta}x_{1}|+|{\Delta}x_{2}|+|{\Delta}x_{3}|+|{\Delta}x_{4}|+|{\Delta}x_{5}|+|{\Delta}x_{6}|{/tex}
=4+3+1+0+1+7=16
Therefore, the sum of deviations of the variate values is 16.

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Pradum Gurjar 6 years, 2 months ago
1Thank You