Find the median, first quartile, third quartile, and interquartile range of the data. 40,33,37,54,41,34,27,39,35

Median=37

First quartile=33.5

Third quartile=40.5

Inter-quartile range=7

Range=27

Step-by-step explanation:

The given data set is ;

40,33,37,54,41,34,27,39,35

We arrange the data set in ascending order to obtain;

{27,33,34,35,37,39,40,41,54}

The median is the middle number {37}

First quartile is the median of the lower half of the ascending set.

{27,33,34,35}

The third quartile is the median of the upper half of the ascending set.

{39,40,41,54}

The inter-quartile range is the the third quartile minus the first quartile.

The range is the maximum value minus the least value.

Range=54-27=27

< xcy ≅ < xby

step-by-step explanation:

in δcax and δbax,

so, δcax ≅ δbax (side-angle-side or sas)

since δcax ≅ δbax, we can conclude < acx ≅ < abx

in δacy and δaby,

so, δacy ≅ δaby (side-angle-side or sas)

since δacy ≅ δaby, we can conclude < acy ≅ < aby

now,

< acy ≅ < aby

=> < acx + < xcy ≅ < abx + < xby

=> < acx + < xcy ≅ < acx + < xby   (since < acx ≅ < abx already proved above)

subtracting < acx from both sides, we get

< acx + < xcy -< acx ≅ < acx + < xby -< acx

cancelling out < acx and -< acx from both sides, we get

< xcy ≅ < xby (proved)