The box plots display the same data for the number of crackers in each snack bag, but one includes the outlier in the data and the other excludes it. 4, 14, 15, 15, 16, 16, 18, 19, 20, 20, 21 number of crackers in each bag, with outlier number of crackers in each bag, without outlier which statement comparing the box plots is true? both the median and the range changed. both the range and the lower quartile changed. both the median and the interquartile range changed. both the interquartile range and the lower quartile changed.

Both the median and the range changed.

Step-by-step explanation:

The data set that contains the outlier is given as:

        4, 14, 15, 15, 16, 16, 18, 19, 20, 20, 21

As we know that:

Minimum value of the data is: 4

The median or middle quartile i.e. is the central tendency of the data and exist in the middle of the data.

The median is: 16

Also, the lower set of data is:

               4,14,15,15,16

Hence, the lower quartile i.e.  is the median of the lower set of data.

Hence, lower quartile i.e. =15

Similarly the upper set of data is:

              18, 19, 20, 20, 21

Hence, the upper quartile i.e.  is the median of the upper set of data.

Hence, upper quartile i.e. =20

Also, the interquartile range is:

Hence, interquartile range is: 5

The maximum value of the set is: 21.

Hence, the range is:

Maximum value-Minimum value

=21-4

=17

Hence, Range=17

As we know that 4 is the outlier of the data as it is the smallest as compared to all the data points.

After removing the outlier the data set is:

          14, 15, 15, 16, 16, 18, 19, 20, 20, 21

Minimum value of set=14

Maximum value of set=21

Range=Maximum value-Minimum value

         =21-14

        =7

Hence, Range=7

Also, we know that median exist in the middle of the data.

Hence, the median i.e. lies between 16,18 and is calculated as:

Hence, Median=17

Also, the lower set of data is:

                   14,15,15,16,16

Hence, the lower quartile i.e. is the median of the lower set of data.

Hence,

Similarly the upper set of data is:

                    18, 19, 20, 20, 21

Hence, the upper quartile i.e. is the median of the upper set of data.

Hence,

Hence, interquartile range is:

20-15=5

Hence, the correct statement is:

Both the median and the range changed.

 ( Since with outlier:

Range=17 and Median=16

Without an outlier:

Range=7 and Median=17 )

step-by-step explanation: i dont

the length of o'p' is .

step-by-step explanation:

it is given that the quadrilateral m'n'o'p' is the image of quadrilateral mnop after dilation.

the center of dilation is origin and the scale factor is 1.5.

in dilation the image and preimge are similar figures. it means the sides of image and preimage are proportional.

the scale factor is defined as

the scale factor is 1.5.

multiply both sides by op.

note: if the value of op is given, then multiply the value of op with 1.5 to get the length of side o'p'.

therefore length of o'p' is .