Find the median for the following list of numbers. 16 20 36 18 22 19 22

Lets analyse if the set is odd or even and divide it:
16 20 36 18 22 19 22
16 18 19 20 22 22 36
that is ordered, and so has 7 elements, that is odd, so we choose the middle element, that is 20 which happens to be the median.

IT IS ASYMMETRIC.

Step-by-step explanation:

Here we will consider all options to determine which of our options holds true.

Given set A;

А - {0, 1, 2}

R {(0,0), (0, 1), (0, 2), (1,2))

Let us not fail to keep the conditions given in the question at heart;

To Check for intransitive;

This condition holds true for all x, y, z belonging to A has -

(1) (x, y)and (y, z) but not (x, z) in R5OR

(2) don't have (x, y) in R5 OR

(3) (x, y)but don't have (y, z) in R5,

for 0, 1, 2 we have (0, 1) and (1, 2) but also have (0, 2) in R5 .

Condition failed here so, no need to check ahead.

⇒ So, It is NOT Intransitive.

To Check for irreflexivity;

This condition holds true for all x belonging to A -

(1) don't have (x, x) in R5,

for 0 we have (0, 0) in R5 so the condition is failed right here So, no need to check further for irreflexivity.

⇒ So, It is NOT Irreflexive.

Now let's Check for asymmetric,

This condition holds true for all x, y belonging to A has -

(1) (x, y) in R5 but not (y, x) OR

(2) don't have (x, y) in R5,

for 0, 1 we have (0, 1) in R5 but do not have (1, 0),

for 1, 0 we don't have (1, 0) in R5

for 1, 2 we have (1, 2) but not (2, 1) in R5

for 2, 1 we don't have (2, 1) in R5

for 0, 2 we have (0, 2) in R5 but not (2, 0)

finally for 2, 0 we don't have (2, 0) in R5

So, Condition satisfied for every г.уеА

⇒ IT IS ASYMMETRIC.

cheers i hope this helps