How many elements are in the union of four sets if each of the sets has 96 elements, each pair of sets share 52 elements, each triple of sets shares 22 elements and there are 6 elements in all four sets.

154 elements

Step-by-step explanation:

Let the four sets be X1, X2, X3, X4.

Each of the four sets has 96 elements

|X1| = 96

|X2| = 96

|X3| = 96

|X4| = 96

They can be paired in the following ways

|X1 n X2| , |X1 n X3|, |X1 n X4|, |X2 n X3|, |X2 n X4|, |X3 n X4|

Each pair of sets contains 52 elements each

|X1 n X2| = 52

|X1 n X3| = 52

|X1 n X4| = 52

|X2 n X3| = 52

|X2 n X4| = 52

|X3 n X4| = 52

The triple sets are |X1 n X2 n X3|, |X1 n X2 n X4|, |X1 n X3 n X4|,|X2 n X3 n X4|

Each triple set has 22 elements

|X1 n X2 n X3|=22

|X1 n X2 n X4|= 22

|X1 n X3 n X4|=22

|X2 n X3 n X4|=22

All the four sets have 6 elements

|X1 n X2 n X3 n X4| = 6

Using the principle of inclusion,

|X1 υ X2 υ X3 υ X4|= |X1|+ |X2|+ |X3|+ |X4|- |X1 n X2| - |X1 n X3| - |X1 n X4| -|X2 n X3| - |X2 n X4| - |X3 n X4|+ |X1 n X2 n X3|+ |X1 n X2 n X4|+|X1 n X3 n X4|+|X2 n X3 n X4- |X1 n X2 n X3 n X4|

= 96 + 96 + 96 + 96 – 52 – 52 – 52 – 52 – 52 – 52 + 22 + 22 + 22 + 22 – 6

= 4(96) -6(52) +4(22) – 6

= 384 – 312 + 88 – 6

= 154

d

step-by-step explanation:

in a an geometric explicit equation, the first number is the starting rate. in this case 10.

median is the value separating the higher half of a data sample, a population or probablity distribution from the lower half.

step-by-step explanation:

say you have 3,8,7,3,1,9,8 order them from lowest to highest

if orderd it would be 1,3,3,7,8,8,9 now you take away from the lowest and the highest until you end up in the middle.

so you would remove the 1 and 9 leaving you with 5 digits remaining

when you've reached the middle that would be your answer so in this case the answer is 7

if there were two numbers left say 7 and 5 you'd and them together so 12 then divide by 2 witch would equal 6 so the answer would be 6

good luck on that project hoped i you at all