Let A = {x1, x2, . . . , x12} be a set of 12 positive integers, not necessarily distinct, such that xi ≤ 150, 1 ≤ i ≤ 12. Prove that there are at least two different 6-element subsets S1 and S2 of A such that the sum of the elements in S1 is equal to the sum of the elements in S2. Use Pigeonhole principal.

By pigeon hole principle, at least 2 subsets of A have same sum

Step-by-step explanation:

Let A = {x1, x2, . . . , x12} where i ≤ xi≤150

for any 6 element subset S of  A, the sum of numbers in S is at least 6 since all integers are positive and it is at most 150+149+148+147+146+145= 885. so we consider number of pigeon holes = number of possible sums= 885 and number of pigeons = number of subsets of A of size 6= ¹²C₆= 924

Since number of pigeons is greater than number of pigeon holes, by pigeon hole principle, at least 2 subsets of A have same sum

see below

step-by-step explanation:

7.   there are 10 data points in the set.   the middle point is between 5 and 6.   the middle point in the top half is 8   and the middle point in the bottom half is 3. to find the iqr , take point 8 and subtract point 3

mid term   point 8 = 100   point 3 = 88

iqr = 100-88 = 12

final   point 8 = 93   point 3 = 78

iqr = 93 - 78 = 15

the final has the greatest iqr

6.   b - there is a high data value that causes the data set to be asymmetrical for the males: the data for males are high and asymmetrical .   there is an outlier for the males that skews the data

5.   the   iqr is a better measure of spread for the movies and basketball games.   since the basketball game has an outlier, we do not want to use standard deviation for measurement.  

3.   the exam median is much higher than the class median.   the median line is farther to the right on the exams

2.   there is no way to tell what the means are from the box and whisker plots.  

answer is x<