Let's represent the eleven integers as A, B, C, D, E, F, G, H, I, J, and K. Let's assume that if we lined them up from least to greatest, the set would be as follows:

A, B, C, D, E, F, G, H, I, J, K

The median in this set is the middle number. In this case, the middle number will always be F, because there are five numbers before it and five numbers after it.

We are told that the median is 39. Thus F = 39, and our set looks like this:

A, B, C, D, E, 39, G, H, I, J, K

Since we know that all the numbers are consecutive odd integers, the set must consist of the five odd numbers before 39 and the five odd numbers after 39. In other words, the set is this:

29, 31, 33, 35, 37, 39, 41, 43, 45, 47, 49

The question asks for the range of the set, which is the difference between the largest and smallest numbers. In this case, the range is 49 – 29 = 20.

The answer is 20.

First, arrange the numbers from highest to lowest, then subtract the highest number from the lowest number.

The range of a data set is the distance between the smallest number and the largest number. To find the range, you subtract the smallest number (5) from the largest number (13). Therefore, the range is 8.

To find the range of a data set, subtract the smallest number in the set from the largest:

The formula for intersection is P(a or b) = P(a) + P(b) – P(a and b).

Now, 30 students out of 100 swim, so P(swim) = 30/100 = 3/10.

40 students run out of 100, so P(run) = 40/100 = 4/10. Notice how we are keeping 10 as the common denominator even though we could simplify this further. Keeping all of the fractions similar will make the addition and subtraction easier later on. 

Finally, 20 students swim AND run, so P(swim AND run) = 20/100 = 2/10. (Again, we keep this as 2/10 instead of 1/5 so that we can combine the three fractions more easily.)

P(swim OR run) = P(swim) + P(run) – P(swim and run)

                       = 3/10 + 4/10 – 2/10 = 5/10 = 1/2.

The formula for intersection is P(a or b) = P(a) + P(b) – P(a and b).

Now, 30 students out of 100 swim, so P(swim) = 30/100 = 3/10.

40 students run out of 100, so P(run) = 40/100 = 4/10. Notice how we are keeping 10 as the common denominator even though we could simplify this further. Keeping all of the fractions similar will make the addition and subtraction easier later on. 

Finally, 20 students swim AND run, so P(swim AND run) = 20/100 = 2/10. (Again, we keep this as 2/10 instead of 1/5 so that we can combine the three fractions more easily.)

P(swim OR run) = P(swim) + P(run) – P(swim and run)

                       = 3/10 + 4/10 – 2/10 = 5/10 = 1/2.

The intersection of two sets contains every element that is present in both sets, so  is the correct answer.

The idea is to draw a Venn Diagram and find the intersection. We have one circle of 70 and another with 40. When we add the two circles plus the 10 students who joined neither, we should get 100 students. However, when adding the two circles, we are adding the intersections twice, therefore we need to subtract the intersection once.

We get , which means the intersection is 20.

This question is asking you to find the intersection of Set A and Set B. The intersection of two sets contains only the things which are already in both Set A and Set B. Because the numbers 5 and 9 are the only numbers that are shared in common between Set A and Set B, the intersection consists of only those two numbers.  : {5, 9}

Draw a Venn diagram with three subsets:  Math, Science, and English.  Start in the center with students that like all three subjects.  Next, look at students that liked two subjects.  Be sure to subtract out the ones already counted in the middle.  Then, look at the students that only like one subject.  Be sure to subtract out the students already accounted for.  Once all of the subsets are filled, look at those students who don’t like any of these subjects.  To find the students who don’t like any of these subjects add all of the students who like at least one subject from the total number of students surveyed, which is 50. 

M = math

S = science

E = English

M∩S∩E = 3        

M∩S = 5 (but 3 are already accounted for) so 2 for M and S ONLY

M∩E = 7 (but 3 are already accounted for) so 4 for M and E ONLY

S∩E = 8 (but 3 are already accounted for) so 5 for S and E ONLY

M = 14 (but 3 + 2 + 4 are already accounted for) so 5 for M ONLY

S = 20 (but 3 + 2 + 5 are already accounted for) so 10 for S ONLY

E = 28 (but 3 + 4 + 5 are already accounted for) so 16 for E ONLY

Therefore, the students already accounted for is 3 + 2 +4 + 5 + 5 + 10 + 16 = 45 students

So, those students who don’t like any of these subjects are 50 – 45 = 5 students