Archive for the ‘Set Theory’ Category

Problem 1156

Write the resulting set using the listing method.

{x | x2=9}

 

Solution:-

 

S = {x | P(x)} means “S is the set of all x such that P(x) is true”

The value of x that satisfy x2 = 9 are 3 and -3.

Thus, {x | x2 = 9} = {3, -3}.

 

Problem 1155

Write the resulting set using the listing method.

{5,6,7}∩ {2,3}

 

Solution:-

 

The intersection of sets A and B, denoted by A ∩ B, is the set of elements in set A that are also in set B. This is stared symbolically below.

A∩B = {x | x ϵ A and xϵ B}

Determine which elements, if any, of {5,6,7} are also elements of {2,3}.

The number 5 is an element if {5,6,7}, but not an element of {2,3}.

The number 6 is an element of {5,6,7}, but not an element of {2,3}.

The number 7 is an element of {5,6,7}, but not an element of {2,3}.

Since the two sets have no elements in common, {5,6,7}∩{2,3} is equal to empty set, denoted { } or ø.

 

Problem 1154

Write the resulting set using the listing method.

{3,9,15} U {9,13,15}

 

Solution:-

 

A U B = {x | x ϵ A or x ϵ B}

Remember that the word ‘or’ is used to mean that x may be an element of set A, set B, or both.

Let A represent the first set. Thus, the elements in set A are 3,9, and 15.

Let B represent the second set. Thus, the elements in set B are 9,13,and 15.

The elements that set A and B have in common are 9 and 15.

When listing the elements in a set, do not list an element more than once.

Finally, list all of the elements that appear in either set A or set B. If an element appears in both of the sets, list it only once.

Therefore , the union of set A and set B, denoted A U B, is equal to {3,9,15,13}.

 

Problem 1153

Write the resulting set using the listing method.

{5,6}∩ {6,7,8}

 

Solution:-

 

The intersection of sets A and B, denoted by A ∩ B, is the set of elements in set A that are also in set B. This is stared symbolically below.

A ∩ B = {x| x ϵ and x ϵ B}

Determine which elements, if any of {5,6} are also elements of {6, 7 ,8}.

The number 5 is not an element of {6,7,8}.

The number 6 is an element of {6,7,8}.

Find{5,6}∩ {6,7,8}.

Since 5 is not an element of both sets and 6 is an element of both sets, the intersection of the two sets is {6}

 

Problem 1152

Indicate whether the statement is true or false.

15 ϵ {2,7,12,20}

 

Solution:-

 

Thus, the statement 15 ϵ {2,7,12,20} means “15 is an element of set {2,7,12,20}”.

The elements of set A are 2, 7,12 and 20. Notice that 15 is not among the elements listed.

Thus, the statement “15 is an element of set {2,7,12,20}” is false.

 

Problem 1137

Slips of paper marked with the numbers 9, 10, 11, and 12 are placed in a box. After being mixed, two slip are drawn simultaneously. Write out the sample space S, choosing an S with equally like. Finally , write the indicated events below in set notation.

a. Both slips are marked with even numbers.

b.One slip is marked with an odd number and the other is marked with an even number.

c.Both slips are marked with the same number.

 

Solution:-

 

The sample space is the set of all people slips of paper. The order of the slips of paper does not matter. Use the number on each slip of paper to write the events.

To write the sample space, list all different possible combinations of two different numbers.

S = {(9,10), (9,11), (9,12),(10,11),(10,12),(11,12)}

The value of n(S) is the number of outcomes in the sample space,

S = {(9,10), (9,11), (9,12),(10,11),(10,12),(11,12)}

 

n(S) = 6

The numbers are all equally likely to be chosen, so the sample space has equally likely outcomes.

 

a. Write the event both slips are marked with even numbers. The sample space is show below. List all the outcomes from the sample space where both slips are marked with even number.

S = {(9,10), (9,11), (9,12),(10,11),(10,12),(11,12)}

 

The event is the set of outcomes that have both slips marked with an even number. Based on the sample space, the event is {(10,12)}.

 

b. Write the event both slips are marked with even numbers. The sample space is shown below.

S = {(9,10), (9,11), (9,12),(10,11),(10,12),(11,12)}

The event is the set of outcomes that have both an odd number and an even number. Based on the sample space, the event is {(9,10), (9,12),(11,10),(11,12)}.

 

c. Write the event both slips both slips are marked with the same number. The sample space is shown below.

S = {(9,10),(9,11),(9,12),(10,11),(10,12),(11,12)}

The event is the set of outcomes that have same number twice. There are no outcomes in the sample space that have the same number twice. Therefore, the event is the  empty set, ø.

 

 

Problem 1136

Use the union rule to answer the question.

If n(B) = 52, n(A ∩ B) = 15 and n(A U B) = 86, what is n(A)?

 

Solution:-

 

Recall that the union rule for sets states that

n(A U B) = n(A ) + n(B) – n(A ∩B)

for any finite set A and B. To find n(A), substitute each of the given quantities into the union rule. Then, solve for n(A).

Substitute the  given values of n(B), n(A ∩ B), and n(A U B) into the union rule.

n(A U B) = n(A) + n(B) – n(A ∩ B)

86 = n(A) + 52 – 15

Now, solve for n(A).

86 = n(A ) + 52 – 15

49 = n(A)

 

Problem 1135

Use the union rule to answer.

If n(A) = 10, n(B) = 15 and n(a ∩ B) = 9, what is n(A U B)?

 

Solution:-

 

The union rule for sets is symbolized n(A U B) = n(A) + n(B) – n(A ∩B). For this problem, you are given all the numbers necessary to apply the formula.

n(A) = 10

n(B) = 15

n(A ∩ B) = 9

Substitute each number into the formula.

n(A U B) = n(A) + n(B) – n(A ∩ B)

n(A U B) = 10 + 15 – 9

=16

 

 

 

Problem 1134

Suppose Kendra Gallegos has appointments with 12 potential customer. Kendra will be ecstatic if all 12 of these potential customers decide to make a purchase from her. Of course, in sales there are no guarantees. How many different sets of customers may place an order with Kendra?

 

Solution:-

 

A set of n distinct element has 2n subsets.

This set has 12 distinct elements.

The set containing 12 elements has 212 subsets.

212 = 4096

Therefore, the set has 4096 sunsets.

 

 

Problem 1133

Let U = {colleges in the United State}, A = {collages in the United State that have the word State in their name}, and B = {colleges in the United States that have the word South in their name}. Describe the set in words.

A ∩ B

 

Solution:-

 

The intersection of A and B, denoted A ∩ B, is the set consisting of those elements belong to both A and B.

A ∩ B is the set of collages in the United State that have the word State and have the word South in their name.