Triangles Venn Diagram

Venn diagrams are used to describe the relationship of the sets. The description of some sets is given and you are asked to draw a Venn diagram to illustrate the sets. You are well known about the triangles that they are having three sides. Based on their sides and the angles, it is classified into different types. The relationship of the triangles can also be described using Venn diagrams.

Triangles and Venn Diagrams

Venn diagrams:

Venn diagrams to determine the relationships between the sets such as subsets and intersections.

Venn diagram for A within B :



Here, all members of A belongs to B or A ⊂ B or A ∪ B = B or A ∩ B = A or n(A ∩ B) = n(A).

Venn diagram for  A overlap B:



Here, some members of A belongs to B or A ∩ B ≠ Ø or n(A ∩ B ) ≠ 0

Venn diagram for disjoint sets of A and B :



Here, no members of A belongs to B or A ∩ B = Ø or n(A ∩ B ) = 0

Triangles:

scalene triangle:

A triangle is said to be scalene triangle, if all the sides are of different lengths.Some scalene triangles are also right triangles.



Right triangle:

A triangle is said to be right triangle, if one interior angle is exactly 90°. Some right triangles are also scalene triangles.



Equilateral triangle:

A triangle is said to be equilateral triangle, if all the sides are of same length.



Isosceles triangle:

A triangle is said to be isosceles triangle, if two sides are of same length and two angles are equal.



Examples

Let U is the set of triangles, A is the set of isosceles triangles, B is the set of equilateral triangles and C is the set of right-angled triangles.

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Draw a Venn diagram.

Solution:

We have to determine the relationships between the sets.

All the equilateral triangles are said to be isosceles triangles, so B ⊂ A. (within)

Some of the right-angled triangles are said to be isosceles triangle. C ∩ A ≠ Ø (overlap)

No right-angled triangles are equilateral triangles. C ∩ B = Ø (disjoint)

Venn diagram for the triangles:



Let U is the set of triangles, S is the set of scalene triangles and R is the set of right triangles.

Draw the Venn diagram

Solution:

The relationship between scalene and right triangle to be defined.

Some of the right triangles are scalene triangles.

Integer Divide by Zero

Integer is the number which is greater than the zero or less than the zero and a number greater than zero is a positive and less than the zero is called negative. Zero has no sign like positive and negative sign. In number line two integers are same distance from the zero in opposite directions are the opposites.In this section we are going to see about any integer which is divided by zero and the integer divided the integers.


Division Property of Integers:

For integers we have some properties, let assume a, b be the two integers for that positive and negative integers. But with the zero it has some special property any integer which divided by the zero we get infinity and the zero divided by the any integer we get zero.


Let a & b be the two integers, where  `a/b` is not always an integer.

Examples: `(-4)/5` ; ` 5/ (-3)`  are not integers ;  `1/2 ` is an integer.

For any integer 'a' is not equal to zero.  `a/a ` = 1; and  `a/1` = a

Examples:
`5/5` = 1

`5/1` = 5

` 1/5`  is not  an integer

For any integer 'a' is not equal to zero.
`a/(-1)` = -a; ` a/(-a)` = -1.

For every non zero integer a; `0/a` = 0.

Examples:

` 3/ (-1)` = -3

`3/ (-3)` = -1

`1/ (-3)` is not an integer

Algebra is widely used in day to day activities watch out for my forthcoming posts on Integers and Absolute Value and greatest integer function. I am sure they will be helpful.

Problems with Integer Division:

Problem 1:
Divide +95 by 5.

Solution:
=    ` 95/5`

=19 is an integer


Problem 2:
Divide  -65 by 0.

Solution:
We know that the any integer divided by 0 is infinity by division property. Here, one twenty five by zero
= ∞

Problem 3:

Divide 106 by 0.

Solution:
We know that the any integer divided by 0  is infinity by division property. Here,thousand divided by zero.

= ∞

Problem 4:

Divide 0 by 55

Solution:
We know that the zero divided by any number is zero by division property

=0

Fraction Chart for Math

Fraction can be defined as part of something bigger than that. Suppose your class has 40 boys and 60 girls thus totaling 100 students. Boys are the part of the class and so are girls. But the class consists both boys and girls and is bigger set. In the class, we say 40 out 100 are boys. We can write this as a Fraction as below



We observe that number 40 is written at the top and then a bar below it and then number 100 below the bar. This is how the fractions are represented.

The boys in the class form a part of the class and they are 40 in number. The class is bigger and has 100 students. So, the part here is the boys and the whole is the class. So, this fraction represents that 40 boys are a part of a class of 100 students. Thus we see, fraction represents the part of a whole.

It must also be noted that there could be fractions where the part can be bigger than the whole.

Fractions Illustration

Let us illustrate how fractions can be formed as a part of the whole. Let us a draw a full circle first as below
 

Now let us divide the circle into two different parts and shade one part green as shown below



So we have total two parts and one part is green. This is represented as ½.

Now let us divide the original circle into three and shade two parts of it green as below



So we have total three parts and two parts are green. This is represented as `(2)/(3)`

 Thus fractions represent a part of the whole.

As we saw a fraction has one number at the top of a bar and one number below it. The top number is called the numerator and the bottom number is called the denominator as shown below.



The chart below shows fractions in pictorial form



Types of Fractions

• If the numerator is less than the denominator, it is called a proper fraction.

All the fractions we saw above like `(1)/(2)` , `(2)/(3)`  etc are proper fractions because the top number or numerator was smaller then the bottom number or denominator

• If the numerator is greater than the denominator, it is called an improper fraction.

Example. `(4)/(3)` , `(7)/(2)`  etc. These fractions represent where the part is greater than the whole. We will illustrate with an example. Suppose your teacher is conducting an examination for you. The examination had 25 questions and the teacher asked you to solve any 20 questions to get the maximum mark of 20. Suppose you managed to solve all the 25 correctly then your score in the test will be `(25)/(20)` . That means that you solved more than the maximum required.

• There are another kind of fractions, which has both a whole number and a proper fraction. These fractions are called mixed fractions.

Example 3 `(1)/(2)` ,4`(2)/(3)`  etc.  Thus these fractions are a mix of whole number and a proper fraction and hence they are called mixed fractions.

Exercises on Fractions

Pro 1: From the figure below, write the region shaded in the circle as a fraction of the whole circle

 i)



ii)





iii)



Ans : (i) `(2)/(10)` (ii) `(5)/(6)`       (iii) `(7)/(10)`

Pro 2: Identify the proper, improper and mixed fractions from the following

`(7)/(8)`  ,7 `(1)/(2)`   , `(16)/(2)` , `(1)/(8)` , `(4)/(3)`

Ans: Proper fractions - `(7)/(8)` ,`(1)/(8)`   - Numerator of these fractions are smaller then the denominator

Improper Fraction - `(16)/(2)` , `(4)/(3)`  - - Numerator of these fractions are bigger then the denominator

Mixed Fraction - 7 `(1)/(2)`   - This fraction has  a whole number and a proper fraction

Probability of a Intersection B

Probability of A Intersection B

Probability is the possibility of the outcome of an event of a particular experiment. Probabilities are occurs always numbers between 0 (impossible) and 1(possible). The set of all possible outcomes of a particular experiment is called as sample space. For example probability of getting a 6 when rolling a dice is 1/6. In this lesson we will discuss about probability problems using intersection rule.

Probability of a Intersection B – Example Problems

Example 1: A jar contains 5 red candies, 4 orange candies. If three candies are drawn at random, find the probability, that 1 is red candy and 2 are orange candies

Solution:

We have to select 3 candies, from 9 (5 + 4) candies.

n(S) = 9C3 = (9!)/(3!xx6!) = (9xx8xx7)/(3xx2xx1) = 84

Let A = Event of getting 1 red candy

B = Event of getting 2 orange candies

n(A) = 5C1 = `(5!)/(1!xx4!) ` = `5/1` = 5

n(B) = 4C2 =` (4!)/(2!xx2!) ` = `(4xx3)/(2xx1)` = 6

P(A) =` (n(A))/(n(S))` = `5/84`

P(B) = `(n(B))/(n(S))` = `6/84`

P(A intersection B) = P(A) ∙ (B) = `5/84` ∙ `6/84` = `30/7056`

P(A intersection B) = `5/1176` .



Example 2: A box contains 6 yellow marbles, 6 orange marbles. If four marbles are drawn at random, find the probability, that 2 are yellow marbles and 2 are orange marbles.

Solution:

We have to select 4 marbles, from 12 (6 + 6) marbles.

n(S) = 12C4 = `(12!)/(4!xx8!)` = `(12xx11xx10xx9)/(4xx3xx2xx1)` = 495

Let A = Event of getting 2 yellow marbles

B = Event of getting 2 orange marbles

n(A) = 6C2 = `(6!)/(2!xx4!)` = `(6xx5)/(2xx1)` = `30/2` = 15

n(B) = 6C2 = `(6!)/(2!xx4!)` = `(6xx5)/(2xx1)` = `30/2` = 15

P(A) = `(n(A))/(n(S)) ` = `15/495` = `1/33`

P(B) = `(n(B))/(n(S)) ` = `15/495` = `1/33`

P(A intersection B) = P(A) ∙ (B) = `1/33` ∙ `1/33` = `1/1089`

P(A intersection B) = `1/1089` .

Probability of a Intersection B – Practice Problems

Problem 1: A jar contains 4 lemon candies, 4 orange candies. If two candies are drawn at random, find the probability, that 1 is lemon and 1 is orange candy.

Problem 2: If P(A) = `1/5` , P(B) = `1/7` , P(A or B) = `1/9` , find (A and B)?

Answer: 1) `1/49 ` 2) `73/315`

Area of a Square Inscribed in a Circle

A square is a four sided figure; all the four sides are equal. If we situate a square inside the circle, all the four edges or vertices may touches the circle. A square is a quadrilateral, if the four sides of a square touch the circle, the four sides act as a four chords of a circle. The area of a square is measured in square units such as feet, inch, meter etc. In this article we shall discuss about the area of a square inscribed in a circle.

Area of a Square Inscribed in a Circle

If a square is inscribed in a circle and four sides of a square touches a circle.



The diagonals of the circle act as a diameter of the circle. We can find the area of a square is by:

Area of square = side times side

Area of the square = side side

= s2

Example:

Find the area of the square, sides of the square is 9ft.

Solution:

Given: Side = 8ft therefore diameter = 9ft

Area of a circle = side × side

= 9 × 9

= 81square ft.

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If the square does not touch the circle but inscribed in a circle, In such case we can also find the area of the square.

Example:

Find the area of a square inscribed in a circle, side of the square is 6ft.

Solution:

Given: side = 6t

Area of a square = side × side

= 6 × 6

= 36square ft.

The one or two sides of a square may touch the circle.

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Example for Area of a Square Inscribed in a Circle:

Example1:

Find the area of the square inscribed in a circle, sides of the square is 12m.

Solution:

Given: Side = 12

Area of a square = side × side

= 12 × 12

= 144m2.

Example2:

Find the area of the square inscribed in a circle, sides of the square is 23ft.

Solution:

Given: Side = 23ft

Area of a square = side × side

= 23 × 23

= 529square ft.

Poisson Probability Table

Introduction:

The allocation was first introduced by Simeon-Denis poisson (1781–1840) and in print, together with his probability theory, in 1838 in his work (Research on the Probability of Judgments in Criminal and Civil Matters). The job focused on sure random variables N that tally, among other things, the number of separate occurrences that take place during a time-interval of given length. In probability hypothesis and statistics, Poisson division is a discrete probability division expresses probability of a number of actions.

Poisson Probability Table Conditions:

Poisson probability distribution states two conditions

1.The amount of successes in two disjoint time intervals is autonomous.

2.The probability of a success during a small time interval is comparative to the entire length of the time interval.

Poisson probability formula:

The Probability distribution of a Poisson random variable, X  instead of the number of successes happening in a given time interval or a specified area of space is given by the formula:

P(X)=e-`nu` `nu`X / X!

where

X=0,1,2,3....

e = expected value

µ = mean figure of successes in the given time interval or area of space.

Mean and Variance for poisson probability:

If µ is the standard number of successes happening in a given time interval or area in the Poisson distribution, then the mean and the variance of the Poisson distribution are both equal to µ.

E(X) = µ

and

V(X) = s2 = µ

Note: In a Poisson distribution, only one factor, µ is needed to determine the probability of an occurrence.
Poission Probability Table:

The below shows the value of poisson probability for the mean and X values of

`lambda` =0.1 to 5.5

X  =0 to 17

poisson probability table

Examples of Poisson Probability Distribution:

Example 1:

A life assurance salesman sells on the usual 3 life insurance policies per week. Use Poisson's law to work out the possibility that in a given week he will sell

(a) a few policies

(b) 2 or supplementary policies but less than 5 policies.

(c) assume that there are 5 working days per week, what is the probability that in a given day he will sell single policy?

Solution:

Here, µ = 3

(a) "Some policies" means "1 or more policies". We can work this out by finding 1 minus the "zero policies" probability:

P(X > 0) = 1 - P(x0)

Now P(X)=e-`nu`    `nu`X / X!    So P(X`@` )= e-3 30 / 3! = 4.9787 `xx` 10-2

So

Probability= P(X > 0)

=1 - P(x0)

=1-4.9787`xx`10-2

=0.95021

(b)     P(2`<=` X `>=` 5) = P(X2)+P(X3)+P(X4)

= (e-3 32 / 2!) + (e-3 33 / 3!) + (e-3 34 / 4!)

=0.61611

(c)    Average number of policies sold per day:  `3/5 ` = 0.6

So on a given day   P(X)=e-0.6 (0.6)1 / 1!

= 0.32929

Example 2:

A business makes electric motors. The probability an electric motor is imperfect is 0.01. What is the probability that a model of 300 electric motors will contain precisely 5 defective motors?

Solution:

The average number of defectives in 300 motors is µ = 0.01 × 300 = 3

The probability of getting 5 defectives is:

P(X ) = e-3 35 / 5!

= 0.10082

Intersection of Two Sets

Set is a fundamental part of the mathematics. This set concept is applied in every branch of mathematics. Sets are used in relations and functions. The application of sets are geometry,  sequences, Probability etc. Sets are used in everyday life such as a volleyball team, vowels in alphabets, various kinds of geometry shapes etc.

There are two main important operations in sets. They are  union and intersection of two sets. Let us learn the concepts and properties of intersection of two sets. We will learn some example problems and give practical problems about intersection of two sets.

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Intersection of Sets:

Let A and B be any two sets.  A intersection B is the set of all elements which are similar to both A and B. The symbol `nn` is used to denote the intersection. A intersection B is the set of all those elements which belong to both A and B. Symbolically, we write A `nn` B = { x : x `in` A and x `in` B }.

Ex:

Consider the two sets X = { 1, 5, 8, 9 } and Y = { 5, 6, 9, 15 } . Find X intersection Y.

Sol:

We see that 5, 9 are the only elements which are similar to both X and Y. Hence  X ∩ Y = { 5, 9 }

Some Properties of Operation of Intersection:

(i) Commutative law:  A ∩ B = B ∩ A

(ii) Associative law: ( A ∩ B ) ∩ C = A ∩ ( B ∩ C )

(iii) Law of identity: φ ∩ A = φ, U ∩ A = A (Law of φ and U).

(iv) Idempotent law: A ∩ A = A

(v) Distributive law: A ∩ ( B ∪ C ) = ( A ∩ B ) ∪ ( A ∩ C ) i. e., ∩ distributes over ∪

Between, if you have problem on these topics union of sets, please browse expert math related websites for more help on Union Set.

Practice Problems on Intersection of Two Sets:

Problem 1:

Find the intersection of each of the following two sets:

1. X = { 1, 3, 5 }   Y = { 1, 2, 3 }

2. A = [ a, e, i, o, u ]   B = { a, b, c }

3. A = { 1 , 2 , 3 }    B = `Phi`

Sol:

1. X `nn` Y = { 1, 3 }

2. A `nn` B = [ a ]

3. A `nn` B = `Phi`

Problem 2:

If A = { 3, 5, 7, 9, 11 }, B = {7, 9, 11, 13}, C = {11, 13, 15}and D = {15, 17}; find

(i) A ∩ B  (ii) B ∩ C (iii) A ∩ C ∩ D

(iv) A ∩ C (v) B ∩ D (vi) A ∩ (B ∪ C)

(vii) A ∩ D (viii) A ∩ (B ∪ D)

Sol:

i) A ∩ B = { 7, 9, 11 }

ii) B ∩ C = { 11, 13 }

iii) A ∩ C ∩ D = Nill

iv) A ∩ C = { 11 }

v) B ∩ D = nill

vi) A ∩ (B ∪ C) = { 7, 9, 11, 13 }

vii) A ∩ D = nill

(viii) A ∩ (B ∪ D) = { 7, 9, 11}