Mode and Median Calculator

Mode: Mode is the value that takes place most repeatedly in the data set. Measure of central tendency is known as mode. If the data’s are given in the form of a frequency table, the class corresponding to the maximum frequency is called the modal class. The value of the variate of the modal class is the mode.

Median: The median is the middle value when the given values are arranged in an ascending order. Let us see the median and mode calculator.

Median and Mode calculator:

In the calculator enter the set of values in first box, after that clcik the median button it will automatically calculate the median value and it will be displayed in answer box. The same process is done for mode.

Examples on Mode calculator:

Example 1:

            Find the mode of 7, 4, 5, 1, 7, 3, 4, 6, and 7.

Solution:

           The above question is entered in the first box. The calculator doing the follwing process,

           Assemble the data in the ascending order, we get

            1, 3, 4, 4, 5, 6, 7, 7, 7.

            The number 7 occurs many times in the above values.
            Mode = 7 will display the answer box after press the mode button on calculator.
Example 2:

            Find the mode for 12, 15, 11, 12, 19, 15, 24, 27, 20, 12, 19, and 15.

Solution:

           The above question is entered in the first box. The calculator doing the follwing process,

           Assemble the data in the ascending order, we get

            11, 12, 12, 12, 15, 15, 15, 19, 19, 20, 24, 27.

            In the above values 12 occurs 3 times and 15 also occurs 3 times.

            ∴ Both 12 and 15 are the modes for the given data. We observe that there are two modes for the given data.The Mode will be displayed in answer box on calculator
Example 3:

            Find the mode of 19, 20, 21, 24, 27, and 30.

Solution:
            Already the above data are in the ascending order. Each value occurs exactly one time in the series. Hence there is no mode in the above given data.
These are the examples on mode calculator.

Examples on Median calculator:

Example 1:    
            Find the median of the following numbers: 12, 45, 62,10,14,31 and 43.
Solution:
           The above question is entered in the first box. The calculator doing the fololwing process,
            Arranging the given numbers in ascending order we get
            10, 12, 14, 31, 43, 45 and 62.
                            `darr`
                    Middle term
            Median = Middle item = 31.     

         The median 31 will display the answer box

Between, if you have problem on these topics Cubic Equation  please browse expert math related websites for more help on ibsat 2013.

Example 2:         
            Find the median of the following numbers: 3, 7, 4, 10, 22, 16, 21 and 5.
Solution:
            The above question is entered in the first box. The calculator doing the following process,
            Arranging the given numbers in ascending order we get
            3, 4, 5, 7, `darr` 10, 16, 21, 22              
                 Median is here
            Median = Item midway between 7 and 10
                       =` (7 + 10) / 2` = `17 / 2` = 8.5
          The Median 8.5 will display on answer box on calculator
These are the examples on mode and median calculator.

Sine Geometry

Trigonometry is the division of geometry dealing among relationships among the sides also angles of triangles. In geometry sine of an angle is the ratio of the length of the opposite side to the length of the hypotenuse. The  ratio does not depend on size of the particular right triangle chosen, as long as it contains the angle A, since all such triangles are similar

(Source: Wikipedia)

I like to share this sine curves with you all through my article. 

Sine geometry

In the above diagram ,
sin A =opposite/hypotenuse
Examples for sine geometry

A - Right angle of the triangle ABC.
The length of AB, BC and CA are frequently represented through c, a, b.
Obtain point B as middle of a trigonometric circle
Circle with radius = 1.
Now sin (B) are comparative to b, c also a.
sin `(B)/b` =`1/a`
sin (B) = `b/a`

Examples for sine geometry

Example 1
Angle of triangle is 20⁰, opposite side of triangle is 12 apply the sine geometry to find the unidentified side of the triangle?
Solution:

Angle of triangle= 20⁰  

Opposite side of triangle = 12.

sin A =opposite/hypotenuse

sin 20⁰ = `12/x`

sin 20⁰ x = 12

x = `12/sin 20^0`

x =`12/0.3420`    {since the value of sin 20 degree is 0.3420}

x=35.08

Hypotenuse side= 35.08

Example 2
Angle of triangle is 78⁰, hypotenuse side of triangle is 20 apply the sine geometry to find the unidentified side of the triangle?
Solution:

Angle of triangle= 78⁰  

Hypotenuse side of triangle = 20.

sin A =opposite/hypotenuse

sin 78⁰ = `x/20`

sin 78⁰ x 20= x

x = sin 78⁰ x20

x =0.97814x20     {since the value of sin 78 degree is 0.97814}

x=35.08

Opposite side = 19.56

Example 3
If hypotenuse side of triangle 40 and opposite side of triangle 20 find the sine angle?
Solution:

Hypotenuse side of triangle = 40.

Opposite side =20

sin A =opposite/hypotenuse

sin A= `20/40`

sin A =` 1/2`

sin A = 0.5    {sin 30 degree is 0.5}

Therefore the angle is 30 degree

Interval Estimates

Interval Estimate:

• Interval estimation is the process of calculate the interval for possible value of unknown parameter in the population.
• It is calculate in the use of sample data and contrast to the point estimation. It is different from the point estimation. It is the outcome of a statistical analysis.

The most common forms of interval estimations as follows:

• A frequents Method or Confidence interval
• A Bayesian method or credible intervals

The other common methods for interval estimations are

• Tolerance interval
• Prediction interval

And another one is known as the fiducial inference.

Construction of interval estimates parameter:

The normal form of interval estimate of the population parameter is,

• Point estimate of parameter and
• Plus or minus margin of error

Margin of error:

• The amount which is subtracted or added from  the point estimate  of the statistic and produce the parameter interval  estimate is known as the margin of error.
• The margin of error size depends on the following factors:
• Sampling distribution type of sample statistics.
• Area under sampling distribution percentage   that includes the researchers      decision.Usually we consider the confident level as 90%, 95%, 99%.
• The interval of each interval estimates are constructed in the region of the point estimate with its confident level.

My forthcoming post is on Set Interval Notation will give you more understanding about Algebra.

Construction of Interval estimate for Population mean

• Take the point estimate of μ  that is  the sample mean`vecx`
• Define  the mean distribution for the sample.When the  value of n is large we  have to use the central limit  theorem. And   is the normal distribution with the,

                      standard deviation `sigma``vecx` = `sigma/sqrt(n)`  

                      and mean μ.

• Choose the most common confident  level as 95%
• Find the margin of  error  which is related with the confidence level.
• The area  under the curve of  the sample means the normal distribution contains the 95%  of the interval from.

                               z= -1.96 to z= 1.96 

• The interval estimate for 95 % is,   

                            `vecx`- 1.96 (`sigma/sqrt(n)` ) to `vecx``sigma/sqrt(n)`

Common Factors

The common factors of two or more whole digits is the biggest whole digit that equally divides all the whole digits. There are two methods to find common factors in math.

The initial method is to list all the factors of each digit. Then decide the biggest factor.

For Example:

Find the common factors of 12 and 18.

The factors of 12 are 1, 2, 3, 4, 6 and 8.

The factors of 18 are 1, 2, 3, 6, 9 and 18.

The common factors of 12 and 18 are 1, 2 and 3.

Methods for Finding Common Factors

There are two methods to find the common factors of numbers. Listed below are the steps to be followed in finding common factors.
Method I:

• List all the factors of the numbers.
• Collect the common factors among each digit.

Method II:

• Find the prime factors of each digit.
• Merge the common terms of each digit.

Examples

Listed Below are some of the examples in finding the common factors.
Example 1:
What are the Common factors of 34 and 36?
Solution:
The factors of 34 are 1, 2, 17, and 34.
The factors of 36 are 1, 2, 3, 4, 6, 9, 12, 18, and 36.
Now, using method 1,
The common factors to the two numbers are 1 and 2.
The Common factors of 34 and 36 are 1 and 2.
Example 2:
What are the common factors of 40, 45 and 50?
Solution:
Prime factors of 40 are 1, 2, 4, 5, 8, 10, 20, and 40.
Prime factors of 45 are 1, 3, 5, 9, 15 and 45
Prime factors of 50 are 1, 2, 5, 10, 25, and 50.
Now, using method 2,
The common factors of 40, 45 and 50 are 1, 2 and 5.

Practice Problems

Listed below are some of the practice problems in finding the common factors.

Problem 1:

What are the Common factors of 8, 14, 18 and 22?

Answer:    

8   `->` 1, 2, 4, and 8

14 `->` 1, 2, 7 and 14

18 `->` 1, 2, 3, 6, 9 and 18

22 `->` 1, 2, 11 and 22

My forthcoming post is on Series Solutions of Differential Equations will give you more understanding about Algebra.

Problem 2:

What are the Common factors of 15, 30, 45 and 60?

Answer:    

15 `->` 1, 3, 5 and 15

30 `->` 1, 2, 3, 5, 6, 10, 15 and 30

45 `->` 1, 3, 5, 9, 15 and 45

60 `->` 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, and 60

Review of Word Problems

Introduction on review of word problems are first used to identify the variables present in the given problem. Also the word problems are used for the unit numbers and unit variables. Word problems are used for finding the phrases. The consideration includes in this is, we have to analyse the problem present in the statements.

I like to share this Fun Math Problems with you all through my article.

Types of word problems:

There are many types review word problems present. They are,
1.Word problem for numbers
2.Word problem for mixtures
3.Word problem for Age
4.Word problem for time
5.Word problem for linear
Word problem for numbers:
          In word problem for numbers, we  first review the relationships are identified. The variables present in the word problems are reduced.
Word problem for mixtures:
       Different types of concentrations are included in this word problem mixtures.
Word problem for age:
       The relationship of ages are calculated on this word problem.
Word problem for time:
     In this word problem, we review  the various types of train problems are included. Time taken for a train to go and coming back are included in this word problem.
Word problem for linear:
      Linear word problem mostly review the cost problems. For example, the cost for an grapes and the apple is 14.etc..





Algebra is widely used in day to day activities watch out for my forthcoming posts on find the prime factorization of 125 and cbse neet 2013. I am sure they will be helpful.

Example for review word problem:

Question: 1. The difference of twice a number 4 is 8. What is that number?
Solution:
Step 1: First we are changing the above said sentence as difference of twice a number 4 equals 8.
Step 2: Next step is to write the above said equation in the equation form.
     2V-4=8.
This is the solution for a word problem.
Question 2. The difference of twice a number 2 is 4. What is that number?
Solution:
Step 1: First we are changing the above said sentence as difference of twice a number 2 equals 4.
Step 2: Next step is to write the above said equation in the equation form.
     2V-2=4.
This is the solution for a word problem.
Practice to review word problem:
Practice 1.The difference of twice a number 12 is 14. What is that number?
Practice 2.The difference of twice a number 20 is 40. What is that number?

Factoring With Fractions

Factoring is the process of finding the factor of the given function. It is otherwise called as divide the given function as two separate part. We can factor the function more than two times. Fractions are also used in factoring process. Fraction means it has some numerator and denominator values. Both the numerator and denominator values are different. Now in this article, we see about ac factoring with fractions and their example problems.

Example problems for Ac factoring with fractions

Ac factoring with fractions example problem 1:

Factorize the given polynomial fraction equation `(1 / 6)` x2 - `(7 / 6)`x  - 20 = 0

Solution:

Given equation is `(1 / 6)` x2 - `(7 / 6)`x  - 20 = 0

For converting the above fraction equation in normal quadratic function,

Multiply the given equation by 6 on both the sides, we get

x2 - 7x - 120 = 0

Factorize the above equation, we get

x2 - 15x + 8x - 120 = 0

Grouping the first two terms and next two terms, we get

(x2 - 15x) + (8x - 120) = 0

x (x - 15) + 8 (x - 15) = 0

(x - 15) (x + 8) = 0

The factors are (x - 15) and (x + 8)

Answer:

The final answer is (x - 15) and (x + 8)

Ac factoring with fractions example problem 2:

Factorize the given polynomial fraction equation `(1 / 9)` x2 - `(4 / 3)` x + ` (11 / 9)` = 0

Solution:

Given equation is `(1 / 9)` x2 - `(4 / 3)` x + ` (11 / 9)` = 0

For converting the above fraction equation in normal quadratic function,

Multiply the given equation by 9 on both the sides, we get

x2 - 12x + 11 = 0

Factorize the above equation, we get

x2 - 11x - x + 11 = 0

Grouping the first two terms and next two terms, we get

(x2 - 11x) + (- x + 11) = 0

x (x - 11) - 1(x - 11) = 0

(x - 11) (x - 1) = 0

The factors are (x - 11) and (x - 1)

Answer:

The final answer is (x - 11) and (x - 1)

Ac factoring with fractions example problem 3:

Factorize the given polynomial fraction equation `(1 / 2)` x2 - `(11 / 2)` x + 14 = 0

Solution:

Given equation is `(1 / 2)` x2 - `(11 / 2)` x + 14 = 0

Multiply the given equation by 2 on both the sides, we get

x2 - 11x + 28 = 0

Factorize the above equation, we get

x2 - 7x - 4x + 28 = 0

Grouping the first two terms and next two terms, we get

(x2 - 7x) + (- 4x + 28) = 0

x (x - 7) - 4 (x - 7) = 0

(x - 4) (x - 7) = 0

The factors are (x - 4) and (x - 7)

My forthcoming post is on Rules for Dividing Integers and class 11 cbse books will give you more understanding about Algebra.

Answer:

The final answer is (x - 4) and (x - 7)

Adding Positive Integers

Let us see about adding positive integers. In mathematics the numbers or integers are very essential. The integers are either may be positive or it may be negative. The positive integers are represented as its normal numbers but the negative integers are represented as (-). Adding positive integers is the one of the operation in arithmetic operation. The integers are may be two digits or four digits or it may be more than that.

Definition:

Addition is the process of adding to integers. The positive integers are easy to add. It is also defined as, the group of objects mutually into a larger collection. It is signified by plus sign (+).Process of the addition is one of the easiest numerical tasks.

Examples:

Let us see about examples of positive integers.

Problem 1:

Adding the integers 24 and 35.

Solution:

The given integers are 24 and 35, both integers are positive integers. So, the addition is easy.

24

35  (+)

-------

59

----------.

This is the answer of the given problem.

Problem 2:

Adding then positive integers 12 and 55.

Solution:

The given integers are 12 and 55, both integers are positive integers. So, the addition is easy.

12

55  (+)

-------

67

----------.

This is the answer of the given problem.

Problem 3:

Adding the positive integers 61 and 76.

Solution:

The given integers are 61 and 76, both integers are positive integers. So, the addition is easy.

61

76  (+)

-------

137

----------.

This is the answer of the given problem.

Problem 4:

Adding the positive integers 90 and 40.

Solution:

The given integers are 90 and 40, both integers are positive integers. So, the addition is easy.

90

40  (+)

-------

130

----------.

This is the answer of the given problem.

Problem 5:

Adding the positive integers 35 and 49.

Solution:

The given integers are 35 and 49, both integers are positive integers. So, the addition is easy.

35

49  (+)

-------

84

----------.


Algebra is widely used in day to day activities watch out for my forthcoming posts on Random Variables Statistics . I am sure they will be helpful.

This is the answer of the given problem.

Problem 6:

Adding the positive integers 78 and 42.

Solution:

The given integers are 78 and 42, both integers are positive integers. So, the addition is easy.

78

42  (+)

-------

120

----------.

This is the answer of the given problem.

Non linear Differential Equations

An ordinary differential equation is an differential equation in For instance (i)dy/dx = x + 5 (ii) (y′)2 + (y′)3 + 3y = x2 which  is a single independent variable enters either explicitly or implicitly are all ordinary non linear differential equations And now we see about the non linear differential equations.

Non linear differential equations in First order

Order and degree of a non linear differential equations:

Definition of non linear differential equations :

The order of an differential equation is the order of the highest order derivative occurring in it. The degree of the differential equation is the highest order derivative which occurs in it, after the differential equation has been made free from radicals and fractions as far as the derivatives are concerned.The degree of a non linear differential equation does not require variables like r, s, t … to be free from radicals and fractions.

Find the order and degree of the following differential equations:

ii) y = 4dy/dx + 3xdx/dy

y = 4dy/dx + 3xdx/dy  y = 4(dy/dx) + 3x 1/(dy/dx )

Making the above equation free from fractions involving

dy/dx we get y .dy/dx = 4(dy/dx)2 + 3x 2Highest order = 1

Degree of Highest order = 2

(order, degree) = (1, 2)

Problem for non linear differential equations:

Formation of non linear differential equations :

Form the differential equation from the following equations.

y = e2x (A + Bx)

ye−2x = A+ Bx … (1)

Since the above equation contains two arbitrary constants, differentiating

twice, we get y′e−2x − 2y e−2x = B

{y′′e−2x − 2y′ e−2x} − 2{y′e−2x − 2y e−2x} = 0

e−2x {y′′ − 4y′ + 4y} = 0 [‡ e−2x ≠ 0]

y′′ − 4y′ + 4y = 0 is the differential equation.

Non linear Differential equations of first order and also first degree :

Solve: 3ex tan y dx + (1 + ex) sec2y dy = 0

`=>` 3 log (1 + ex) + log tan y = log c

`=>` log [tan y (1 + ex)3] = log c

`=>`   (1 + ex)3

required solution to the problem is tan y = c

Non linear Differential Equations

An ordinary differential equation is an differential equation in For instance (i)dy/dx = x + 5 (ii) (y′)2 + (y′)3 + 3y = x2 which  is a single independent variable enters either explicitly or implicitly are all ordinary non linear differential equations And now we see about the non linear differential equations.

Non linear differential equations in First order

Order and degree of a non linear differential equations:

Definition of non linear differential equations :

The order of an differential equation is the order of the highest order derivative occurring in it. The degree of the differential equation is the highest order derivative which occurs in it, after the differential equation has been made free from radicals and fractions as far as the derivatives are concerned.The degree of a non linear differential equation does not require variables like r, s, t … to be free from radicals and fractions.

Find the order and degree of the following differential equations:

ii) y = 4dy/dx + 3xdx/dy

y = 4dy/dx + 3xdx/dy  y = 4(dy/dx) + 3x 1/(dy/dx )

Making the above equation free from fractions involving

dy/dx we get y .dy/dx = 4(dy/dx)2 + 3x 2Highest order = 1

Degree of Highest order = 2

(order, degree) = (1, 2)

Problem for non linear differential equations:

Formation of non linear differential equations :

Form the differential equation from the following equations.

y = e2x (A + Bx)

ye−2x = A+ Bx … (1)

Since the above equation contains two arbitrary constants, differentiating

twice, we get y′e−2x − 2y e−2x = B

{y′′e−2x − 2y′ e−2x} − 2{y′e−2x − 2y e−2x} = 0

e−2x {y′′ − 4y′ + 4y} = 0 [‡ e−2x ≠ 0]

y′′ − 4y′ + 4y = 0 is the differential equation.

Non linear Differential equations of first order and also first degree :

Solve: 3ex tan y dx + (1 + ex) sec2y dy = 0

`=>` 3 log (1 + ex) + log tan y = log c

`=>` log [tan y (1 + ex)3] = log c

`=>`   (1 + ex)3

required solution to the problem is tan y = c

Radicals and Exponents

The opposite operation to the exponent is known as radical. A radical is an expression which contains the square roots, cube roots etc. For example the expression √49 can also be called as square root of 49 or root of 49. Radicals have the same property of the numbers. The simplification is to reduce the numbers and reduce the power of variables inside the roots.

Exponent of a number shows you how many times the number is to be used in a multiplication.  It is shown as a small number to the right and above the base number.  In this example: 22 = 2 × 2 = 4 ( Another name for exponent is index or power ).

Please express your views of this topic Fraction Exponents by commenting on blog.

Examples for Radicals:

Example 1 for radical problem:

Simplify the radical:    √ (144) / (√36)

Step 1: Factors of 144 = 12 × 12

√ (144) = √ (12 × 12)

= √122

Step 2: Square root of 122 = 12

Step 3: Factors of 36 = 6 × 6

√ (36) = √ (6 × 6) = √62

Step 4:  √62 = 6

Step 5: so, √ (144) / √36) = 12 / 6

The answer of the given radical is 12 / 6
Example 2 for radical problem:

Simplify the radical:    3√8 × √12

Step 1: Find the factor of 8 = 2 × 2 × 2

Step 2: Take cubic root of 8 = 2

Step 3: Find the factor for 12 = 2 × 2 × 3

Step 4: Take square root of 12 = 2 √3

Step 5: so, 3√8 × √12 = 2 × 2 √3

The answer for the given radical is = 4√3

Examples for exponents:

EXAMPLE-1:

453 . 454 = 453+4 = 457

EXAMPLE-2:

148 / 145  = 14 8-5 = 143

EXAMPLE-3:

( 92 )3  = 9 2x3  = 96

EXAMPLE-4:

( 11 / 13 )2  = 112 / 122

Algebra is widely used in day to day activities watch out for my forthcoming posts on Triple Integral Spherical Coordinates and Calculus Integration by Parts. I am sure they will be helpful.

EXAMPLE-5:

70 = 1

EXAMPLE-6:

5-3 = 1/ 53

Inverse Variation

To understand the inverse variation,let is correlate it with our real life of inverse variation examples such as the total time taken with the speed traveled in a trip. Assuming distance of a trip would be fixed, let us say the distance of a trip would be 250 mile.

I like to share this Inverse Sine Function with you all through my article. 

Here  inverse variation equation will be used as T=250/r , according to the time will be  T be time of the trip will be equal to 250  divided by the Rate in miles per hour which will called r .so if rate is 50 miles per hour it would take,  5 hours  .

The next of the inverse example is total time taken to spread the soil and the number of people working on a part of land. so let us say if the total time of the job will take 12 hours the total time T to complete the job would be  12 hours divided by n, ( n is the number of workers).

Here the equation will be T= 12/n.  so if have 1 worker , we take 12 hours, 3 workers we  take 4 hours and so on.

Total amount of cost required by per person for petrol and the number of people in a cab. So if we know that it will cost 45 dollars’ worth of petrol to go for a picnic trip. The cost represents C= 45 dollars divided by number of people in the cab.

That is C= 45/no of people. So if n will increases the total cost decreases.
Here in inverse variation definition is as the ones quantityvalue gets bigger and the others value gets smaller in such that there product remains the same or proportional. Let us understand with one more example, Y varies inversely as X. where Y= 4 and when X=2.

Algebra is widely used in day to day activities watch out for my forthcoming posts on math problem solver online and math problem solver for free. I am sure they will be helpful.

Find the value of the inverse-variation equation. Then determine Y when X= 16.  By performing substitution we can solve this question.  Y= k/x.  by doing cross multiplication , we get 4= k/2 which is k=8.  So our inverse-variation will be  Y=K/x that is 8/x. Now we can answer the second query on determining the y when x is 16. So, y= 8/x , substituting the value of x as 16. So y= 8/16. And we get  1/2 . y= ½ and x= 16.

Solve Laplace Equation

In mathematics, Laplace equation is one of the most interesting topics in second order partial differential equations. The two dimensional heat equations in Cartesian form in unsteady state is

`(delu)/(delt)` = `alpha^2 [(del^2u)/(delx^2) + (del^2u)/(dely^2)]`

I like to share this Laplace Transform Pairs with you all through my article.

For the steady state `(delu)/(delt)` = 0, then the above equation becomes

`[(del^2u)/(delx^2) + (del^2u)/(dely^2)]` = 0

(or)

`grad^2 u` = 0

This is the two dimensional heat equation in unsteady state (or) Laplace equation. In this article, we shall discuss about how to solve the Laplace equation. The following are the example problem in solve Laplace equation.

Solve Laplace equation - Laplace equation:

Finite difference solution for the two dimensional heat equation in steady state (or) Laplace equation is

`grad^2 u` = 0

(or)

`(del^2u)/(delx^2)` + `(del^2u)/(dely^2)` = 0

Here we know the values of `(del^2u)/(delx^2)` , `(del^2u/dely^2)`

`(del^2u)/(delx^2)` = `(u_(i+1, j) + u_(i-1, j)-2u_(ij))/(h^2)`

`(del^2u)/(dely^2)` = `(u_(i, j+1) + u_(i, j-1)-2u_(ij))/(h^2)`

substitute the values of `(del^2u)/(delx^2)` , `(del^2u)/(dely^2)` in the Laplace equation, then we get

`4u_(ij)` = `(u_(i+1, j) + u_(i-1, j) + u_(i, j+1) + u_(i, j-1))`

`u_(ij)` = (`1/4` ) `(u_(i+1, j) + u_(i-1, j) + u_(i, j+1) + u_(i, j-1))`

This is the finite difference solution for the Laplace equation, which is also called as the standard five-point formula.

Solve Laplace equation

Solve Laplace equation - Example problem:

Example 1:

Solve the laplace equation for the given region

Solve Laplace equation - Example

Solution:

The finite difference sheme for the Laplace equation is

Solve Laplace equation - Example

u(ij) = (`1/4` ) [`(u_(i+1, j) + u_(i-1, j) + u_(i, j+1) + u_(i, j-1))` ]

Using this we can solve this,

u1 = `1/4` [200 + 50 + u2 + u4]

= `1/4` [250 + u2 + u4]                                → (1)

u2 = `1/4` [u1 + 0 + 0 + u3]

= `1/4` [u1 + u3]                                         → (2)

u3 = `1/4` [u4 + u2 + 100 + 50]

= `1/4` [150 + u2 + u4]                                → (3)

u4 = `1/4` [100 + 200 + u1 +  u3]

= `1/4` [300 + u1 + u3]                                → (4)

My forthcoming post is on Images of Acute Angles and cbse cce syllabus for class 10 will give you more understanding about Algebra.

Solve the equation  by using the gauss seidal method

u1 = `1/4` [250 + u2 + u4]

u2 = `1/4` [u1 + u3]

u3 = `1/4` [150 + u2 + u4]

u4 = `1/4` [300 + u1 + u3]

Let the intial approximation is

Learn Slope Form

The ratio of rise value of the line equation to the run value of the line equation is called as slope. Slope form learning is one of the important part of algebra. In other words, it can be defined as the ratio of the change of  Y values to the change of the X value. Generally, the slope of the equation is denoted as "m". It is necessary to learn the slope form. learn slope form is mainly used in linear equation.

learn slope form of the linear equation

The slope form of the linear equation can be written as,

y = mx + b

Where, y is the line equation,

m is the slope of the equation and

b is the Y intercept value.

In point form, the line equation can be written as,

Y - Y1 = m (X - X1)

Where, X1 and y1 are the points of the line

m is the slope of the equation.

I like to share this Math Slope Formula with you all through my article.

Given two points, we could find the slope using the formula

m = (Y2 - Y1) / (X2 - X1)

Where, (X1, Y1 ) and  (X2, Y2 ) are any two points in the line.

Example problems for learn slope form

Ex:1 Find the slope of the given equation Y = 7X + 6.

Sol: In general form, the line equation is given as,

y = mx + b

where, m is the slope of the equation.

In the above equation, the slope is coefficient of x,

slope = 7

Ex:2 Find the slope of the given points (1, 3) and (3, 6).

Sol:

The given points are (1, 3) and (3, 6).

Here, X1 = 1, Y1 =3

X2 = 3, Y2 = 6

The formula for finding the slope is given as,

m = (Y2 - Y1) / (X2 - X1)

Substitute all  values in the above equation,

m = (6 – 3) / (3 -1)

= 3 / 2

Ans:

The slope of the given equation is m= 3/2

Ex:3 Find the slope of the given points (2, 6) and (6, 14).

Sol: The given points are (2, 6) and (6, 16)

Here, X1 = 2, Y1 = 6

X2 = 6, Y2 = 16

The formula for finding the slope is given as,

m = (Y2 - Y1) / (X2 - X1)

Substitute all  values in the above equation,

m = (16 – 6) / (6 -2)

= 5 / 2

Ans:

The slope of the given equation is m = 5 / 2

Expressions With Rational Exponents

A rational number is a number represented as p/q where p and q are integers and q != 0.Exponent means  power.

The large numbers are very difficult to read and write and understand. To make them simpler we can use exponents, with the help of this many of the large numbers are converted to simpler forms. The short notation 74 stands for the product 7 x 7 x 7x 7. Here ‘7’ is called the base and ‘4’ the exponents. The number 74 is read as 7 raised to the power of 4 or simply as fourth power of 4. 74 are called the exponential form of 2401.

The above example is of a natural number as exponents.

Ex 5 ^(3/2) here 3/2 is a rational number so this is an example of rational exponents.

Rules invovled for solving rational exponents:

The Rational exponents rules are:

p,q are any real numbers except zero and m,n are positive integers.

Rule I:The base value is same under multiplication so we can directly add the power values. Then the exponent take the following form

pm  × pn = pm+n

Solving the expression 73  × 75 = 73+5 =78

Rule II:The base value is same under division so we can directly subtract the power values. Then the exponent take the following form such that m > n then

(p^m)/(p^n) = pm-n

Solving the expression  (4^4)/(4^2)= 44-2=42

Rule III:If m < n, (p^m)/(p^n) =(1)/(p^(n-m))

Solving the expression 8^3/8^2 =1/8

Rule IV:A real number to the power of power

(pm )n  = pm*n

Solving (82 )3  = 82*3=86

Rule V:The power of 0 is

p0 = 1. Anything power zero is equals to 1.

p^m/p^m   = pm-m =p0 = 1

Rule VI:p to the power of negative number is

p-m =1/p^m
Solving the expression 2-5 =  1/2^5

Rule VII:Two numbers to the same power can be written as

pm  × qm = (p*q)m

Solving the expression 32 × 82 = (3*8)2=242

Rule VIII A number to rational power is written as

p^(m/n) = root(n)(p^m)

Examples of expressions having rational exponents:

Ex : 1Solve the exponent    2x^(1/3)xx8^(5/3)xxx^(2/3)

Sol:Step 1:Given

2x^(1/3)xx8^(5/3) xx x^(2/3)

Step 2:Exponents are added

=2x^(1/3+2/3) (2^3)^(5/3)

Step 3: 8 is written as 2 power 3

Step 4 : Simplify=2x2^5

=2xx32 x

=64 x

Ex 2: Solve x^(-1/3)8^(-2/3)

x^(-1/3)8^(-2/3)

Solu: Step 1  x^(-1/3) 8^(-2/3)

Step2: 8^(-2/3) = (2^3)^(-2/3) = 2^(3 xx-2/3) = 2^-2 = 1/2^2 = 1/4

My forthcoming post is on  Define Arithmetic Mean will give you more understanding about Algebra.

Step3 Simplify the variable with negative exponent

=1/(x^(1/3)) 1/4

Step 4 Simplify:=1/(4x^(1/3))

=1/(4root(3)(x))

Power Coefficient Equation

A coefficient is a multiplicative factor in a few phrase of an expression. It is typically a number, but in any case does not occupy some variables of the equation. For example in equation 7x2 − 3xy + 1.5 + y the first three stipulations correspondingly have coefficients 7, −3, and 1.5. In the third expression there are no variables, so the coefficient is the idiom itself called the power constant term or constant power coefficient of this expression

I like to share this Sample Correlation Coefficient with you all through my article.

Example Problem for the Coefficient:

Find the coefficient of the given equation:

X2 + 5y3 +2xy2 + 5yx2

Solution:

Given that X2 + 5Y3+ 2XY2 + 5YX2

Here X2+ 5Y3+ 2XY2+ 5YX2 the expression with the coefficient

The coefficient of x2 is 1

The coefficient of  5Y3 is 5

The coefficient of 2xy2 is 2

The coefficient of 5yx2 is 5.

Coefficient Of Varience

Coefficient of variation is a relative determine for standard deviation.

It is defined as 100 times the coefficient of spreading base in the lead standard difference is called coefficient power of variation.

Less C.V. designate the fewer variability or additional power.

More C.V. indicate the more variability or less power.

According to Karl Pearson who suggested this compute the equation, C.V. is the percentage dissimilarity in the mean, standard deviation being considered as the total variation in the mean.

With the help of C.V. we can find which salesman is more consistent in making sales, which batsman is more consistent in scoring runs, which student is more consistent in scoring marks, which worker is more consistent in production.

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Coefficients Of Dispersion

Whenever we want to compare the variability of two equations which differ widely in their averages or which are measured in different units.

We do not merely compute the measures of dispersions in the equations but we calculate the coefficients of dispersion which are pure numbers autonomous of the units of measurement.

Quotient Rule for Integration

The quotient rule is a method of finding the integration of a function that is the quotient of two other functions for which derivatives exist.

By the Product Rule,

if f (x) and g(x) are differentiable functions, then
d/dx[f (x)g(x)]= f (x)g'(x) + g(x) f' (x).
Integrating on both sides of this equation,

∫[f (x)g'(x) + g(x) f '(x)]dx = f (x)g(x),
which may be rearranged to obtain

∫f (x)g'(x) dx = f (x)g(x) −∫g(x) f' (x) dx.     (A)
Letting U = f (x) and V = g(x)

then differentiating it  we get

dU = f '(x) dx and dV =g'(x) dx,

pluging these values in (A), we get

∫U dV = UV −∫V dU.                  (1).


By the Quotient Rule,

if f (x) and g(x) are differentiable functions, then

d/dx[f (x)/g(x)]= g(x) f '(x) − f (x)g'(x)/[g(x)]2 .
Integrating both sides of this equation, we get
[f (x)/g(x)]=∫g(x) f '(x) − f (x)g'(x)/[g(x)]2 dx.
That is,

f (x)/g(x)=∫f '(x)/g(x)dx -∫f (x)g'(x)/[g(x)]2 dx,
which may be rearranged to obtain

∫f '(x)g(x)dx = f (x)g(x)+∫f (x)g'(x)/[g(x)]2 dx.      (B)

Letting u = g(x) and v = f (x) and then differentiating it , we get

du = g'(x) dx and

dv = f '(x) dx,
we obtain a Quotient Rule Integration by Parts formula:

plugging values of u , v , du and dv in B we get
∫dv/u= v/u+∫(v/u²) du.                                  (2)

quotient rule for integration-Application

∫[sin(x−½)/x²] dx.
Let
u = x½, du=1/2(x-½)

v=2cos(x-½),dv = sin(x−1/2)/x³/²

Then

∫sin(x−½)/x² dx = 2 cos(x-½)/x½+∫2 cos(x-½)/x• 1/2(x-½) dx

= 2 cos(x-½)/x½− ∫2cos(x-½) ·•[−(x-³/²)/2]dx

= 2 cos(x-½)/x½− 2 sin(x-½) + C,
which may be easily verified as correct.

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quotient rule for integration -Illustration


The Quotient Rule Integration by Parts formula (2) results from applying the
standard Integration by Parts formula (1) to the integral

Let ∫dv/u

with

U = 1/u,V = v,then differentiating ,we get

dU = − 1/u² du,dV=dv
plugging these values

∫dv/u=∫U dV

= UV −∫V dU

= 1/u·( v) −∫v (− 1/u²)du
= v/u+∫v/u² du

Terminating Decimals are Rational Numbers

Rational numbers are contrast among irrational numbers like Pi and square roots and sins and logarithms of information. On rational numbers and the ending of the article you be able to click on a linkage to maintain studying about irrational numbers. A number is rational but you can note down it in a form a/b where a and b are integers, b not zero. Visibly all fractions are of that form.

Terminating decimals are rational numbers:

A rational number is several number that be able to be expressed as the section a/b of two integers, among the denominator b not equivalent to zero. While b could be equal to 1, every one integer is a rational number. The set of every rational in sequence is frequently denoted by a boldface Q stands for quotient.

The terminating decimal extension of a rational numbers forever whichever terminating follows finitely several digits or begins to replicate the similar sequence of digits over and over. Moreover, any repeating or terminating decimal represent a rational number. These statements hold true not now meant for base 10, but as well for binary, hexadecimal, or any further integer base. Terminate decimal numbers be able to simply be written in that form: for example 0.67 is 67/100, 3.40938 = 340938/100000.

A real number so as to be not rational is called irrational. Irrational numbers contain square root 2, pi, and e. The decimal extension of an irrational number continues evermore without repeating. Because locate of rational numbers is countable, and the position of real numbers is uncountable, just about each real number is irrational.

In terminating conceptual algebra, the rational numbers shape a field. This is the representative field of characteristic zero, with is the field of fractions for the ring of integers. Finite extensions of Q are called algebraic number fields, and the algebra finality of Q is the field of algebraic number.

In mathematical study, the rational numbers shape a dense separation of the real numbers. The real numbers is able to be constructed from the rational numbers by achievement, using either Cauchv sequences or Dedekind cuts.

Zero separated by any other integer equals zero, consequently zero is a rational number even though division by zero itself is indeterminate.

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Solving Positive Direction Coordinates

A coordinate is a number that determine the position of a point the length of a number of line or arc. A list of two, three, otherwise additional coordinates can be use to establish the position of a point on a surface, volume, otherwise higher-dimensional area.

The Positive direction represents the positive value for both the x axis and y axis. The x axis value and the y axis value should be positive to move towards the positive direction. Any of the negative change in the axis will not lead to positive direction.

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Solving positive direction examples:

Solve the following equation:

Y = x2 +2

Solution :

Now we have to substitute values for the variable x in the given equation.

When we substitute any positive value for the given equation we get only positive output values.

That is when  x = 1

Y = (1)2 +2

Y = 1+2

Y = 3

The co ordinates will (1,3) which lies on the positive direction.

when we substitute x =2

We get

Y = (2)2 + 2

Y = 4 + 2

Y = 6

The co ordinates are (2,6) which lies on the positive direction.

Similarly when we substitute any positive value for the given equation the output will be positive.

Even though we give negative values, the output will be (that is y) positive. But we need to give positive values to move in the positive direction.

Solving positive direction example:

Solve the following equation:

Y = 3x +4

Solution:

Now we have to substitute values for the variable x in the given equation.

When we substitute any positive value for the given equation we get only positive output values.

That is when x = 4

Y = 3(4) +4

Y = 12+4

Y = 16

The co ordinates will (4,16) which lies on the positive direction.

when we substitute x =6

We get

Y = 3(6) + 4

Y = 18 + 4

Y = 22

The co ordinates are (6,22) which lies on the positive direction.

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Similarly when we substitute any positive value for the given equation the output will be positive.

Here when we substitute negative values for the variable x, we get negative values as output also. So we need to substitute positive values for the equation.

Poisson Distribution

If n is large, the evaluation of the binomial distribution can involve considerable computation. In such a case a simple approximation to the binomial probability could be considerable use. The approximation of binomial when n is large and p is close to zero is called the Poisson Distribution Mean.

Definition:

A random variable is said to follows Poisson distribution if it assumes only non-negative values and its probability mass function is given by

P(x,lambda ) =P(X=x) = elambda lambda x  /  X!    

x=0, 1, 2…

O            

otherwise

lambda is known as parameter of Poisson distribution .X~p() denotes it.

Characteristic function of Poisson distribution

phi x ( t )  =E[ei t x ]

=sum_(x=0)^oo ei t x   e-lambda lambda x / x!

=  e-lambda sum_(x=0)^oo ei t x lambda x  / x!

=  e-lambda [1+(lambda eit ) + (lambda eit)2 / 2! +(lambda ei t)3  /3!  + ............. ]

=e-lambda elambda eit

=  elambda (ei t -1)

Additive property of Poisson distribution:

Independent Poisson variate is also a Poisson variate xi (i=1,2,......n). xi follows Poisson with parameter lambda i.

xi ~ P(lambda i )    (i=1,2.....n)

then sum_(i=1)^n    xi ~ P(sum_(i=1)^n lambda i )

Proof:

Mxi(t) = elambda i(e^t -1)

Mx1+x2+x3+.....Xn(t) = Mx1 (t) Mx2 (t). . . Mxn (t)

= elambda 1(e^t-1). elambda 2(e^t-1) ..........elambda n(e^t-1)

=e(lambda 1 +lambda 2 +lambda 3+.......lambda n )(et-1)

=esum_(i=1)^n lambda i(et-1)

M.G.F of Poisson distribution:

Mx ( t ) = E [ etx ]

= sum_(x=0)^oo etx   e-lambda lambda x  / x!

= e-lambda sum_(x=0)^oo etxlambda x / x!

= e-lambda [ 1+ et lambda +  (lambda et )2 /2! + (  (lambda et )3  / 3! +...... ]

=e-lambda elambda

= elambda (et -1)

Applications of Poisson distribution

The following are some instance where the distribution is applicable

Number of deaths from a disease
Number of suicide reported in a particular city.
Number of defective materials in packing manufactured by good concern.
Number of printing mistakes at each page of the book.
Number of air accidents in some unit of the time.

Needs Assessment Definition

Assessment is an important and essential part of teaching. If teachers are to ask themselves whether what they are teaching and how they are teaching has the desired outcomes, they will needs to assess what children are able to do according to a set of criteria.

The criteria, which indicate what children, should be able to do and think in mathematics by the end of the foundation phase.

Teachers need to ask themselves with assessment:

The following are two very important questions that teachers have to ask when teaching.

Is what I am doing helping children to develop a desire to learn mathematics?

Is what I am doing teaching children to become numerate?

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Why we needs assessment?

Some reasons for assessing learners are so that we can measure whether a lesson has been effective for each learner (whether we have taught the lesson effectively enough). We assess learners to see what each one can do, for example, which learners are able to calculate change, or which learners are able to solve relevant word problem. We assess learners to see which of the children are ready for a new challenge and which must still practice what has already been taught. We assess learners so that we can plan further lessons that suit the needs of the children.

Types of assessment:

There are two main ways in which to assess children

Formative assessment

Summative assessment.

Formative assessment is assessing a learner while the learner is forming the new knowledge.

Example for formative assessment:

An example of formative assessment would be sitting with a learner while he or she is doing a task (say using a number line to count in groups), watching how the child goes about the task and asking the child to explain how and what he or she is doing. In this way, you find out what strategies the child is using and developing and what strategies you should be helping the child with; you are getting direct and instant feedback on hoe the child is coping and you are able to respond to the situation immediately through re-teaching and explaining again, asking anther learner to help, or planning another lesson on that needs for the next day.

Summative assessment is assessing a learner at the end of the lesson, section, topic, quarter or year as a summing up of what the learner knows. Therefore, tests and exams are summative versions of assessment.

When both formative and summative assessments are used, that is continuous assessment. In an outcomes-based education system, continuous assessment is used. The teacher studies the learning outcomes required of the learners and then plans lessons to teach to achieve these outcomes. During the lessons, the teacher observers what children are doing and saying and how children are doing a task. The teacher asks for explanations from the children as to what and how they are doing a take. The teacher helps those learners who are confused and continually monitors which learners are gaining control of the skills and concepts. Once a child can do something independently within the number range for that learner, a teacher can say that that child has learnt what was intended by the lesson and so can record that performance as a desired learning outcome for that child.

Learn Horizontal Line Test

Definition of function:

The mathematical concept of a function expresses the intuitive idea that one quantity (the argument of the function, also known as the input) completely determines another quantity (the value, or the output). A function assigns a unique value to each input of a specified type. The argument and the value may be real numbers, but they can also be elements from any given sets: the domain and the codomain of the function. An example of a function with the real numbers as both its domain and codomain is the function f(x) = 2x, which assigns to every real number the real number with twice its value. In this case, it is written that f(5) = 10.

A relation f:A`->B ` is said to be a function if every element in domain(A) has an unique image in codomain(B).

Definition of One-to-One Function:

A function is said to be One-to-One if no two different elements in domain have same image in codomain.The definition of one-to-one function can be written algebraically as follows:

Let a  and b any elements of domain.

A function f(x) is said to be one-to-one

1.if a is not equal to b then f(a) is not equal to f(b)

OR contra positive of the above

2.if f(a)=f(b) then a=b

Horizontal Line Test::

The horizontal line test is used to determine if a function is one-to-one.The lines used for this test are parallel to x axis.

If the function is one-to-one, then it can be visualized as one whose graph is never intersected by any horizontal line more than once.



If and only if  f is onto, any horizontal line will intersect the graph at least at one point (when the horizontal line is in the codomain).



If f is bijective, any line horizontal or vertical will intersect the graph at exactly one point.


Graph of one-to-one function:

If a line is drawn parallel to x axis (horizontal line) to this curve  then it will cut the curve at only one point so it is an one-to-one function.

Graph of a function which is not one-to-one:

If a line is drawn parallel to x axis (horizontal line) to this curve  then it will cut the curve at more than  one point so it is not an one-to-one function.

Algebra is widely used in day to day activities watch out for my forthcoming posts on answers to math problems for free and cbse 9th class science book. I am sure they will be helpful.

Meaning of Correlation Coefficient

Two variables are related in such a way that:

(i) if there is an increase in one accompanied by an accompanied by a decrease in the other, Then the variables are said to be correlation The value of  the correlation will be in the interval [1, -1].

If the value of the correlation is positive then it is direct and if it is negative, then it is inverse.

If  the value of correlation is 1, it is said to be perfect positive correlation. If it is -1, it is said to be perfect negative correlation. If  the correlation is zero, then there is no correlation.

The formula to find the correlation is [sum dx dy] /[ sqrt [sum d_x^2 . sum d_y^ 2]]

where dx = x – barx ,  dy = y – bary

Now let us see few problems on correlation.

Example problems on meaning of correlation coefficient:

Ex 1: Find the correlation between two following set of data:

Cor_Tab1

Soln: barX = 180 / 9 = 20,

barY = 360 / 9 = 40.

Cor_SolTab1

Therefore the correlation =  [sum dx dy] /[ sqrt [sum d_x^2 . sum d_y^ 2]] = 193 / [sqrt [120 xx346]] = 0.94

This value shows that there is a very high relationship between x and y.

More example problem on meaning of correlation coefficient:

Ex 2: Find the correlation between the following set of data.

Cor_Tab2

Soln: barX = 36 / 6 = 6,  barY = 60 / 6 = 10

Cor_SolTab2

Therefore the correlation is   [sum dx dy] /[ sqrt [sum d_x^2 . sum d_y^ 2]]  = -67 / [sqrt [50 xx 106]]

= -0.92

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This value shows that there is a very low relationship between x and y.

By now the meaning of correlation coefficient will be more clearer. I believe that these examples would have helped you to do problems on correlation coefficient.

Average Velocity

Before going to the concept of average velocity,you should know about the concept of velocity.So, here is the introduction to velocity.

Velocity - the rate of change of position of an object. Velocity is a vector physical quantity because it requires both magnitude and direction to define. Speed is an absolute scalar value (magnitude) of velocity. Velocity is measured in meter/second (m/s).

In mathematics, velocity is the ratio of  displacement and  during the time taken to the interval. As it is associated to how fast waves travel between layers of the earth, it also tells us of how compact they are in between.

For example, 10 meter per second is a scalar value of velocity, but 10 m/s east is a vector. The rate of change in the velocity is called as acceleration.

Average velocity:

The Average velocity topic is dealt under mathematics. Students get to learn how to calculate the average velocity when the rest of the values are given. Students get to learn to practice various word problems on the concept of average velocity and also they get ample support from the online tutors regarding their homework problems and examination preparation.They can get all the answers for homework problems.

The “Rate of change(derivative) of position at which an object ” is known as velocity . It is a vector physical quantity; both magnitude and direction are essential to define it. The scalar complete value (magnitude) of velocity is speed, a quantity that is measured in meters per second (m/s or ms-1) when using the SI (metric) system.

The average velocity v of an object affecting through a displacement  (?x) during a time interval (?t) is described by the formula.Here is the Formula for average velocity:

V = Final displacement / total time taken

=  ?x / ?t

Students can get more detailed explanation on the topic on the Physics help page.

Finding average velocity

Example 1: If an object is thrown from the ground at an initial velocity of 10 meter per second, after t seconds the height of the object “h” in meter is given by,   h = t2 + 10t.  Find the time taken to the object to reach a height of 200 meter.

Solution:

Given, Height, h = 200 meter

h = t2 + 10t --> (1)

t =? (At 200 meter)
Plug h = 200 in equation (1)

200 = t2 + 10t

t2 + 10t – 200 = 0

By solving the quadratic equation,

t2 + 20t -10t -200 = 0

t(t+20) – 10(t+20) = 0

(t+20) (t -10) = 0

t = -20 or t = 10

We have two values for time “t”. Since time cannot be negative, any problem that gives a negative answer for one of its answers is always false, so we just go for the positive value.
So the final answer is 10 seconds.

So, the object reaches the height (200 meters) in 10 seconds.

Students can get more solved examples and problems to practice upon on the mastering physics page.

Calculate average velocity

Here is one more example of how to Calculate average velocity.

Example :  A particle moving along the x axis is located at 17.2 m at 1.79 s and at 4.48 m at 4.28 s. what is the average velocity in the particular time interval?

Solution:

Distance traveled by particle beside x-axis = Final location  - Initial location

= 17.20 - 4.48

= 12.72 meters

Time taken to cover up this distance = Final time - Initial time

= 4.28 - 1.79

= 2.49 seconds

Average velocity    = (Distance traveled)/(Time taken)

= 12.72 / 2.48

= 5.1084 (approximately)

Average velocity of the particle is    5.1084 m/s

Bijective Function Examples

In mathematics, a bijection, or a bijective function, is a function f from a set X to a set Y with the property that, for every y in Y, there is exactly one x in X such that f(x) = y and no unmapped element exists in either X or Y.

A one - one onto function is said to be bijective or a one-to-one correspondence.

A few examples of bijective function is given below which helps you for learning bijective function.

(Source: Wikipedia)

Examples of bijective function:

Example 1:

Show that the function f : R → R : f(x) = 3 - 4x  is one-one onto and hence bijective .

Solution:

We have

f(x1) = f(x2)

3 - 4x1 = 3 - 4x2

x1 = x2

Therefore, the function f is one-one.

Now, let y = 3 - 4x. Then, x = (3 - y)/4

Thus, for each y ε R (codomain of f), there exists x = (3 - y)/4 ε R

such that f(x) = f((3 - y)/4)

= {3 - 4 (3 - y)/4 }

= 3 - (3 - y)

= y

This shows that every element in codomain of f has its pre-image in dom(f).

Therefore, the function f is onto.

Hence, the given function is bijective.

Example 2:

Let A = R - {3} and B = R - {1}. Let f : A → B : f(x) = (x - 2)/(x - 3) for all values of x ε A.

Show that f is one-one and onto.

Solution:

f is one-one, since

f(x1) = f(x2)

(x1 - 2)/(x1 - 3) = (x2 - 2)/(x2 - 3)

(x1 - 2)(x2 - 3) = (x2 - 2)(x1 - 3)

x1x2 - 3x1 - 2x2 + 6 = x1x2 - 2x1 - 3x2 + 6

x1 = x2

Let y ε B such that y =  (x - 2)/(x - 3) .

Then, (x - 3)y = (x - 2)

x = (3y - 2)/(y - 1)

Clearly, x is defined when y ≠ 1.

Also, x = 3 will give us 1 = 0, which is false.

Therefore,

x ≠ 3.

And, f(x) = ((3y - 2)/(y - 1) - 2)/((3y - 2)/(y - 1)- 3) = y

Thus, for each y ε B, there exists x ε A such that f(x) = y.

Therefore, f is onto.

Hence, the given function is one-one onto.

These examples of bijective function help you to solve the following practice problems.

My forthcoming post is on icse board question papers and Nonlinear Partial Differential Equations will give you more understanding about Algebra.

Practice problems of bijective function:

Following examples of bijective function is given for your practice which helps you to learn more about bijective function.

1) Show that the function f : R → R : f(x) = x3  is one-one and onto.

2) Let R0 be the set of all non zero real numbers. Show that f : R0 → R0 : f(x) = 1/x is a one-one onto function.

Probability Distribution

In probability theory and the statistics, a probability distribution identifies either the probability of each value of a random variable (when the variable is discrete), or the probability of the value falling within a particular interval (when the variable is continuous). The probability distribution describes the range of the possible values that a random variable can get and the probability that the value of the random variable is within any subset of that range.

I like to share this Non Central T Distribution with you all through my article.

The Normal distribution is often called as the "bell curve", when the random variable takes the values in the set of real numbers. Let us see some sample problems on probability distribution statistics.

Examples

Given below are some of the examples on Probability Distribution Statistics.

Example 1:

A continuous random variable X has probability distribution function f(x) = 3x2, 0 ≤ x ≤ 1, Find a and b such that. (i) P(X ≤ a) = P(X > a) and (ii) P(X > b) = 0.05. Calculate probability distribution for this function.

Solution:

(i) Since the total probability is 1, [Given that P(X ≤ a) = P (X > a)

P(X ≤ a) + P(X > a) = 1

i.e., P(X ≤ a) + P(X ≤ a) = 1

⇒ P(X ≤ a) =1/2⇒ `int`3x2dx = 12

i.e.,[3x3/3]a0=1/2 ⇒ a3 =1/2

i.e., a = 1/213

(ii) P(X > b) = 0.05

∴ `int` f(x) dx = 0.05 ∴ `int` b1 3x2  dx = 0.05

[3x3]31b= 0.05 ⇒ 1 − b3 = 0.05

b3 = 1 − 0.05 = 0.95 =95

100 ⇒ b = 19/2013

Example 2:  A random variables X has probability mass function as in the probability distribution tables given below
X
0 1 2 3 4 5 6
P(X = x) k 3k 5k 7k 9k 11k 13k



(1) Find k.

(2) Evaluate P(X < 4), P(X ≥ 5) and P(3< X ≤ 6)

Solution:

(1) Since P(X = x) is a probability mass function `sum_(n=0)^6` P(X = x) = 1

ie.,P(X=0) + P(X = 1) +P(X = 2) +P(X = 3) +P(X = 4) +P(X = 5)+P(X = 6) = 1.

⇒ k + 3k + 5k + 7k + 9k + 11k + 13k = 1 ⇒ 49 k = 1 ⇒ k =1/49

(2) P(X < 4) = P(X = 0) + P(X = 1 ) + P(X = 2) + P(X = 3) =1/49 +3/49 +5/49 +7/49 =16/49

P(X ≥ 5) = P(X = 5) + P(X = 6) =11/49 +13/49 =24/49

P(3 < X ≤ 6) = P(X = 4) + P(X = 5) + P(X = 6) =9/49 +11/49 +13/49 =33/49

∴ The smallest value of x for which P(X ≤ x) > 1/2 is 4.

Example 3:

In a class, the average number of marks obtained by student in Physics is 0.52, chemistry is 0.48 and in both Physics and chemistry they obtained 0.37. Find the total average obtained in either Physics or chemistry.

Solution:

P(A) = Average number of marks in Physics = 0.52

P(B) = Average number of marks in Chemistry = 0.48

P(A and B) = Average number of marks in both physics and chemistry = 0.37

P(A or B)    = P(A) + P(B) – P(A-B)

= 0.52 + 0.48 – 0.37

= 0.63

Calculating Interquartile Range

Calculating interquartile range is referred as a measure of variability, spread or dispersion, H-spread. It is the difference sandwiched between the 75th percentile (often called Q3 or third quartile) and the 25th percentile (often-called Q1 or first quartile). The primary step is to find the interquartile range is to arrange the given set of numbers in ascending order. The standard formula to find the interquartile range is given as

Interquartile ranges = Quart3 – Quart1.    ----- Standard formula

Whereas Q1 = first quartile.

Q3 = third quartile.

Example Problems for calculating interquartile range:

Ex : Determine the interquartile range of following set of numbers

34, 15, 8, 26, 22, 9, 19

Sol :  The following are the steps to find the interquartile range of a set of numbers.

Step 1:  Arranging of numbers

The initial step is to modify the given data in order, from smallest to biggest.

9, 8, 15, 19, 22, 26, 34

Step 2:    Calculating 1st quartile Q1.

The next step is to find the lower median (1st quartile Q!).

This is the middle of the lower three numbers.

1st quartile Q1 is 8.

Step 3:    Calculating 3rd quartile Q3.

Now find the upper median (The 3rd quartile Q2).

This is the middle of the upper three numbers.

The 3rd quartile Q3 is 26.

Step 4:     Calculating interquartile range.

The formula used to find the interquartile range is

Interquartile range = Q3 – Q1.

Q1 = first quartile.

Q3 = third quartile.

Here

Q1 the first quartile = 8

Q3 the third quartile = 26

Interquartile range = Q3 – Q1. ----- Standard formula

Plug in the Q1 and Q3 values in the standard formula Q3 – Q1.

Interquartile range = 26 – 8.

Interquartile ranges = 18.

Practice Problems for calculating interquartile:

Pro 1:  Find the interquartile range of following set of numbers

42, 81,56,21,63,12,5

Ans : Interquartile range = 51.

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Pro 2:  Find the interquartile range of following set of numbers

56, 23, 14, 25, 86, 45, 63, 15, 49, 18, 16

Ans:   Interquartile range = 40.

Pro 3:   Find the interquartile range of following set of numbers

12, 8, 5, 22, 15, 45, 2

Ans :   Interquartile range = 17

Permutations in Math

In mathematics, the notion of permutation is used with several slightly different meanings, all related to the act of permuting (rearranging in an ordered fashion) objects or values. Informally, a permutation of a set of values is an arrangement of those values into a particular order. Thus there are six permutations of the set {1,2,3}, namely [1,2,3], [1,3,2], [2,1,3], [2,3,1], [3,1,2], and [3,2,1]. (Source – Wikipedia.)

Example problems for permutations in math:

Example 1

1) How many permutations of 5 apparatus are probable from 2 apparatus. what is a permutation in given math problem?

Solution:

By means of the formula we can evaluate the answer,

P (n, r) = n! / (n – r)!

Here, n = 5, r = 2

P (5, 2) = 5! / (5–2)!

P (5, 2) = 5! / 3!

=   `(5xx4xx3xx2xx1)/ (3xx2xx1)`

Now 3, 2, 1 gets crossed out

= 5 x 4

= 20

Answer is: 20

Example 2

How many permutations of 9 files are possible from 5 books. what is a permutation in given math problem?

Solution:

Formula for calculate permutations,

P (n, r) =n! / (n – r)!

Now, n = 9

r = 5

P (9, 5) = 9! / (9 – 5)!

P (9, 5) = 9! / 4!

= `(9 *8 * 7 * 6 *5 * 4 *3 *2 *1)/ (5 * 4 *3 *2 *1)`

Here 5, 4, 3, 2, 1 gets crossed out

= 9x 8 x 7x 6

= 3,024

The answer is: 3, 024

Example 3

How many ways can 6 graduates from group of 15 are lined up for a function. what is a permutation in given math problem?

Solution:

There are 15P6 possible permutations from a group of 15.

15P6 = (15!)/(15-6!)

= (15!)/ (9!)

= 15x 14x 13x 12x 11x 10

= 3, 603, 600 different lineups.

The answer is: 3, 603, 600 different lineups.

Example 4

How many ways can 3 kids from group of 13 are lined up for a photograph. what is a permutation in given math problem?

Solution:

There are 13P3 possible permutations of 3 students from a group of 13.

13P3 = (13!)/ (13- 3!)

= (13!)/ (10!)

= 13 x 12 x 11

= 1716 different lineups.

The answer is: 1716 different lineups.

My forthcoming post is on Divide Fractions by Whole Numbers and cbse sample papers for class 9 science will give you more understanding about Algebra.

Practice problems for what is a permutations in math:

Practice problem -1

1) How many permutations of 9 components are possible from 2 elements. what is a permutation in given math problem?

Ans: 72

Practice problem -2

2) How many ways can 3 girls from group of 14 are lined up for a photograph?

Ans: 2184

Practice problem -3

3) How many permutations of 17 books are possible from 4 books?

Ans: 57, 120

Practice problem -4

4) How many 4-digit numbers can be ordered from the digits 1, 2, 3, 4, and 5, if each digit is unique. what is a permutation in given math problem?

Ans: 120.