In Math What Does Range Mean

Introduction to range in math:

• In math, Data set is a collection of data, which is usually presented, in tabular form. Each column represents a variable. In math, Range is generally defined as the value we obtained as a result of difference between a greater value and a smaller value.
• In other words, range is defined as the difference between a maximum and minimum value.

Steps to learn the range in math:

     In order to learn the range of data set the following steps are necessary.
     Step 1: Arrange the numbers from ascending to descending order.
     Step 2: Identify the greater value in the given set
     Step 3: Identify the smaller value in the given set
     Step 4: Find the difference between the greater value and the smaller value to identify the range of the given data set.

Worked Examples for range in math:

Example 1:
Find the range of the data set given below:
                    8 , 15 , 13 , 7 , 24 , 37 , 6.
    Step 1: Arranging the numbers given in data set from least to greatest.
                    6 , 7 , 8 , 13 , 15 , 24 , 37.
    Step 2: Identify the greater value in the given set
                  From the given set, we can identify 37 as the greater value.
    Step 3: Identify the smaller value in the given set
                  From the given set, we can identify  6 as the smaller value.
    Step 4: Find the Range.
                          Range     =   Greater value – Smaller value
                                           =   37 - 6
                                           =  31
                   Hence, the range of the given data set is 31.
Example 2:
Find the range of the data set shown below:
                    33 , 12 , 79 , 24 , 34 , 53 , 27 , 61
     Step 1: Arranging the numbers given in data set from least to greatest.
                    12 , 24 , 27 , 33 , 34 , 53 , 61 , 79.
     Step 2: Identify the greater value in the given set
                    From the given set, we can identify 79 as the greater value.
     Step 3: Identify the smaller value in the given set
                    From the given set, we can identify  12 as the smaller value.
    Step 4: Find the Range.
                              Range    =   Greater value – Smaller value
                                               =   79 - 12
                                               =   67
                        Hence, the range of the given data set is 67.

Practice problems for range in math:

1) Find the range of the data set shown below:
                    7 , 9 , 19 , 31 , 37 , 43 , 6 , 37
     Answer: 37
2) Find the range of the data set shown below:
                    14 , 11 , 32 , 10 , 32 , 77 , 27 , 43
     Answer: 67

Geometrical Meaning of the Zeroes of a Polynomial

We know that  a real number k is a zero of the polynomial p(x) if p(k) = 0 in a geometrical way.

The x-coordinate of the point, where the graph of a polynomial intersect the x-axis is called the zero of the polynomial.

An nth - degree polynomial intersects the x-axis of n points and therefore, has a maximum of n zeros in geometrical graph

In a quadratic polynomial ax2 + bx +c,

If a>0, then the graph is a parabola that open upwards.

If a<0, then the graph is a parabola that open downwards.

Special case to the zeroes of a polynomial in geometrical meaning


In a geometrical way the evaluation of zeroes with polynomial evaluation of the equation in geometrical meaning is follows

Case (i) :

Here, the graph cuts x-axis at two distinct points A and A′.
The x-coordinates of A and A′ are the two zeroes of the quadratic polynomial ax2 + bx + c in this case is shown in the following figure



Case (ii) :

The graph given here cuts the x-axis at exactly one point, i.e., at two coincident points. So, the two points A and A′ of Case (i) coincide here to become one point A is shown in the following figure.



The x-coordinates  of the  A is the only zero for the quadratic polynomial ax2 + bx + c in this case.

No zero case

The graph given here is either completely above the x-axis or completely below the x-axis. So, it does not cut the x-axis at any point

So, the quadratic polynomial ax2 + bx + c has no zero in this case.
So, you can see geometrically that a quadratic polynomial can have either two distinct zeroes or two equal zeroes (i.e., one zero), or no zero. This also refers that the polynomial of degree 2 has at-most two zeroes is shown in the following figure.

Concept of the Derivative Online Help

In online, concept of derivative is clearly explained with the help of solved example problems. In online so many websites explain the concepts with the help of math sites. The derivative concept is clearly explained in calculus whereas derivative concept helps to find the rate of change for the given function with respect to change in the input function. The following are the example problems with detailed solution helps to explain the concept of derivative in online.


Concept of derivative online help example problems:


The following example problems explain the concept of derivative in online.

Example 1:

Determine the derivative by differentiating the polynomial function.

f(t) = 3t 3 +4 t 4  + 5t

Solution:

The given function is

f(t) = 3t 3 +4 t 4  + 5t

The above function is differentiated with respect to t to find the derivative

f '(t) = 3(3t 2 )+4(4t 3  ) + 5

By solving above terms

f '(t) = 9t 2 +  8t 3 + 5

Example 2:

Determine the derivative by differentiating the polynomial function.

f(t) = 6t6 + 5 t5 + 4 t4 + t

Solution:

The given equation is

f(t) = 6t6 + 5 t5 + 4 t4 + t

The above function is differentiated with respect to t to find the derivative

f '(t) =  6(6t 5)  +5 (5 t4 ) +4(4 t3) + 1

By solving above terms

f '(t) =  36t 5  + 25 t4  + 16 t3 + 1

Example 3:

Determine the derivative by differentiating the polynomial function.

f(t) = 2t 2 +4 t 4  + 15

Solution:

The given function is

f(t) = 2t 2 +4t 4  + 15

The above function is differentiated with respect to t to find the derivative

f '(t) = 2(2t  )+4(4 t 3 ) + 0

By solving above terms

f '(t) = 4t +16t3

Example 4:

Determine the derivative by differentiating the polynomial function.

f(t) = 5t5 +4t 4 +3t 3  + 2

Solution:

The given function is

f(t) = 5t5 +4t 4 +3t 3  + 2

The above function is differentiated with respect to t to find the derivative

f '(t) = 5(5t 4 )+4(4t 3 ) +3( 3t 2) +0

By solving above terms

f '(t) = 25t 4 +16t 3  +9 t 2


Concept of derivative online help practice problems:


1) Determine the derivative by differentiating the polynomial function.

f(t) = 2t 3 +3 t 4  + 4 t 5

Answer: f '(t) = 6t 2 +12 t3 + 20 t 4

2) Determine the derivative by differentiating the polynomial function.

f(t) = t 3+t5 + 4 t 6

Answer: f '(t) = 3t2 + 5t4 + 24 t 5

Solve Online Scale Factor

Scale is a thing which is used to measure the dimensions of the object. The terms scale factor is used to mention the dimensions of the geometric shapes. In geometry, the term scale factor is usually used to change the dimensions of the image. Using the scale factor, any shapes dimension can be increased or decreased. In online, students can learn about various topics. In online, students can learn about scale factor clearly. Online learning is very interesting and interactive. Moreover online learning is different from class room learning in different ways. In online, students have one to one learning.


More about scale factor:


A scale factor is a figure used as a figure by which another figure is multiplied in scaling.

A scale factor is new to scale shape in 1 to 3 proportions.
Scale factor can be established by the following scenario
Size Transformation:     In size change, the scale factor is the ratio of express the amount of reduction.

Scale Drawing:    In scale diagram a diagram that is like but either superior or lesser than the real object, for example, a road map or a building blueprint...

Comparing Two Similar Geometric Images:      The scale factor when compare two parallel geometric imagery is the ratio of lengths of the equivalent sides.






Solve Scale factor Problem:


Q 1:   The area of a rectangle is 5 and it is enlarged a scale factor by 5. Solve for the area of an enlarged rectangle?

Sol :  The Area of a rectangle(R) is 5 and its Enlarged by a scale factor of 5.

that means the length L and the width W are equally multiply by 5.
As well note that the area of Rectangle is the length L times the width W, so Rectangle = LW.

Because the area of a rectangle is given by Rectangle(R) = LW, and L turn out to be 4L and W turn out to be 4W,

Solve for new triangle,

the new area of a enlarged rectangle  = (5L)(5W) = 25LW = 25(R).
Now we know that Rectangle(R) = 5, so 25(R) = 125.

Q 2:  The side of the square is 6 cm and it is enlarged a scale factor by 4. Solve for the area of an enlarged square?

Sol :  The Area of a square = a2

=  6 * 6

= 36 cm2.

Scale factor is 4 cm.

New side of square  = 4 * 6

= 24 cm.

Area of enlarged square = 242.

= 576 cm2.

Answer is 576 cm2.

Study Online Interest

To study online interest is useful to calculate the interest easily. We have two types of interest Simple interest and compound

interest. Formula for simple interest is,    Interest = Principle `xx` Rate `xx` Time

Where  I = Total amount of interest paid

P = Amount lent

R = Percentage of principle charged as interest each year.

Example: rate = 2% means then 2/100 =0.02. Use this value in formula.

Formula for compound interest is,       A = p`(1+r)^(n)`


Examples of study online interest:


Ex:1

To find simple interest if principle= 3000, rate= 3%, time= 5

Sol:

To study online interest we have to write the formula first

That is Simple interest = Principle `xx` Rate `xx` Time

Apply the given values in the formula,

Interest = 3000 `xx` 3%`xx` 5               R = 3% = 3/100 = 0.03

= 3000`xx` .03`xx` 5

= 450

The answer is 450

Ex:2

Shan invested Rs.6000 for 5 years. He received the simple interest of Rs.1,500.Find the rate of interest.

Sol:

To study online interest we have to write the interest formula first

That is Simple interest = Principle `xx` Rate `xx` Time

Here principal, P= Rs.6000

Number of years, n= 5

Simple interest, I = Rs.1500

Rate of interest, r =?

We know that rate of interest, r =`(100 xx 1500)/(6000 xx 5)`

=5

Therefore Rate of interest = 5%

Ex:3

If the principal amount is $7000, the periodic rate of interest is 3% and the number of compounding periods is 3, what is the value at maturity?

Sol:

To study online interest we have to write the formula first

A = p`(1+r)^n`

Here principal, P= $7000

Rate of interest, r= 3%

= 3/100 = .03

n = 3

A = P (1 + r) n

A = 7000 (1 + .03) 3

A = 7649.089

Ex:4

If the principal amount is $2000, the periodic rate of interest is 2% and the number of compounding periods is 5, what is the value at maturity?

Sol:

To study online interest we have to write the formula first

A = p`(1+r)^n`

Here principal, P= $2000

Rate of interest, r= 3%

= 2/100 = .02

n = 5

A = P (1 + r) n

A= 2000 (1 + .02) 5

A= 2208.1616

These are the some examples to study online interest.

Learning Examples of Histograms

Introduction on learning examples of histograms:

In statistics, a histogram is a graphical display of tabular frequencies, shown as adjacent rectangles. Each rectangle is erected over an interval, with an area equal to the frequency of the interval. The height of a rectangle is also equal to the frequency density of the interval, i.e. the frequency divided by the width of the interval. The total area of the histogram is equal to the number of data. A histogram may also be based on the relative frequencies instead. It then shows what proportion of cases fall into each of several categories (a form of data binning), and the total area then equals 1.

Procedure and learning example for drawing the histogram:

To draw the histogram for a given frequency table
                  Marks    : 10-20   20-30    30-40   40-50   50-60   60-70
Number of students:    4           6            8         10          7         5  
Sol:
Histograms are drawing the first quadrant .because the first quadrant frequencies and class intervals are positive. Class intervals are marked along x axis similarly the frequency are marked along the y axis. Identify the maximum frequency in the table and then chose a suitable scale along the y- axis.
                     Draw first rectangle on the intervals 10 -20 as the base width and height of 4 units. Draw all rectangles taking the consecutive class intervals as base of the rectangles and their heights corresponding to the frequencies. The closed figure obtained is called histogram

Histogram:

This is the histogram for given frequency table.
Example: 2
To draw the histogram for a given frequency table,
 
Sol:

• Here in the frequency table, data range and frequency are given
• We take a data range in x axis and frequency in y axis

Histogram:

This is the histogram for given frequency table

More examples for learning histograms

To draw the histogram for following frequency table,

Sol:

• Here the given data deals with score(group) and frequency(count)
• We are taking score in x-axis
• Similarly the frequency is in y axis

Histogram:

This is the histogram for given frequency table.

Trigonometric Unit Circle Learning

A circle, whose radius is equal to one unit, is called as unit circle. The concept of unit circle is frequently used in trigonometry. In trigonometry, a circle with center (0, 0) and a radius of one unit is a unit circle. The equation of a circle is (x-h)2 + (y-k)2 = r2

For a unit circle, the center is (0, 0) and radius is 1, so the equation of a unit circle is x2 + y2 = 1


Learning - Properties of unit circle


Consider a point (x, y) in a unit circle.



The right triangle in the unit circle in the above diagram, Pythagoras theorem satisfies the equation of unit circle.

x2 + y2 = 1

Learning - Forms of unit circles points

Exponential form:            eit

Trigonometric form:        z = cos(t) + i sin(t)

Learning - Trigonometric functions

In a unit circle, consider a point (x,y) on the circle. If the angle formed between line joining the center (0,0) and the point (x,y) and the horizontal axis is `theta`,



Then the trigonometric functions for the angle `theta` is given by,

sin `theta` = opposite side/hypotenuse

cos `theta`= adjacent side/hypotenuse

tan `theta`= opposite side/adjacent side

csc `theta`= 1/sin`theta` = hypotenuse/opposite side

sec `theta`= 1/cos `theta` = hypotenuse/adjacent side

cot `theta`= 1/tan `theta`= adjacent side/opposite side


Example for trigonometric unit circle learning


Find the value of each of the 6 trigonometric functions for an angle theta that has a terminal side containing the point (3, 4).



By Pythagoras theorem, x2 = 32 + 42

x2 = 9 + 16

x2 = 25

x = 5

So, hypotenuse = 5, opposite side = 4 and adjacent side = 3

Then the trigonometric identities are given by,

sin `theta` = opposite side/hypotenuse = 4/5

sin `theta` = 4/5

cos  `theta` = adjacent side/hypotenuse = 3/5

cos `theta` = 3/5

tan `theta` = opposite side/adjacent side = 4/3

tan `theta` = 4/3

csc `theta` = 1/sin `theta`   = hypotenuse/opposite side = 5/4

csc `theta` = 5/4

sec `theta` = 1/cos `theta` = hypotenuse/adjacent side = 5/3

sec `theta` = 5/3

cot `theta` = 1/tan `theta` = adjacent side/opposite side = 3/4

cot `theta` = 3/4

A Factorization is the method for finding the variable factor value of the given expression. A Factor number is multiply with other number. If the factor value is a prime number then it is called prime factor value. The Factor is nothing but a value it is gives the answers to the given problem. If a polynomial factor might be written as the product of two or more expressions, then each expression is called the factor of the given polynomial. If a polynomial factor value might be written as the product of two or more expressions, then each expression is called the factor of the given polynomial. In this article we shall discuss about factoring solver.

Sample problem for factoring solver:

The following problem will help you understand the factoring solver.

Example 1:
Find the factoring value of the given function. 2x²- 12 x +16 = 0

Solution:
In the first step we find pattern factor value of the given numerical values of the x coefficients.

                                                                       32   (product)
                                                                     /    \    
                                                               - 4        - 8 
                                                                     \    / 
                                                                     -12    (sum)

2x²- 12 x + 16 = 2x² - 4x - 8x + 16
                      = 2x (x-2) - 8(x-2)
                     = (2x - 8) (x – 2)
                      = (x -4) ((x – 2)

So the factor of the given function is 2 and 4.

Example 2:
Find the factoring value of the following trinomial. 3x²- 9 x +6

Solution:
In the first step we find pattern factor value of the given numerical values of the x coefficients.

                                                                      18   (product)
                                                                     /    \    
                                                                 - 3      -6
                                                                     \    / 
                                                                      -9     (sum)
3x²- 9 x + 6 = 3x² - 6x - 3x + 6
                        = 3x (x-) - 3(x-2)
                        = (3x – 3) (x – 2)
                        = (x-1) (x-2)

So the factor of the given trinomial function is 2 and 1.

Practice problem for factoring solver:

• Find the factors of the given function. 6x² - 18 x + 12

            Answer: x = 1, 2.

• Find the pattern factors of the given function. 4x²- 24 x +32 = 0

            Answer: x = 2, 4.

study about triangles area

In mathematics, triangle is one of the important topics in geometry.  A geometrical figure which is formed by connecting three points by straight line segment is defined as a triangle. The three sides of a triangle are called as a three-sided polygon. Its can be identified by their sides, by their angles or by the combination of both sides and angles. Here we study about the triangles area.


Study about triangles area – formula:


Triangle area formula:

              If the base and the height of the triangle are given, then the formula to find the area of the triangle is

                                    Area of triangle is A = `(bh)/2` (or) `1/2` bh square. Units

                                          b denotes the base of triangle.

                                          h denotes the height of triangle



In a triangle, base and height can be perpendicular to each other.


Study about triangles area – Example problems:


Here we study some example problems about the triangles area

Example 1:

Determine the area of the triangle whose base side is 4cm and the height of the triangle is 8cm.



Solution:

             Given

              Base side of triangle is 4cm

              Height of the triangle is 8cm

       Triangle area Formula:

                Area of the triangle is 1/2bh square. Units

                                              A = `1/2` bh

                                                = `1/ 2` (4*8)

                                                = `32/2`

                                                = 16

                Area of the triangle = 16 cm2

Answer:

                Area of triangle is 16 cm2

Example 2:

Determine the area of the triangle whose base side is 8m and the height of the triangle is 12cm.



Solution:

             Given

              Base side of triangle is 8m

              Height of the triangle is 12m

       Triangle area Formula:

                Area of the triangle is 1/2bh square. Units

                                              A = `1/2` bh

                                                = `1/ 2` (8*12)

                                                = `96/2`

                                                = 48

                Area of the triangle = 48 m2

Answer:

                Area of triangle is 48 m2

Example 3:

Determine the area of the triangle whose base side is 13cm and the height of the triangle is 17cm.



Solution:

             Given

              Base side of triangle is 13cm

              Height of the triangle is 17cm

       Triangle area Formula:

                Area of the triangle is `1/2` bh square. Units

                                              A = `1/2` bh 

                                                = `1/ 2` (13*17)

                                                = `221/2`

                                                = 110.5

                Area of the triangle = 110.5 cm2

Answer:

                Area of triangle is 110.5 cm2

column of a matrix

Let T be a n × n real matrix. It is known that when T is singular, then its unique generalized inverse T (known as the Moore-Penrose inverse) is defined. In the case when T is a real m×n matrix, Penrose showed that the column matrix satisfying the four Penrose equations, called the generalized column of T. A lot of work concerning generalized with the column has been carried out, in finite and infinite dimension. Having problem with Matrix Solver Read my upcoming post, i will try to help you.


Definition of Column matrix:


A matrix with a one column is called a column matrix. In other words geometric vector may possibly be represent with a listing of numbers are known as column matrix. A column matrix is an ordered list of numbers given in a column.

Example

          Column matrix is an m × 1 matrix, i.e. a matrix consisting of a single column of m elements.

                     [x1]    

                     [x2]

                X= [ . ]

                     [ . ]

                     [xm]

 For example of column matrix:     

                                  `[[2.3],[5]]`


Column matrix product:


          Let T be a n × n real matrix. It is well-known that as soon as T is singular, then its exceptional generalized inverse T (known as the Moore-Penrose inverse) is defined. In the case after T is a real m×n matrix, Penrose showed that the column matrix satisfying the four Penrose equations, called the generalized column of T. A lot of work concerning generalized with the column has been carried out, in finite and infinite dimension.

                3x4 matrix                                 4x5 matrix                                 3x5 matrix

                [      .  .   .]                                      [ . . . a . ]                                        [. . . . . ]

                [.    .  .   .]                                       [. . . b . ]                       =               [. . . . . ]

                [1 2 3 4]                                        [. . . c .]                                        [. . . x3,4 .]

                                                                      [. . . d .]

           The element x3,4 of the above matrix product is computed as follows

                              x3,4 = (1,2,3,4) . (a,b,c,d) = 1xa + 2xb + 3xc + 4xd.

Dividing Integers Solving Online

The natural number and negative numbers together with zero are called integers. The natural numbers are 1, 2, 3, 4….., the whole numbers are 0, 1, 2, 3… and the negative numbers are -1,-2,-3…. Therefore the set of integer can be -4,-3,-2,0,1,4,6……….. Here we have to learn about how to solve and divide the integers in online and its operations.



Solving methods dividing integers online:


The operation of dividing integers in an online performed by four different ways, they are following,

Positive integer divided by positive integer = positive integer.
Negative integer divided by negative integer = positive integer.
Positive integer divided by negative integer = negative integer.
Negative integer divided by positive integer = negative integer.

Problems of Dividing Integers Solving Online :


Online example: 1

To solve the following integer 8/4

Solution:

Here the both numerator and denominator are positive.

So, = 8 / 4

=2

So the result is positive =+2

Example: 2

To solve following 66/9

Sol: Here also both numerator and denominator are positive so the answer will be positive

=divide both numerator and denominator by 3

=22/3

Answer is = +22/3

Example: 3

To the following, -12 divide by -6

Solution:

Given both integers are negative so the answer will be positive

Both numerator and denominator by -6

=2

Answer is positive =+2

Example: 4

To solve the following -76 divide by -4

Solution:

Both numerator and denominator are negative so the result is positive

Divide -4 by numerator and denominator

= 19

Answer is positive = +19

Example: 5

To solve the following integers

25 divide by -5

Solution:

Here the numerator is positive and denominator is negative so the answer will be negative

Divide both numerator and denominator by 5

=-5

The answer is negative = -5

Example: 6

Divide the following,-81 divide by 9

Solution:

Here the numerator is negative and denominator is positive so the answer will be negative

= divide both numerator and denominator by 9

= -9

The answer is negative = -9

Example: 7

-121 divide by 11

Solution:

Here numerator is negative and denominator is positive

= divide by 11

= -11

The answer is negative =-11

So in these solving and dividing integers online if any one that is numerator or denominator will be negative means the answer will be negative.

Similarly both are positive are negative the answer will be positive.

Geometric Distributions

Introduction:

Let we will discuss about the geometric distributions. The geometric distributions should be either of two discrete probability distributions in statistics and probability theory. The probability allocation of number X of Bernoulli trials desired to obtain one success, beard on the set { 1, 2, 3, ...}. The probability allocation of number Y = X − 1 of failures previous to the first success, maintained on the set { 0, 1, 2, 3, ... }  


More about geometric distributions:

• The two distinct geometric distributions does not mystified with each other.
• Most commonly, name shifted geometric allocation is accepted for the former one.
• But, to keep away from ambiguity, it is measured wise to point out which is planned, by mentioning the range explicitly.
• If the probability of success on every check should be p, then the probability that kth check is the first success is,

Where, k = 1, 2, 3, ....

• Consistently, if the probability of success on each check is p, then the probability that there are k failures before the first success is

Where, k = 0, 1, 2, 3, ....

• In both case, the series of probabilities is a geometric series.

Example:

Assume a normal die is thrown frequently until the first time a "1" appears. The probability allocation of the number of times it is thrown is holded on the endless set { 1, 2, 3, ... }. They should have a geometric allocation with p = 1/6.

Moments and cumulants:

• The predictable value of a geometrically allocated random variable X is 1/p , variance is (1 − p)/p2

• Likewise, the expected value of the geometrically dispersed random variable Y is (1 − p)/p, and its variance is (1 − p)/p2

• Let μ = (1 − p)/p be the expected value of Y and then the cumulants κn of the probability distributions of Y satisfy the recursion

Outline of proof:

That the expected value is (1 − p)/p can be shown in the following way. Let Y be as above. Then

Percentage Change Calculator

A percentage change is a way to express a change in a variable. It represents the relative change between the old value and the new one.

The formula used to calculate the percentage change is

Percentage change = `((V2 - V1) / (V1)) * 100` .

V1- represents the old value

V2 - the new one.                                                              Source: - Wikipedia



some percentages as fractions:

a.       `5%=5/100=1/20` .

b.      `10%=10/100=1/10` .

c.      `25%=25/100=1/4` .

d.     `75%=75/100=3/4` .

e.      `125%=125/100=5/4` .

f.        `175%=175/100=7/4` .

g.       `(3 1/8)%=25/800=1/32` .

h.    ` (6 1/4)%=25/400=1/16` .

i.        `(8 1/3)%=25/300=1/12` .

j.        `(16 2/3)%=50/300=1/6` .

k.       `(66 2/3)%=200/300=2/3` .

l.        `(87 1/2)%=175/200=7/8` .

Calculation of Percentage:


Calculation of percentage:

The percent symbol can be treated as being equivalent to the pure number constant `1/100=0.01,`  while performing calculations with percentage.

If a number is first changed by`P% ` and then changed by `Q%` , then the net change in the number `=[P+Q+((PQ)/100)]` . Remember that any decreasing value in the formula should be taken as ‘negative’ and increasing value should be taken as ‘positive’.

Similarly, if A’s salary is `P%`  less than B’s salary, then the percentage by which B’s salary is more than A’s salary is`(100P)/(100-P)` .

If expenditure also, then percentage change in expenditure or revenue`=[P+Q+((PQ)/100)]` . Where ‘P’ is the percentage change in price and ‘Q’ is the percentage change in consumption.


Percentage change calculator - Example problems:


Percentage change calculator - Problem 1:-

Ram scored 86 runs in the cricket match on  Monday.  On Friday he scored 95 runs.  Calculate the Percentage of change?

Solution:-

Given

V2 = new value = 95 runs.

V1 = old value = 86 runs.

Percentage of change =  ?.

The formula used to calculate the percentage change is

Percentage change = `((V2 - V1) / (V1)) * 100` .

By plugging in the given values in to the formula we get

Percentage change  =` (95 - 86) / (86) * 100` .

The difference between  95 and 86 is  9.

By plugging in it to the formula we get the answer as

=`9 / 86` * 100.

The fraction 9/ 86 gives us 0.1046.

=0.1046 * 100

=10.46

The percentage of change is 10.46.

Percentage change calculator - Problem 2:-

Mary bought 40 Compact disks last month.  He bought only 30 this year.  Calculate is the percent of change.

Solution:-

Given :-

Here,

V2 = new value =  35 Compact disks

V1 = old value = 40 Compact disks

Percentage of change =?

The formula used to calculate the percentage change is

Percentage change = `((V2 - V1) / (V1)) * 100` .

V2 = new value.

V1 = old value

By plugging in the given values in to the formula we get

Percentage change =  `(35 -40) / (40)` * 100.

The difference between 35 and 40 is 5

By plugging in the given values in to the formula we get,
33333
= `5 / 40` * 100

The fraction 8/ 40 gives us  1/5.

`= 1/ 5 ` * 100.

= 20%

The percentage of change is 20 %


Percentage change calculator - Practice Problem:


Ex1 : John scored 16 runs in the cricket match on  Monday.  On Friday he scored 32 runs.  Calculate the Percentage of change?

Answer:-

The percentage of change is 50%

Ex2 : Calculate `40%`  of `625` .

Sol: `40% ` of a number `=2/5`  of the number `=2/5`  of `625=(2/5)(625)=250` .

Temperature Conversion

In this page we are going to discuss about temperature Conversion concept . Measurement is one of the important terms in day to day life. Temperature is usually measured in terms of Fahrenheit and Celsius. Temperature of is generally in terms of these two names.

Fahrenheit:

The degree Fahrenheit is usually represented as (F). Fahrenheit is names after the German scientist Gabriel Fahrenheit, who invented the Fahrenheit measurement. The zero degree in the Fahrenheit scale represents the lowest temperature recording.

Celsius:

The degree Celsius is usually represented as (C). Celsius is names after the Swedish astronomer Ander Celsius, who proposed the Celsius first. In Celsius temperature scale, water freezing point is given as 0 degrees and the boiling point of water is 100 degrees at standard atmospheric pressure.

Formula for Fahrenheit and Celsius Conversion


Fahrenheit to Celsius Conversion Formula:

The formula for converting Fahrenheit to Celsius conversion is given as,

Tc = (5/9)*(Tf-32)

Where,

Tc = temperature in degrees Celsius,

Tf = temperature in degrees



Celsius to Fahrenheit Conversion Formula:

The formula for converting Celsius to Fahrenheit conversion is given as,

Tf = (9/5)*Tc+32

Where,

Tc = temperature in degrees Celsius,

Tf = temperature in degrees Fahrenheit.


Examples on temperature Conversion

Below are the examples on fahrenheit to celsius conversion problems :

Example 1 : Convert 68 degree Fahrenheit to degree Celsius.

Solution:

The formula for converting Fahrenheit to Celsius conversion is,

Tc= (5/9)*(Tf-32)

Tc= 68

Tc= (5/9) * (68 – 32)

Tc= (5/9) * 36

Dividing 36 by 9, we get 4,

Tc= 5 * 4

Tc= 20 degree.

The answer is 20 degree Celsius

Example 2 :Convert 132 degree Fahrenheit to degree Celsius.

Solution:

The formula for converting Fahrenheit to Celsius conversion is,

Tc= (5/9)*(Tf-32)

Tf = 132

Tc= (5/9) * (132 – 32)

Tc= (5/9) * 100

Dividing 100 by 9, we get 11.11,

Tc= 5 * 11.11

Tc= 55.11 degree.

The answer is 55.11 degree Celsius.



Now see the Celsius to Fahrenheit conversion problems:

Example 1 :  Convert 50 degree Celsius to degree Fahrenheit.

Solution:

The formula for converting Celsius to Fahrenheit is given as,

Tf= (9/5)*Tc+32

Tf= (9/5) * 50 + 32

Tf= 9 * 10 + 32

Multiplying 9 and 10 we get 90,

Tf= 90 + 32

Tf= 122 degree Fahrenheit

The answer is 122 degree Fahrenheit.


Example 2 :  Convert 42 degree Celsius to degree Fahrenheit.

Solution:

The formula for converting Celsius to Fahrenheit is given as,

Tf= (9/5)*Tc+32

Tf= (9/5) * 42 + 32

Tf= 9 * 8.4 + 32

Multiplying 9 and 8.4 we get 75.6,

Tf= 75.6 + 32

Tf= 107.6 degree Fahrenheit.

The answer is 107.6 degree Fahrenheit.

Introduction to Probability and Statistics

Introduction for Probability:

Introduction for Probability is the possibility that rather will happen - how to be expected it is that some event will happen. Now and again you can measure a probability with a number: "10% chance of rain", or you know how to use words such as impossible, unlikely, and possible, even chance, likely and certain. when a coin is tossed there is a probability of getting head and tail possible outcomes for an experiment occur is sample space.

Example: "our team may won match today"

Here is a probability formula:

`P(A) = (The Number Of Ways Event A Can Occur)/(The Total Number of possibLe outcomes)`

Introduction to Statistics:

Statistics is the branch which is applied by mathematics, which deals with the scientific analysis of data. The word ‘Statistics’ is derived from Latin word ‘Status’ which means ‘political state’. In statistics introduction datas are in two types which are primary and secondary datas. Sometimes an investigator uses the primary data of another investigator collected for a different purpose. Such data are called statistics secondary data.

From the data we have learnt to calculate the measures of the central tendency like mean, median and mode. These central measures do not give us all the details about the distribution.Further descriptions of the data called measures of dispersion are necessary. According to A.L. Bowley, “Dispersion is the measure of the variation of the individual item”. That is the dispersion is used to indicate the extent to which the data is spread.

Statistics properties:

Statistics deals with three properties, which are  Mean, Median , Mode

Examples:
A cancer patient wants to identify the probability that he will survive for at least 5 years. By collecting data on survival rates of people in a similar situation, it is possible to obtain an empirical estimate of survival rates. We cannot know whether or not the patient will survive, or even know exactly what the probability of survival is. However, we can estimate the proportion of patients who survive from data.

Common Pythagorean Triples

Introduction to Pythagoras online study:

The Pythagoras Theorem is a statement relating the lengths of the sides of any right triangle.

 The theorem states that:

For any right triangle, the square of the hypotenuse
is equal to the sum of the squares of the other two sides.

Mathematically, this is written:

c^2 = a^2 + b^2

We define the side of the triangle opposite from the right angle to be the hypotenuse, c. It is the longest side of the three sides of the right triangle. The other two sides are labelled as a and b.





Pythagoras generalized the result to any right triangle. There are many different algebraic and geometric proofs of the theorem. Most of these begin with a construction of squares on a sketch of a basic right triangle. We show squares drawn on the three sides of the triangle. For a square with a side equal to a, the area is given by:

A = a * a = a2

So the Pythagorean theorem states the area c2 of the square drawn on the hypotenuse is equal to the area a2 of the square drawn on side a plus the area b2 of the square drawn on side b.

pythagoras online study-Pythagorean triplets


A knowledge of Pythagorean triplets will also help the student in working the problems at a faster pace.

 The study of these Pythagorean triples began long before the time of Pythagoras.

There are Babylonian tablets that contain lists of such triples, including quite large ones.

There are many Pythagorean triangles all of whose sides are natural numbers. The most famous has sides 3, 4,

and 5. Here are the first few examples:

32 + 42 = 52;

52 + 122 = 132;

82 + 152 = 172;

282 + 452 = 532

There are infinitely many Pythagorean triples,that is triples of natural numbers (a; b; c) satisfying the equation a2 + b2 = c2.

If we take a Pythagorean triple (a; b; c),and multiply it by some other number d, then we obtain a new Pythagorean triple

(da; db; dc). This is true because,

(da)2 + (db)2 = d2(a2 + b2) = d2c2 = (dc)2 :

Clearly these new Pythagorean triples are not very interesting. So we will concentrate our attention on triples with no common factors.They are primitive Pythagorean triples

A primitive Pythagorean triple (or PPT for short) is a triple of numbers

(a; b; c) so that a, b, and c have no common factors1 and satisfy

a2 + b2 = c2:

There are 16 primitive Pythagorean triples with c ≤ 100:

( 3 , 4 , 5 )

( 5, 12, 13)

( 7, 24, 25)

( 8, 15, 17)

( 9, 40, 41

(11, 60, 61)

(12, 35, 37)

(13, 84, 85)

(16, 63, 65)

(20, 21, 29)

(28, 45, 53)

(33, 56, 65)

(36, 77, 85)

(39, 80, 89)

(48, 55, 73)

(65, 72, 97)

 One interesting observation in a primitive  Pythagoras triple is  either a or b must be a multiple of 3.


pythagoras online study-Solved examples


The Pythagorean Theorem must work in any 90 degree triangle. This means that if you know two of the sides, you can always find the third one.

 



In the right triangle, we know that:

c^2 = 6^2 + 8^2

Simplifying the squares gives:

                                                   c2= 36 + 64

                                                  c2 = 100    

                                                   c = 10       

                                      (taking the square root of 100)



In this example, the missing side is not the long one. But the theorem still works, as long as you start with the hypotenuse:



                                                15^2 = a^2 + 9^2

Simplifying the squares gives:

                                                225 = a2 + 81

                                         225 - 81 = a2                 

                                                144 = a2        

                                                  12 = a  

                                                  a   = 12

                              (Notice that we had to rearrange the equation)

Learn Intercept Formula

Learn Intercept formula is nothing but slope intercept formula. Before going to learn intercept formula we need to know why it is called so.

It is called slope intercept form because the equation includes slope and the y-intercept. So now we know the reason why it is called slope intercept form. Now coming to actual concept.....

The general form of slope intercept form is:

y= mx^+b

Where m--> slope of the line.

b --> y-intercept.

y --> y-coordinate.

x -->x-coordinate.

Formula - learn intercept formula


Slope intercept form is the simplest of all forms as we just need to plug in the values of slope(m) and y-intercept(b).

Now how are we going to get the slope??. In problems the slope might be mentioned directly or two points through which the line passes might be given. When the second case occurs slope can be find out using formula
slope(m)= (change in x)/(change in y)

Now let us consider that two points are (x1,y1) and (x2,y2)  then slope is given by

m=(y2-y1)/(x2-x1)

We have learnt how to find slope in the slope intercept form. now the next one to be calculated is y-intercept(b).
For this we need to plug the point through which the line passes.

This point will be mentioned in the question. If slope is m and the point is (p,q) then

plugging these values in slope intercept form we get

q= m*p+b

==> b= q-mp.

Now we have slope and y-intercept substituting these values we get slope intercept form.

I think to learn intercept form is very easy  and... cool

Examples on learn intercept formula


Ex1:  What is slope intercept form of line having slope 2 and y-intercept 3?
Sol1:

Given slope (m)=2, and y-intercept (b)=3

Plug these values in slope intercept form y=mx+b

Then        y= 2*x+3

y=2x+3.

So slope intercept form of a line having slope 2 and y-intercept 3 is

y=2x+3

Ex2:  What is slope intercept form of line passing through the point (2, 3) and having a slope of 4.
Sol 2:

Here we have slope = 4 and y-intercept is not mentioned but the point through which line passes is given.

Plug m=4 and (x,y)= (2,3) in the slope intercept form y=mx+b.

Then we get

3=4*2+b.

==>b+8 =3

==>  b=3-8=-5

So y-intercept is b=-5 and we have slope as m=4.

Plugging these in slope intercept form we get

y= 4x -5.

Ex3:  Express the equation 3x+4y+5=0 in slope intercept form.
Sol 3:

The given equation is 3x+4y+5=0.

In order to convert it in to slope intercept form bring y terms on one side and the remaining terms to other side.

For this subtract with 3x and 5 on both sides

3x+4y+5-3x-5=-3x -5

By doing this 3x and 5 get cancelled and the equation becomes

4y= -3x-5

Divide both sides by 4

4y /4 = (-3x-5)/4

y=-(3/4)x  - 5/4

Thus the slope intercept form of 3x+4y+5=0 is y=-(3/4)x  - 5/4

Mean Value Theorem Derivatives

In calculus, the mean value theorem states, roughly, that given an arc of a smooth continuous (derivatives) curve, there is at least one point on that arc at which the derivative (slope) of the curve is equal (parallel) to the "average" derivative of the arc. It is used to prove theorems that make global conclusions about a function on an interval starting from local hypotheses about derivatives at points of the interval. I like to share this Anti derivative with you all through my article.

Formal Statement of Mean Value Theorem Derivatives:

Let f: [x, y] → R be a continuous function on the closed interval [x, y] , and differentiable on the open interval (x, y), where x < y. Then there exists some z in (x, y) such that,

f'(z) = `(f(y) - f(x))/(y - x)`

The mean value theorem is a generalization of Rolle's Theorem, which assumes f(x) = f (y), so that the right-hand side above is zero.

Only one needs to assume that f: [x, y] → R is continuous on [x, y] , and that for every m in (x, y) the limit,

`lim_(h ->0) (f(m + h) - f(m))/(h)`

exists as a finite number or equals + ∞ or − ∞. If finite, that limit equals f′(m). An example where this version of the theorem applies is given by the real-valued cube root function mapping m to m1/3, whose derivative tends to infinity at the origin.

Note that the theorem is false if a differentiable function is complex-valued instead of real-valued.

For example, define f(m) = eim for all real m. Then

f (2π) − f(0) = 0 = 0(2π − 0)

while, |f′(m)| = 1.

- Source Wikipedia


Proof of Mean Value Theorem Derivatives:


Let g(m) = f(m) − rm, where r is a constant, because f is continuous on the closed interval [x, y] and differentiable on the open interval (x, y), now we want to select r, so as g satisfies the conditions of the Theorem,

g (x) = g (y) `hArr` f (x) − rx =  f (y) − ry

`hArr` ry − rx = f (y) − f(x)

`hArr` r(y − x) = f (y) − f(x)

`hArr` r = `(f(y) - f(x))/(y - x)`

By Rolle's theorem, g is continuous on the closed interval [a, b] and g(a) = g(b), there is some c in (a, b) for which g′(c) = 0, and it follows from the equality g(x) = f(x) − rx that,

f' (c) = g' (c) + r = 0 + r =  `(f(b) - f(a))/(b - a)`