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