Ways of Analyzing Data

Introduction to ways of analyzing data:

The analysis of data is the process of  inspecting and modelling the data which is used to highlight the useful information.and used to suggest some conclusion and it used to help in decision making.the analysis of data includes multiple methods to collect data and it goes through several phases of  analysing.

The pre-stage of analysis data integration..The data analysis includes Exploratory and confirmatory data analysis.

Ways of Analyzing Data-process of Data Analysis

We can distinguish the data analysis in to ,

Data cleaning
Initial data analysis
Main data analysis
Final data analysis
Data cleaning:

It is the process of cleaning the data,that is the process of including the preferable data and avoiding errors in the data.,and if possible correcting the data.This process can be done in the stage of data entry itself.

Initial data analysis:

In this stage we are refraining the data depends up on whether the data is having ht equality enough,depends upon the quality of measurements.

And it is in this stage ,checks for the missing data.and checked for any data whether it disturbs  the collected data.

For the analysis of data uni variate and bi variate ,and graphical methods are used ,to see whether the data collected are relevant.

And checks for whether the intension of the data collections are met,with the data.

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Ways of Analyzing Data-final Stage

For the final analysis of data computer simulations and data envelopment strategies are used.

For the final analysis of data some decision support systems are also used .For which some separate decision theory is also there.

As per the part of the final analysis some influence diagrams are made to view a clear picture for the datas going to be documented.

As a last step of analysing the data,the pareto analysis and the stakeholder analysis is also done.

In this step ,the stakeholders are examines whether the data collected are relevant and whether it will meet the objective of the datacollection.

During the final stage the datas are documented.

Pre Algebra B Calculator

Introduction to pre algebra b calculator:

Pre algebra is a branch of mathematics. Pre algebra plays an vital role in our day to day life. Pre algebra b calculator will do the four basic operations such as addition, subtraction, multiplication and division. The most important terms are variables, constant, coefficients, exponents, terms and expressions are handled by pre algebra b calculator. By pre algebra b calculator, we will know the use the symbols and alphabets in the place of unknown values, to form a statement. Hence, pre algebra b calculator gives the leads of Arithmetic. Therefore, students are getting pre algebra b calculator for their studies.

Examples by Pre Algebra Calculator:

Example 1:

Solve the equation x + 22 = 235 for x.

Solution:

x + 22 = 235 ( Now we have to add -22 on both sides. So we get,)

x + 22 –22 = 235 – 22

x = 213

Example 2:

Solve the equation x – 29 = 240 for x.

Solution:

x – 29 = 240 (Now we have to add 29 on both sides. So we get,)

x – 29 + 29 = 240 + 29

x = 269

Example 3:

Solve the equation 24x = 232 for x.

Solution:

24x = 234 (Now we have to divide both sides by 24. So we get,)

`(24x)/24` = `234/24`

x = 9.75

Example 4:

Solve the equation x ÷ 25 = 49 for x.

Solution:

x ÷ 25 = 49 ( This statement can be written as below.)

x/25= 49 ( Now we have to multiply both sides by 25, so we get)

`x/25` × 25 = 49 × 25

x = 1225

Example 5:

Solve the equation (x `xx` 23) – 22 = 25 for x.

Solution:

(x `xx` 23) – 22 = 25

( First we have to evaluate the expressions within the parenthesis. So, x `xx` 23 becomes 23x. Therefore, the given equation becomes like below.)

23x - 22 = 25 ( add 23 on both sides. So we get,)

23x - 22 + 22 = 25 + 22

23x = 25 ( divide both sides by 23)

`(23x)/23` = `25/23`

x = 1.09

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Practice Problems to Pre Algebra B Calculator:

Problem 1:

Solve the equation x + 23 = 28 for x.

Solution is, x = 5

Problem 2:

Solve the equation x - 23 = 29 for x.

Solution is, x = 52

Problem 3:

Solve the equation 3x = 242 for x.

Solution is, x = 80.67

Problem 4:

Solve the equation x ÷ 2 = 252 for x.

Solution is, x = 504

Steps to Solving Percentages

Introduction to solving percentages:

Percentages are used to express how large/small one quantity is, relative to another quantity. The first quantity usually represents a part of, or a change in, the second quantity, which should be greater than zero. Percentage was can represent the number in the form of percentage (%). For example 5% mean it was can written like as 5/100=0.05.

(Source –Wikipedia)

In this article steps to solving percentages, we see about some percentages example problems with detail steps.

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Example Problem for Solving Percentages with Steps:

Example problem for solving percentages with steps:

Example problem 1:

solving 24 percentages of 120?

Solution: Steps for fining percentages

Step 1: Given 24% of 120

Step 2: 24% it can written like as =>24/100

Step 3: 24% of 120=> 24/100*120.here we have to multiply 24/100 and 120

Step 4: 0.24(120)

Step 5:  24 percentages of 120 is 28.8



Example problem 2:

What percents of 43 is 9?

Solution:

Step 1: Given x% of 43=9

Step 2: Here Find the value of x

Step 3: x% of 43=9.It can written as like  x/100(43)=9

Step 4: x/100*43=9

Step 5: X=9*(100/43)

Step 6: X=900/35

Step 7: X=22.71%

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Example problem 3: Solving decreased percentages problem

A mobile was last year sold at the rate of  35000.now current year glass was sold at 2100 rate .solving the reduced (amount)percentage of mobile?

Solution:

Step 1: Given data:

Price of mobile (old price) =35000

Current price of mobile is (new price) =21000

Step 2: Reduced amount=old price of mobile – new price of mobile

Step 3: Reduced price= 35000-21000

Reduced price=14000

Step 4: Reduced percentage= (14000/35000)*100=40%

Step 5: 40% of the amount was reduced in the New Year.

Example Problem for Solving Percentages with Steps:

Example problem 4:

A glass shop pays some amount $70 to client, and then sells glass at the price of $120.Find the percentage of markup rate?

Solution:

Step 1:Given data paid amount =$70

Sales price =$120

Step 2: First we have to find the absolute amount( difference)

Markup amount=120-70=50

Step 3: Let us consider as x is percentage (unknown value)

Step 4: Now we have to find the x% of 70 is 50

Step 5: That amount is original markup rate of glass

X% of 70=>50

Step 6: x/100*70=>50

Step 7: x=50* (100/70)

Step 8: x=71.42%

Percentage of mark rate= 71.42%

Two Equations Two Unknowns

Introduction to two equations and two unknowns:

Two linear equations in the same two variables (unknowns) are called a pair of linear equations in two variables. The most general form of a pair of linear equations is

a1x+b1y+c1=0

a2x+b2y+c2=0

An example of linear system involves two equations and two unknowns:

x+ y=3 and x - y=2

There are the methods used to solve two unknowns with two equations:

Substitution method

Elimination method

Graphing method

Here, we are going to see the problems on two equations and two unknowns by substitution and elimination method.

Two Equations Two Unknowns-solving

Example problem 1:

Solve for the two variables x and y from the following two equations:

3x+y=5
y+5x=2

Solution:

Here, we have to solve the pair of equations with two unknowns by substitution method.

Step 1: We pick any one of the equations and write one variable in terms of the other.

Let us consider the Equation (1):

3x+y=5

Subtract 3x on both sides of the equation

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

y=5-3x------------------------Equation (3)

Step 2: Substitute the value of y in Equation (2). We get

y+5x=2

5-3x+5x=2

5+2x=2

Subtract 5 on both sides

2x=2-5

2x=-3

Divide by 2 on both sides of the equation

2x/2=-3/2

x=-1.5

Step 3: Plugging this value of x in Equation (3), we get

y=5-3x

y=5-3(-1.5)

y=5+4.5

y=9.5

So, the solution of two unknowns is (-1.5, 9.5).

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Two Equations Two Unknowns- by Elimination Method:

Example problem 2: Solve for the two variables x and y from the following two equations:

9x – 4y = 2000----------Equation (1)

7x – 3y = 2000----------Equation (2)

Solution:

We have to solve the pair of equations by Elimination method.

Step 1: Equation (1) is multiplied by 3 and Equation (2)is multiplied by 4 to make the coefficients of y equal. Then we get the equations:

27x – 12y = 6000------Equation (3)

28x – 12y = 8000------Equation (4)

Step 2: Equation (3) is subtracted from Equation (4) to eliminate y, because the coefficients

of y are the same. So, we get

(28x – 27x) – (12y – 12y) = 8000 – 6000

i.e., x = 2000

Step 3: Substituting this value of x in (1), we get

9(2000) – 4y = 2000

i.e., y = 4000

So, the solution of two unknowns is (2000, 4000).

Integer Linear Problem

Introduction to integer linear problem:

Integer linear problem can be solved under linear algebra category. The linear expression with integer term is called as integer linear function. The linear problem associates with the families of vectors called linear spaces, and the expression has the general form as input one vector and output one vector, based to certain rules.

Linear problem has the demonstration in analytic geometry and their functions with integer can be generalized in operator theory. The following are the examples for linear integer problem.

Example Problems in Linear Integer:

Example 1:

Solve the linear expression -2(c - 3) – 4c - 1 = 3(c + 4) - c

Solution:

Given expression is
-2(c - 3) – 4c - 1 = 3(c + 4) - c

Multiplying the integer terms
-2c + 6 – 4c - 1 = 3c + 12 - c

Grouping the above terms
-6c + 5 = 2c + 12

Subtract 5 on both sides
-6c + 5 - 5 = 2c + 12 -5

Grouping the above terms
-6c = 2c + 7

Subtract 2x on both sides
-7c – 2c = 2c + 7 -2c

Grouping the above terms
-9c = 7

Multiply -1/9 on both sides
C = - 7/9

C= - 7/9 is the solution for the given equation

Example 2:

Solve the linear expression     -5(k + 2) = k + 9

Solution:

Given expression is
-5(k + 2) = k + 9

Multiplying the factors in left term
-5k - 10 = k + 9

Add 10 on both sides
-5k - 10 + 10 = k + 9 + 10

Grouping the above terms
-5k = z + 19

Subtract k on both sides
-5z - k = k + 19 -k

Grouping the above terms
-6k = 19

Multiply -1/6 on both sides
K = -19/6

K = -19/6 is the solution for the given equation

Practice Problems for Linear Integer:

1) Solve the linear expression -5(z - 3) – 2z - 3 = 2(z + 1) – 3z

Answer: z = 13/4 is the solution for the above given equation

2) Solve the linear expression     -7(b - 2) – 2b - 2 = 5(b + 2) – 5b

Answer: b = 2/9 is the solution for the given equation

Mixed Number Percent

Introduction to mixed number percentage:

Numbers are the basic blocks of mathematics. Numbers are of different types, they are natural numbers, whole numbers, rational numbers, irrational numbers, mixed numbers, fraction numbers, decimal numbers etc.  Mixed number percentage is defined as converting a mixed fraction into percentage, for this the first step is to convert the mixed number into improper fraction, then  the improper fraction is converted into a decimal value and multiplied by  a value of hundred.

Problems for Mixed Number Percentage:

Problem 1:

Find mixed number percentage of 4 ½.

Sol:

Step 1: convert mixed number to improper fraction,

Step 2: then improper fraction to decimal value and then multiply by hundred,


4 1/2 = ((4*2) +1)/2 = 9/2 = 4.5

Multiply 100 with this to get the percent,

4.5 X 100 = 450%

So the result is 450%

Problem 2:

Find mixed number percent of 6 ½.

Sol:

Step 1: convert mixed number to improper fraction,

Step 2: then improper fraction to decimal value and then multiply by hundred,


6 1/2 =((6 X 2)+1)/2 = 13/2 = 6.5

Multiply 100 with this to get the percent,

6.5 X 100 = 650%

so the result is 650%

Problem 3:

Find mixed number percent of 8 ½.

Sol:

Step 1: convert mixed number to improper fraction,

Step 2: then convert  improper fraction to decimal value and then multiply by hundred,


8 1/2 = ((8 X 2) +1)/2 = 17/2 =8.5

Multiply 100 with this to get the percent,

8.5 X 100 = 850%

so the result is 850%

More Problems for Mixed Number Percentage:

Problem for mixed number percentage to fraction form:

Ex 1:

Find mixed number percentage of 8 ½.

Sol:

Step 1: convert mixed number to improper fraction,

Step 2: then improper fraction value divides by hundred,


8 1/2 = ((8 X 2) +1)/2 = 17/2 %

Multiply 100 with this to get the percent,

(17/2) * 100)

so the result is 850%

Ex 2:

Find mixed number percentage to fraction:

=7 ½ (mixed number)

Sol:

Step 1: convert mixed number to improper fraction,

Step 2: then improper fraction value divides by hundred,


7 1/2 = ((7 X 2) +1)/2 = 15/2 %

Multiply 100 with this to get the percent,

(15/2) X 100

so the result is 750%

Solve Simultaneous Linear Equations

Introduction :

An equation which has only one variable and its degree (power) is 1 called a simple equation.  A linear equation with only one variable is of the form ax + b=0.  A linear equation includes  two variables is in  the formation of  ax +by +c =0. Here the variables  x and y,and the  constants are a,b,c. When two variables in the linear equations are satisfied by the same pair of values of the variables, the equations are called simultaneous linear equations.

Methods of solving simultaneous linear equations:

(a)   Substitution method

(b)   Elimination method


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Steps for Solving the Simultaneous Linear Equation

Substitution method:

This involves the following steps,

1: Simplify the equations. 

2:  Solve one equation for any variable.

3: Substitute what you get for step 2 into the next equation.

4:  Solve for the next variable.

Example for solving simultaneous Linear equation by using Substitution method:

2x+3y= -4   ------------(1)

y=x-3  

Solution:

Plug y=  x-3 in equation 1

2x+3y= -4

2x+3(x-3)= -4

2x+3x-9= -4

5x-9= -4

5x=-4+9

5x=5

x=5/5

x=1

Plug in x=1 in y=x-3

y=x-3

y=1-3

y= -2

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Elimination Method for Solving Simultaneous Linear Equation:

The second method for  Solving simultaneous  equation  is Elimination method. It is also known as either addition or subtraction method. It is the concept of eliminating any one of the variable in the given equation either by adding or subtracting the equations.

In other words solving the simultaneous equation by making the co-efficient of any one of the variables in both equations has the same value. After adding or subtracting those two equations to form a new equation contains only one variable that is known as the eliminating the variable.

Example:

x+2y=3

2x+3y=4

Solution:

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

2x + 3y = 4          ---------------------------------(2)

Subtracting equation 2 from equation 1.

(1)*(2)=>     2x + 4y = 6

(2)*(1)=>     2x+ 3y =  4

----------------------------------------------------------

y = 2

Now plug in y=2 in equation (1)

x+2y=3

x+2(2)=3

x+4=3

x=3-4

x=-1