About Square Root

Square root of a number it can be represented by `sqrt(y)` . The square root of number is also written as exponent form `sqrt(y)` which is equal to y^1/2
                Square root of a number 25 which is equal to `sqrt(25)` =25^1/2=5.
                In this article we shall discuss about the some square roots of the numbers.

Square root table

                The following table shows the square root of the some numbers.
                Number (x)        square (x²)          Square root (x^1/2)
                1                              1                              1.000
                2                              4                              1.414
                3                              9                              1.732
                4                              16                           2.000
                5                              25                           2.236
                6                              36                           2.449
                7                              49                           2.646 
                8                              64                           2.828 
                9                              81                           3.000 
                10                           100                           3.162

Problem 1:

Find the Square root value of 150
Solution:
                Write prime factors for 150 = 2 * 3 * 5 * 5
                `sqrt(150)` = `sqrt(2 * 3 * 5 * 5)`
                                   = `sqrt(2 * 3 * 5^2)`
             `sqrt(2 * 3 * 5^2) ` we can written as (2 * 3 * 5²)^1/2
                (2 * 3 * 5²)^1/2      By using algebraic property (a*b)^m =a^m * b^m
                        (2 * 3)^1/2 * (5^2*1/2)
                5 (2 * 3)^1/2
                        Square root value of 150= 5 `sqrt(6)`

Problem 2:
Find the Square root value of 300
Solution:
                Write prime factors for 300 = 2 * 2 * 3 * 5 * 5
                `sqrt(300)` = `sqrt(2 * 2 * 3 * 5 * 5)`            
                                   = sqrt(2^2 * 3 * 5^2)
                `sqrt(2^2 * 3 * 5^2)` we can written as (2 * 3 * 5²)^1/2
                (2² * 3 * 5²)^1/2     By using algebraic property (a*b)^m =a^m * b^m
                        (3)^1/2 * (2^2*1/2 * 5^2*1/2)
                5 * 2 (3)^1/2
                        Square root value of 300=10 `sqrt(3)`

Problem 3:
Find the Square root value of 625
Solution:
                Write prime factors for 625 = 5 * 5 * 5 * 5
                `sqrt(625) ` = `sqrt(5 * 5 * 5 * 5)`   
                                   = `sqrt(5^2 * 5^2)`
                `sqrt(5^2 * 5^2)` we can written as (5² * 5²)^1/2          
                (5² * 5²)^1/2            By using algebraic property (a*b)^m =a^m * b^m
                        (5²)^1/2 * (5^2*1/2)
                5 * 5
                        Square root value of 625= 25

Problem 4:
Find the Square root value of 75
Solution:
                Write prime factors for 75 = 3 * 5 * 5
                `sqrt(75)` = `sqrt(3*5*5)`
                                   = `sqrt(3*5^2)`
                `sqrt(3*5^2)` we can written as (3*5²)^1/2 
                (3*5²)^1/2               By using algebraic property (a*b)^m =a^m * b^m
                        (3)^1/2 * (5^2*1/2)
                5 (3)^1/2
                        Square root value of 75= 5 `sqrt(3)`

Divide Complex Fractions

Complex fractions are having both the numerator and the denominator values. The complex fractions are also consists of number of rational fractions. The other names for the complex fractions are compound fractions. For example, `(1/3)/(2/3)` is called as the division complex fractions.  Complex fractions are also used for many of the arithmetic operations.

Example problem for divide complex fractions

The example problem for divide complex fractions is,

Problem 1: Divide the following complex fractions, `(2/3)/(3/8)` .

Solution:

Step 1: In the divide complex fractions, the first step is to we have to convert the given fraction into the following,

`(2/3)/(3/8)` = `(2)/(3)` `-:``(3)/(8)`

Step 2: In this step we have to multiply the fraction, so that we have to take reciprocals to the second fractions,

= `(2)/(3)` `xx` `(3)/(8)`

=`(2)/(8)`

=`(1)/(4)`

This is the required solution to the divide complex fractions.

Problem 2: Divide the improper fraction into the complex fractions, `(5 (3/4))/(2/4)` .

Solution:

Step 1: In the first step, we are convert the following into the proper complex fraction, we get,

`(23/4)/(2/4)` = `(23)/(4)` `-:` `(2)/(4)`

Step 2: In the next step, we are multiplying the complex fractions,

`(23)/(4)` `xx` `(4)/(2)`

=`(23)/(2)`

This is the required complex fractions.

Problem 3: Divide the improper fraction into the complex fractions, `(6 (2/8))/(5/8)` .

Solution:

Step 1: In the first step, we are convert the following into the proper complex fraction, we get,

`(50/8)/(5/8)` = `(50)/(8)` `-:` `(5)/(8)`

Step 2: In the next step, we are multiplying the complex fractions,

`(50)/(8)` `xx` `(8)/(5)`

=`(50)/(5)`

`This is the required solution for the complex fractions.`

Problem 4: Divide the following complex fractions, `(2/8)/(4/6)` .

Solution:

Step 1: In the divide complex fractions, the first step is to we have to convert the given fraction into the following,

`(2/8)/(4/6)` = `(2)/(8)` `-:``(4)/(6)`

Step 2: In this step we have to multiply the fraction, so that we have to take reciprocals to the second fractions,

= `(2)/(8)` `xx` `(6)/(4)`

=`(3)/(8)`

`(1)/(4)`

This is the required solution to the divide complex fractions.

Practice problem to divide complex fractions

Problem 1: Divide the following complex fractions, `(2/3)/(4/6)` .

Answer: `(2)/(3)`

Problem 2: Divide the improper fraction into the complex fractions, `(4 (2/4))/(1/4)` .

Answer: 32

Solve Stata Box Plots

Stata box plot is to construct the box and whisker plot for the given statistical records. Statistical data are very large in numbers, so for dividing and conquer the given data, understanding on the given data is necessary, and to find the relationship between the given data, studying stata box plot is more helpful.

Looking out for more help on Anova in Stata in algebra by visiting listed websites.

Solving stata box plots:

While studying stata box plots the data is to be detached. The values obtained from the given datas are Q1, Q2, and Q3, these are median values. Studying stata box plot facilitate for proportional learning of a mixture of sections of box plots.

Steps to study stata box plots:

Step 1: The first step is to arrange the given data.

Step 2: The second step is to find the median (Q2)  for the given data, if there should be two central numbers means find the mean value, it should be the median, if the count is to be odd means, median is the middle value.

Step 3: Find the values present before Q2 is called as lower quartile, the values present after Q2 is called as upper quartile.

Step 4: Find Q1 and Q3 percentiles (middle values of lower quartile and upper quartile).

Step 5: Find the subordinate and superior values.

Step 6: Mark the positions of Q1 , Q2 and Q3 percentiles in the lattice.

Step 7: Draw a box from lower to upper quartile through Q2

Step 8: Find the inter quartile range.

Step 9: Draw a line from lower to higher series.
Example to study stata box plots:

Draw a stats box plots for the given data

Cricket :    10, 20, 90, 20, -10, 80, -20, 25, 65, 85, 5

Footbal:    20, -30, 50, 60, 90, 10, 70, 65, 35, 55, 95

Solution for learning stata box plots is:

Sort out the data in ascending order.

Cricket :  -20, -10, 5, 10, 20, 20, 25, 65,  80, 85,  90

Football:  -30, 10, 20, 35, 50, 55, 60, 65, 70, 90, 95

Median:

Median (Cricket)   Q2  = 20
Median (Football) Q2  = 55
Lower Median(Q1)

Cricket :  -20, -10, 5, 10, 20, 20, 25, 65,  80, 85,  90

Football:  -30, 10, 20, 35, 50, 55, 60, 65, 70, 90, 95

Lower numbers are

Cricket :  :  -20, -10, 5, 10, 20

Football   :   -30, 10, 20, 35, 50

Middle (Cricket ) Q1= 5

Middle (Football) Q1= 20

Upper Median

Upper numbers are

Cricket :  -20, -10, 5, 10, 20, 20, 25, 65,  80, 85,  90

Football:  -30, 10, 20, 35, 50, 55, 60, 65, 70, 90, 95

Upper three numbers are

Cricket  :  25, 65,  80, 85,  90

Football :   60, 65, 70, 90, 95
Middle (Cricket  ) Q3= 80

Middle (Football) Q3 = 70


Maximum

Cricket :  -20, -10, 5, 10, 20, 20, 25, 65,  80, 85,  90

Football:  -30, 10, 20, 35, 50, 55, 60, 65, 70, 90, 95


Maximum (Cricket) = 90

Maximum (Football) =95


Minimum

Cricket :  -20, -10, 5, 10, 20, 20, 25, 65,  80, 85,  90

Football:  -30, 10, 20, 35, 50, 55, 60, 65, 70, 90, 95
Minimum (Cricket) = -20

Minimum (Football) =-30


box plot

Study stata box plots

Practice problem for learning stata box plots:

Draw stata box plot for the given set of report

Brand X      : 22, 23, 39, 27, 44, 22, 53, 32, 12, 56, 67

Brand Y      : 22, 34, 76, 45, 98, 35, 19, 49, 12, 32, 34

Multiply and Simplify

Multiplication is an one of the arithmetic operation which extending one number by another. In other words, multiplication is a product of one number to another number. Simplifying or reducing means to make a fraction as simple as possible. In other words, simplifying means both the denominator and numerator divided by the common number. For example `3/6` =`1/2` . Here 3 is a common divisor.

Example problems of multiply and simplify:

Multiplication problem 1:

Jack delivers 350 newspapers in a day. How many newspapers does he deliver in 8 days?

Solution:

We can find the count of newspapers by using the multiplication.

Jack delivers 350 newspapers per day.

Therefore delivering a newspapers in 8 days= 350*8

Here we are Multiplying the 350 and 8. Then we get the final answer.

350*8= 2800

Answer: 2800 newspapers

Multiplication problem 2:

A lodge charges $200 per night for a room. What is the cost of booking the room for 3 nights?

Solution:

We can find the count of cost by using the multiplication.

A lodge charges $200 per night for a room.

Therefore cost of booking room for 3 days = 200*3

Here we are multiplying the 200 and 3. Then we get the final answer.

200*3= 600

Answer: $600

Simplification problem 1:

Simplify the fraction `6/15`

Solution:

We can simplify the given fraction by using the following method.

In this problem, first we can find Greatest common divisor.

6, 15 = 3 is a GCF

Therefore, we are dividing by 3 on both numerator and denominator.

Then we get,

6/15= `2/5`

Answer: `2/5`

Simplification problem 2:

Simplify the fraction `4/18`

Solution:

We can simplify the given fraction by using the following method.

In this problem, first we can find Greatest common divisor.

4, 18 = 2 is a GCF

Therefore, we are dividing by 2 on both numerator and denominator.

Then we get,

`4/18` = `2/9`

Answer: `2/9`

Practice problems of simplify and multiply:

Multiply 25*56
Simplify 5/50
Simplify `14/18`

Answer: 1) 1400 2) `1/10` 3) `7/9`

How to Compute Variance

In probability theory and statistics, the variance is used as one of several descriptors of a distribution. In particular, the variance is one of the moments of a distribution. The variance is a parameter describing a theoretical probability distribution, while a sample of data from such a distribution can be used to construct an estimate of this variance: in the simplest cases this estimate can be the sample variance. (Source: Wikipedia)

I like to share this Calculate Sample Variance with you all through my article. 

How to Compute Variance - Examples

Example 1: In class 7 student’s height are as follows 144, 154, 175, 180, 165, 160, 170 centimeters. Compute the variance of given data. 

Solution:

Mean = Sum of all the elements in a data set / total number of elements in a data set

by adding and dividing by 7 to get

       `barx` = 1148 / 7 = 164

Table for getting the variance

x	x – 164	(x - 164 )2
144	-20	400
154	-10	100
175	11	121
180	16	256
165	1	1
160	-4	16
170	6	36
Total		930

Formula to compute the variance is

`s^2 = 1/(N-1) sum_(i=1)^n (x_i-barx)^2`

Put all the values in the formula to get

930 / (7-1) = 155

Therefore variance is 155.

Example 2: In a class 9 student’s weight are 45, 50, 61, 85, 62, 72, 66, 75, 78 kilograms. Compute the variance of given data. 

Solution:

Mean = Sum of all the elements in a data set / total number of elements in a data set

Mean by adding and dividing by 9 to get

        x = 594 / 9 = 66

Table for getting the variance:

x	x – 66	(x - 66 )2
45	-21	441
50	-16	256
61	-5	25
85	19	361
62	-4	16
72	6	36
66	0	0
75	9	81
78	12	144
Total		1360

Formula to find the variance is

`s^2 = 1/(N-1) sum_(i=1)^n (x_i-barx)^2` 

Put all the values in the formula to get

1360 / (9-1) = 170       

Therefore variance is 170

How to Compute Variance - Practice

Problem 1: In class 7 student’s height are as follows 154, 164, 185, 190, 175, 170, 180 centimeters. Compute the variance of given data. 

Answer: 155

Problem 2: In a class 9 student’s weight are 55, 60, 71, 95, 72, 82, 76, 85, 88 kilograms. Compute the variance of given data. 

Answer: 170

Problem 3: 9 person's age are 25, 35, 30, 42, 45, 60, 39, 14, 52 years. Compute the variance of given data. 

Answer: 195.5

Subdivision In Math

Algebra is a subdivision in math, which comprises of infinite number of operations on equations, polynomials, inequalities, radicals, rational numbers, logarithms, etc. Sketching graphs for algebraic equations or function is also a part of algebra. Graphing is nothing but the pictorial view of the given function or equation to study their characteristics, it may be a line, parabola, hyperbola, curve, circle, etc. Graphing negative square root is also done in algebra. The procedure to graph negative square root  function with the graph is explained in the following sections.

                                             

Is this topic Negative Correlation Graph hard for you? Watch out for my coming posts.

Procedure for graphing negative square root:

The procedure for graph negative square root function is given below,

Step 1: Consider the equations as y = `sqrtx` , 

Step 2:  y =`sqrtx` . Since the given equation is a function of x, let y =f(x).

Step 3: Therefore f(x) =` sqrtx`

Step 4: Substitute various values for ‘x’ and find corresponding f(x).

Step 5: Tabulate the values as columns x & f(x).  The values of x as -6, -5, -4, -3, -2, -1, 0, and for f(x), their corresponding  values. Positive values cannot be used, since they tend to become imaginary.

Step 6: The values in the table are the co-ordinates, graph them.

Step 7: Connect the points to find the shape of the function.

Graph for negative square root:

The graphs for the negative square root function is shown below,
1. Convert the given equation as

y = `sqrt(-x)`

Since the given equation is a function of y,

Let y =f(x).

Therefore

f(x) =`sqrt(-x)`

Substitute various values for ‘y’ and find corresponding f(y).

When x= -9

f(-3) = `sqrt(-(-9))`

3

 therefore the co-ordinates are (-9,3)

When x= -4

f(-2) = `sqrt(-(-4))` ,

2,

therefore the co-ordinates are (-4, 2)

When x= -1,

f(-1) = `sqrt(-1)` ,

1,

1, therefore the co-ordinates are (-1, 1)

I am planning to write more post on Selection Bias Example and What is Percentage?. Keep checking my blog.

When x= 0

f(0) =` sqrt(0)` ,

0,

0, therefore the co-ordinates are (0, 0)

Following the above procedure and the co-ordinates are,

                      
The graph for the above table is shown below,

Variable Selection

The variable selection represents that the process of assigning the variable to some values and functions. The differentiation process sometimes involve big functions on that time the reference variable is used to assign and then it process into the simple calculations. In this way some confusions exist for the students for which variable we can ssign and for which functions. In this article we are going to discuss about the variable selection process in detail with the applications of the differentiation and integration process.

Examples for the variable selection in integration

• Review on the variable selection with the integrable function  `int tanh^-1 2theta ` `d theta`    

Solution:

Variable Selection:

Here the confusion exist for the variable selection.

Only one function exist  `tanh^-1 2theta`        

The another function is the constant term   1 d`theta`                   

Another one confusion

For which term we can take it as u and dv

If we choose u then it should be easily differentiable.

If we choose dv then it should be easily integrable.

u = `tanh^-1 2theta` and `(du)/(d theta)` = `theta` ;   dv = 1 d`theta`

du =` 1/(1-(2theta)^2) 2 d theta = 2/(1-4theta^2) d theta ` and  v = `theta` .

Therefore,

`int tanh^-1 2theta ` = `theta`   `tanh^-1 2theta` – `int theta 2/(1-4theta^2)` d`theta`   = `theta`   `tanh^-1 2theta` – `2int theta 1/(1-4theta^2)` d`theta` .

Let     u = `1-4theta^2`  

 `(du)/(d theta)` = -8`theta` ,

`(-1/8)` du = `theta` d`theta` .

 `int tanh^-1 2theta ` = `theta`   `tanh^-1 2theta` – `2int theta 1/(1-4theta^2)` d`theta` .

 = `theta`   `tanh^-1 2theta` `-` 2`int 1/(u) (-1/8)` du

= `theta`   `tanh^-1 2theta` `+`    `(1/4) int 1/(u)` du

 =  `theta`   `tanh^-1 2theta` `+`   ` (1/4) int ` `u^(-1)du`     

=    `theta`   `tanh^-1 2theta` `+` `(1/4) (log u)` + C

 =   `theta`   `tanh^-1 2theta` `+` `1/4 log(1-4theta^2)` + C 

Problems for variable selection in differentiation

• Review on the variable selection with the differentiable function `y = e^x [sin x + cos x]` calculate   `y'' - 2y' + y `         

Solution:

Given     `y = e^x [sin x + cos x]`           

The formula is given as

    `d/dx [f(x). g(x)] = f'(x).g(x) + f(x).g'(x)`      

or      `d/dx [u.v] = u'.v + u.v'`      

Variable Selection:

Here sometime one problem exist, which variable we can take to apply in the formula.

We can take any of them.

Take v = `sinx + cosx`  u =`e^x`     

     `v'`   = `cos x - sinx`   `u'` =`e^x`    

First Differentiation:

      `y' = u'.v + u.v'`

      `y' = e^x [sin x + cos x] + e^x[cos x - sin x]`

      `y' = e^x sin x + e^x cos x + e^x cos x - e^x sin x`    

      `y' = 2 e^x cos x `    

Second  Differentiation:

      `y'' = 2e^x [-sin x] + 2e^x[cos x ]`    

      `y'' = 2e^x cos x - 2e^x sin x`      

  `y'' - 2y' + y `  = ` 2e^x cos x - 2e^x sin x``+` `[-2( 2 e^x cos x)] ` `+ e^x sin x + e^x cos x`             

  `y'' - 2y' + y `  = ` 2e^x cos x - 2e^x sin x``-` `4 e^x cos x ` `+ e^x sin x + e^x cos x`             

  `y'' - 2y' + y `  = ` [2 - 4 + 1]e^x cos x + [1-2]e^x sin x`                    

  `y'' - 2y' + y `  = ` -e^x cos x - e^x sin x`   is the required solution.