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.

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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.

                                             

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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)

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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.   

Basic Math Facts

 Basic math facts article deals with the basic facts behind the various arithmetic operations. The method or steps need to be followed for solving the basic math problem.
                           The Basic math facts are:

• Addition Basic facts.
• Subtraction Basic facts
• Multuiplication Basic facts.
• Division Basic facts.

Basic math facts for addition and subtraction:

 Addition Basic facts:
           The basic math facts for the addition of two numbers.

1. Symbols for the both number should be same.
2. negative number  (+) negative number = negative number
3. Positive number  (+) postive number = positive number

Model problems:

1. 15+15

             Solution:
                          Here the sign for both are same. So add the two numbers
                                        = 15+15
                                        = 30 (positive number)
2. -15 -15
               Solution:
                       Here the both numbers are negative numbers, so add the two numbers.
                                       = -15-15
                                       = -30 (negative number)
Subtraction Basic facts:
           The basic math facts for the subtraction of two numbers.

1. Symbols for the numbers should be different.
2. postive number  (-) negative number = postive  number
3. Negative number  (-) postive number = negative number

Model problems:

1. 15-10

 Solution:
                          Here the sign for both are different. So subtract the two numbers
                                        = 15-10
                                        = 5 (positive number)
2. -15 +10
 Solution:
                       Here the both numbers are having different sign, so subtract the two numbers.
                                       = -15+10
                                      = -5 (negative number)

Basic math terms for multiplication and divsions:

Mulitplication Basic facts:
           The basic math facts for the multiplication of two numbers.

• Symbols for the both number may  be same or different
• negative number  (*) negative number = positive number
• Positive number  (*) postive number = positive number
• postive number (*) negative number = negative number

Model problems:
1.15*15
             Solution:
                          Here the sign for both are same. Therefore, answer is positive numbers
                                        = 15*15
                                        = 225 (positive number)
2.-15 *15
         Solution:
                       Here both numbers are  not of same sign

                                       = -15*15
                                       = -225 (negative number)
3.-15 * -15
Solution:
                      Here the sign for both are same. Therefore, answer is positive numbers
                                        = -15* -15
                                        = 225 (positive number)

Division Basic facts:
           The basic math facts for Division of two numbers.

1. Symbols for the both number may  be same or different
2. (negative number') /( negative number) = positive number
3. Positive number  / postive number = positive number
4. postive number / negative number = negative number

Model problems:
1.15/15
 Solution:
                          Here the sign for both are same. Therefore, answer is positive numbers
                                        = 15/15
                                        = 1 (positive number)
 2.-15 /15
 Solution:
                       Here both numbers are not of same sign
                                       = -15/15
                                       = - 1 (negative number)

3.-15 / -15
Solution:
                      Here the sign for both are same. Therefore, answer is positive numbers
                                        = -15/ -15
                             =  1 (positive number)