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)

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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)

Solve Multiplying Trinomials

In Algebra trinomial is a polynomial having three terms .otherwise it consisting of three monomial. For an example.

8x + 5y + 3z having x, y z variable

4t + 6s2 + 3y3 Here t, s, y is a variable

5ts + 9t + 6s with t, s variables

Here we are going to study about how to solve multiply a trinomial and its example problems.
Solve example problems for multiplying trinomial

Example: 1

Multiplying trinomial with binomial

Solve (x2+2x+3) (3x+3)

Solution:

First we have to take 3x multiply with trinomials

3x3+6x2+9x

Next we take a 3

3x2+6x+9

Now combine the like terms

3x3+6x2+3x2+9x+6x+9

3x3+10x2+15x+9

This is the multiplying answer.

Example: 2

Multiplying trinomial with another trinomial

Solve (x2 +3x+3) (2x2 +6x+4)

Solution:

Here there are two trinomial is given

First we have to multiply the x2  term to the other trinomial we get

2x (2+2) +6x (2+1) + 4x2

= 2x4+6x3+4x2

Next we take the 3x term we get

3*2x(1+2)+ 3*6x2+12x

= 6x3+18x2+12x

Finally we take a constant term 3

= 6x2+18x+12

Now we combine all the terms we get

= 2x4+6x3+4x2 + 6x3+18x2+12x + 6x2+18x+12

Combine the like term x3

6x3+6x3 = 12 x3

Combine x2 term

= 4x2+18x2+6x2

= 28x2

Combine x term

12x+18x=30x

Therefore the final answer is

2x4+12x3+28x2+30x+12

Multiplying trinomial with same trinomial

Example: 3

Solve (x2+3x+2) (x2+3x+2)

Solution:

Now we have to expand the polynomial

Take x2 and multiply with all the terms we get

X4+3x3+2x2

Similarly take 3x and multiply with all the term we get

3x3+9x2+6x

Finally take constant 2 multiply with all the term we get

2x2+6x+4

Now we combine all the term

X4 +3x3 + 2x2 + 3x3 + 9x2 + 6x + 2x2 + 6x + 4

Combine the like terms we get

3x3+3x3= 6x3

Combine the x2 terms

2x2+2x2+9x2 = 13x2

Similarly combine the x term

6x+6x=12x

Finally constant 4

Combine all the terms

x4+6x3+13x2+12x+4

This is the multiplying answer for the given two trinomials.

Math Game Absolute Value

In math, the absolute value is also known as the modulus |x| of a real number x is x's arithmetic value without consider to its sign. So, for example, 5 is the absolute value of both 5 and −5.

A simplification of the absolute value for real numbers occurs in an extensive selection of math settings.

Properties of the math game absolute value:

The absolute value has the following four fundamental properties:

`|x| = sqrt(x^2)`                                         (1) Basic

`|x| \ge 0 `                                                  (2)     Non-negativity

`|x| = 0 \iff x = 0`                                  (3)     Positive-definiteness

`|xy| = |x||y|\,`                                        (4)     Multiplicativeness

`|x+y| \le |x| + |y|`                                  (5)     Subadditivity

Other important properties of the absolute value include:

` |-x| = |x|\, `                                             (6)     Symmetry

`|x - y| = 0 \iff x = y `                            (7)     Identity of indiscernible (equivalent to positive-definiteness)

`|x - y| \le |x - z| +|z - y|`                        (8)     Triangle inequality (equivalent to sub additivity)

`|x/y| = |x| / |y| \mbox{ (if } y \ne 0) \,`                          (9)      Preservation of division (equivalent to multiplicativeness)

`|x-y| \ge ||x| - |y||`                                  (10)     (equivalent to sub additivity)

If y > 0, two other useful properties concerning inequalities are:

`|x| \le y \iff -y \le a \le y`

` |x| \ge y \iff x \le -y \mbox{ or } y \le x`

Math game absolute value – Games:

Math game absolute value – Game 1:

Arrange the order of ascending -|-15|, |12|,|7|,|-99|,|-5|,|-8|, |-65|, |6|

Solution:

First we remove the modulus symbol

-15, 12, 7, 99, 5, 8, 65, 6

Then to arrange the given order

-15, 5, 6, 7, 8, 12, 65, 99
Math game absolute value – Game 2:

Arrange the order of descending -|-16|, |13|,|8|,|-9|,|-6|,|-7|, |-66|, |63|, -|21|, |-68|

Solution:

First we remove the modulus symbol

-16, 13, 8, 9, 6, 7, 66, 63, -21, 68

Then to arrange the given order

68, 66, 63, 13, 9, 8, 7, 6, -16, -21

Math game absolute value – Game 3:

Absolute Value

Arrange the descending order

Solution:

First we remove the modulus symbol

-81, -28,67,-59,98,71,38

Then to arrange the given order

98, 71, 67, 38, -28, -59, -81

Evaluate Each Expression

Evaluate each expression is very simple concept in math. When you substitute a particular value for each variable and then act the operations called evaluating expression. The mathematical expression is algebraic, it involves a finite sequence of variables and numbers and then algebraic operations. Evaluation is simplifying the expression.

Algebraic operations are:

Addition

subtraction

multiplication

division

raising to a power

Extracting a root.

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Evaluating algebraic expressions are using PEMDAS method.

Rules for PEMDAS method:

1. First it performs operations inside the parenthesis.

2. Second exponents

3. Third multiplication and division from left to right

4. Finally addition and subtraction from left to right

Evaluate each expression for x=-5 and y=8

2x4-x2+y2

Solution:

Given expression is 2x4-x2+y2

Substitute x and y values in given expression

So,

=2(-5)4-(-5)2+ (8)2

=2(625)-(25) +64

=1250-25+64

=1314-25

=1289

Therefore answer is 1289

Evaluate the expression x3-x2y+9 for x=3, y=9

Solution:

Given expression x3-x2y+9

Plugging the x and y values in given expression

=33-329+9

=27-9*9+9

=27-81+9

=36-81

=-45

Therefore answer is -45

example problems for evaluating the expressions:

Evaluate each expression using PEMDAS method

3 * (9 + 12) - 62 `-:` 2 + 8

Solution:

Remember the PEMDAS rules

First parenthesis: (9+12) = (21)

3*21-62`-:` 2+8

Then exponents: 62 = 6 *6 = 36

3*21-36`-:` 2+8

Then multiplication and division: 3 * 21 = 63 and 36`-:` 2 = 18

63-18+8

Then addition: 63+8=71

71-18

Finally subtract the term

53

Therefore the answer is 53

Evaluate each expression for a = –2, b = 3, c = –4, and d = 4.

bc3 – ad

Solution:

= (3) (–4)3 – (–2) (4)

using PEMDAS exponent rule (-4)3=-4*-4*-4 = -64

= (3) (–64) – (–8)

then multiply the numbers 3*(-64) = -192

= –192 + 8

Add the numbers

= –184

(b + d)2

Solution:

= ((3) + (4))2

Add the inside numbers 3+4=7

= (7)2

using exponent rule 72=7*7=49

= 49

so, the answer is 49

a2b

Solution:

= (–2)2(3)

= (4) (3)

= 12

a – cd

Solution:

= (–2) – (–4)(4)

= (-2)-(-16)

= (-2) +16

= 16-2

= 14

b2 + d2

Solution:

= (3)2 + (4)2

= 9+16

= 25

Evaluate each expression For x = –3.

3x2 – 12x + 4

Solution:

= 3(–2)2 – 12(–2) + 4

= 3(4) + 24 + 4

= 12+24+4

=40

x4 + 3x3 – x2 + 6

Solution:

= (–3)4 + 3(–3)3 – (–3)2 + 6

= 81 + 3(–27) – (9) + 6

= 81 – 81 – 9 + 6

= –3

Solving Proportions

We will be solving proportions for x, by using our property that says the cross products of a proportion are equal to each other. If you remember the cross product we find them by multiplying a numerator one ratio time the denominator of the other. So just by looking over one proportion problems we will get the idea how exactly it is solved.

Say, x/15=21/45 here we will do the cross multiplication as 45 times x = 15 times 21 that is 45x= 315, now divide 45 both sides and we get x= 7. And this makes sense, 7 over 15 and 21/45 are equal fractions which is basically what proportions are.

One more question, 17/20=x/25. Here 17 times 25 equals to 20 times x. 425=20x, again dividing both the sides by 20 and we get the solution as x= 21.25. These types are nothing but theproportion solver.

We can learn this method only by practicing more by solving proportion examples. Thus the proportion examples are as follows, solving the proportion let us start with x/15=6/10. So here we have two ratios which are equal to each other. We need to find the value of x which will make the whole fraction proportionate to another.

We are going to find what this x number is going to represent so then this ratio in fact is equal to another ratio.  Using the cross product method will be used to solve this problem. we multiply two numbers which are diagonal from each other.10 times x=15 times 6.we get 10x=90, kindly remember there is no sign between the number and the variable it means they both have to multiply that here in this question will be 10 times x as 10 x. now by dividing 10 no both the sides,

we get the value of x as 9. Make sure if our solution is correct, simplify the problem by putting the value of x into the fraction, we will find the both fractions are proportionate to each other.

Let us take one more example, 3/15= y/50. Now again we have to solve this for the variable known as y. to do this we are going to find cross products. So 15 times y equals to 3 times 50, will be written as 15y=150, now dividing both the sides by 15 we get the value for y. and the value of y is 10.thus our final answer is y=10. And we can check that by plugging the value of y and simplifying shows that the fractions are in proportion.

Finctions Substituting

Substitution is the procedure of replace a changeable in a term with its real worth. If you are given an equation like 6x + 7 = 6+7, told that x = 1, and ask to find the value of the function what do you do? The first step is to replace with 1 for each 'x' in the problem. We get the expression as, 6(1) + 7 = 13

Functions substituting Example problems:

Problem 1: Find the functions substituting given below.

f(x) = 2x + 1

Substitute x=1, 2,3,4,5

Solution for  Functions substituting:

Step 1

x=1 f(1) =(2*1)+1

The answer is

f(1) =3

Step 2

x=2 f(2) =(2*2)+1

The answer is

F (2) =5

Step 3

x=3 f (3) =(2*3)+1

The answer is

F (3) =7

Step 4

x=4 f(4) =(2*4)+1

The answer is

F (4) =9

Step 5

x=5 f(4) =(2*5)+1

The answer is

F (4) =11

The substituting functions is 1, 2,3,4,5

F (1) = 3

F (2) =5

F (3) =7

F (4) = 9

F (5) = 11

Problem 2: Find the functions substituting given below.

f(x) = 3x +2

Substitute x=1, 2,3,4,5

Solution for  Functions substituting:

Step 1

x=1 f(1) =(3*1)+2

The answer is

f(1) =5

Step 2

x=2 f(2) =(3*2)+2

The answer is

f(2) =8

Step 3

x=3 f(3) =(3*3)+2

The answer is

f(3) =11

Step 4

x=4 f(4) =(3*4)+1

The answer is

f(4) =13

Step 5

x=5 f(5) =(3*5)+1

The answer is

f (5) =16

The substituting functions is 1, 2,3,4,5

F (1) = 5

F (2) =8

F (3) =11

F (4) =13

F (5) =16

Functions substituting Example problems:

Problem 3: Find the functions substituting given below.

F(x) = 2x +2

Substitute x=1, 2,3,4,5

Solution for  Functions substituting:

Step 1

x=1 f (1) =2*1+2

The answer is

f(1) =4

Step 2

x=2 f(2) =(2*2)+2

The answer is

f(2) =6

Step 3

x=3 f(3) =(3*2)+2

The answer is

f(3) =8

Step 4

x=4 f(4) =(2*4)+2

The answer is

f(4) = 10

Step 5

x = 5 f(5) = (2*5)+2

The answer is

f(5) = 12

The substituting functions  is 1, 2,3,4,5

F (1) = 4

F (2) = 6

F (3) = 8

F (4) = 10

F (5) =12

Problem 4.Find the functions substituting given below.

F(x) = 4x +4

Substitute x = 1, 2,3,4,5

Solution for Functions substituting

Step 1:

x=1 f (1) = (4*1)+4

The answer is

f(1) = 8

Step 2

x=2 f(2) = (4*2)+4

The answer is

f(2) =12

Step 3

x=3 f(3) =(4*2)+4

The answer is

f(3) =12

Step 4

x=4 f(4) =(4*4)+4

The answer is

f(4) =20

Step 5

x=5 f(5) =(4*5)+4

The answer is

f(5) =24

The substituting functions is 1, 2,3,4,5

F (1) =8

F (2)=6

F (3) =12

F (4) =20

F (5) =24

Math Word Problems Rate

Definition

Solving the Time, Speed, and Distance Triangle

The following formulas have been used if the speed is measured in knots, the distance in nautical miles, and the time in hours and/or tenths of hours (0.1 hour = 6 minutes).

Distance = Speed x Time

Speed = Distance ÷ Time

Time = Distance ÷ Speed

Math word problems rate - Examples

Math word problems rate - Example 1

A person crosses a 600 m long street in 5 minutes. What is his speed in km per hour?

Explanation:

Speed=`(600/(5*60))` = 2 m/sec.

Converting m/sec to km/hr (see important formulas section)

=`2*18/5`

=7.2 km/hr.

Math word problems rate - Example 2

The ratio between the speeds of two trains is 7: 8. If the second train is runs at 400 kms in 4 hours, then the speed of the first train is:

Explanation:

Let the speed of two trains be 7x and 8x km/hr.

Than, 8x=`400/4`

8x=100

X=12.5

Speed of first train = (7 x 12.5) km/hr = 87.5 km/hr.

Math word problems rate - Example 3

An aero plane covers with a certain distance at a speed of 240 kmph in 5 hours. To cover the same distance in 1 hours it must travel at a speed of:

Explanation:

Distance =` (240 x 5)` = 1200 km.

Required speed=`(1200*3/5)` km/hr

=720km/hr

Math word problems rate - Example 4

A man completes a journey in 10 hours. He travels the first half of the journey of the rate of 21 km/hr and second half at the rate of 24 km/hr. Find the total journey in km.

Explanation:

`(1/2)` `(x/21)` +`(1/2)` `(x/24)` =10

`(x/21)` +`(x/24)` =20

15x = 168 x 20

X(`168*` `20/15` `)` =224km

Math word problems rate - Example 5

A farmer travelled a distance of 61 km in 9 hours. He travelled partly on foot at 4 km/hr and partly of bicycle at 9 km/hr. The distance travelled on foot is:

Explanation:

Let the distance travelled on foot be x km.

Then, distance travelled on bicycle = (61 -x) km.

So, `x/4` + `(61-x)/9` =9

9x + `4(61 -x)` = 9 x 36

5x = 80

x = 16 km.

Math word problems rate - Practice Problem

Example

A man on tour of the travels at first 160 km in 64 km/hr and the next 160 km in 80 km/hr the average speed for the first 320 km of the tour is:

Answer=71.11 km/hr

Step by Step Differentiation

In calculus (a branch of mathematics) the derivative is a measure of how a function changes as its input changes. Loosely speaking, a derivative can be thought of as how much one quantity is changing in response to changes in some other quantity; the process of finding a derivative is called differentiation. The reverse process is called antidifferentiation (Source: Wikipedia)

General formula for differentiation:

`(d / dx)` (xn) = nx(n - 1)

`(d / dx)` (uv) = u `((dv) / (dx))` + v `((du) / (dx))`

Example problems for step by step differentiation

Step by step differentiation example problem 1:

Differentiate the given function u = 4x3 + 3x2 + 245x. Find the second derivative value of the given function.

Solution:

Given function is u = 4x3 + 3x2 + 245x

Step 1:

Differentiate the given function u with respect to x, we get

`((du) / (dx))` = (4 * 3)x2 + (3 * 2)x + 245

= 12x2 + 6x + 245

Step 2:

Differentiate the above value `((du) / (dx))` with respect to x, we get the second derivative value

`((d^2u) / (dx^2))` = (12 * 2)x + 6

= 24x + 6

The second derivative value of the given function is 24x+ 6

Answer:

The final answer is 24x + 6

Step by step differentiation example problem 2:

Differentiate the given function v = 9x2 + 12x. Find the second derivative value of the given function.

Solution:

Given function is v = 9x2 + 12x

Step 1:

Differentiate the given function u with respect to x, we get

`((dv) / (dx))` = (9 * 2)x + 12 + 0

= 18x + 12

Step 2:

Differentiate the above value `((dv) / (dx))` with respect to x, we get the second derivative value

`((d^2v) / (dx^2))` = 18 + 0

= 18

The second derivative value of the given function is 18

Answer:

The final answer is 18

Step by step differentiation example problem 3:

Differentiate the given function v = 11x4 + 41x3. Find the third derivative value of the given function.

Solution:

Given function is v = 11x4 + 41x3

Step 1:

Differentiate the given function u with respect to x, we get

`((dv) / (dx))` = (11 * 4)x3 + (41 * 3)x2

= 44x3 + 123x2

Step 2:

Differentiate the above value ((dv) / (dx)) with respect to x, we get the second derivative value

`((d^2v) / (dx^2))` = (44 * 3)x2 + (123 * 2)x

= 132x2 + 246x

Step 3:

Differentiate the above value with respect to x, we get the third derivative value

`((d^3v) / (dx^3))` = (132 * 2)x + 246

= 264x + 246

The third derivative value of the given function is 264x + 246

Answer:

The final answer is 264x + 246

Practice problems for step by step differentiation

Step by step differentiation practice problem 1:

Find the first derivative of the given function f (x) = 10x2 - 782x

Answer:

The final answer is f' (x) = 20x - 782

Step by step differentiation practice problem 2:

Find the Second derivative of the given function f (x) = 1.7x3 + 82x2 + 37

Answer:

The final answer is f'' (x) = 10.2x + 164

Gaussian Integer

A Gaussian integer is a complex number  whose real and imaginary part are both integers . That is  aGaussian integer is a complex number of the form a +ib where a and b are  integers.The Gaussian integers, with ordinary addition  and multiplication  of complex numbers, form an  integral domain, usually written as Z[i].

Formally, Gaussian integers are the set

\mathbb{Z}[i]=\{a+bi \mid a,b\in \mathbb{Z} \}.

Thse absolute value of   Z= a+ib  is √a2 + b2    .The square of  the  absolute value  is  called  the numbers complex norm.

Norm (Z)=a2 + b2

For example, N(2+7i) = 22 +72 = 53.

The norm is multiplicative  i.e.

N(z\cdot w) = N(z)\cdot N(w).
The only Gaussian integers which are invertible in Z[i] are 1 and i.

The units  of   Z[i] are therefore precisely those elements with norm 1, i.e. the elements
1, −1, i and −i.
Divisibility in Z[i] is de ned in the natural way: we say β divides α if
α = βγ for some
γ ε Z[i]. In this case, we call a divisor or a factor of .

A Gaussian integer = a + bi is divisible by an ordinary integer c if and
only if c divides  a and c divides b in Z.
A Gaussian integer has even norm if and only if it is a multiple of 1 + i.

Historical background

The ring of Gaussian integers was introduced by  Carl Friedrich Gauss    in his second monograph on (1832).  The theorem of  quadratic reciprocity   (which he had first succeeded in proving in 1796) relates the solvability of the congruence x2 ≡ q (mod p) to that of x2 ≡ p (mod q). Similarly, cubic reciprocity relates the solvability of x3 ≡ q (mod p) to that of x3 ≡ p (mod q), and biquadratic (or quartic) reciprocity is a relation between x4 ≡ q (mod p) and x4 ≡ p (mod q). Gauss discovered that the law of biquadratic reciprocity and its supplements were more easily stated and proved as statements about "whole complex numbers" (i.e. the Gaussian integers) than they are as statements about ordinary whole numbers (i.e. the integers).

Skew Normal Distribution

The skew normal probability distribution refers the normal probability distribution. It is also called as the Gaussian distribution. In normal distribution the mean is μ and the variance is `sigma^2` . Normal distribution is the close approximation of a binomial distribution. The limiting form of Poisson distribution is said to be normal distribution probability. This article has the information about the skew normal probability distribution.

Formula used for skew normal distribution:

The formula used for plot the standard normal distribution is

Z = `(X- mu) /sigma`

Where X is the normal with mean `mu` and the variance is `sigma^2` , `sigma` is the standard deviation.

Examples for the skew normal distribution:

Example 1 for the skew normal distribution:

If X is normally distributed the mean value is 1 and its standard deviation is 6. Determine the value of P (0 ≤ X ≤ 8).

Solution:

The given mean value is 1 and the standard deviation is 6.

Z = `(X- mu)/ sigma`

When X = 0, Z = `(0- 1)/ 6`

= -`1/6`

= -0.17

When X = 8, Z = `(8- 1)/ 6`

= `7/6`

= 1.17

Therefore,

P (0 ≤ X ≤ 4) = P (-0.17 < Z < 1.17)

P (0 ≤ X ≤ 4) = P (0 < Z < 0.17) + P (0 < Z < 1.17) (due to symmetry property)

P (0 ≤ X ≤ 4) = (0.5675- 0.5) + (0.8790 - 0.5)

P (0 ≤ X ≤ 4) = 0.0675 + 0.3790

P (0 ≤ X ≤ 4) = 0.4465

The value for P (0 ≤ X ≤ 4) is 0.4465.

Example 2 for the skew normal distribution:

If X is normally distributed the mean value is 2 and its standard deviation is 4. Determine the value of P (0 ≤ X ≤ 5).

Solution:

The given mean value is 2 and the standard deviation is 4.

Z = `(X- mu)/ sigma`

When X = 0, Z = `(0- 2)/ 4`

= -`2/4`

= -0.5

When X = 5, Z = `(5- 2)/ 4`

= 3/4

= 0.75

Therefore,

P (0 ≤ X ≤ 6) = P (-0. 5 < Z < 0.75)

P (0 ≤ X ≤ 6) = P (0 < Z < -0.5) + P (0 < Z < 0.75) (due to symmetry property)

P (0 ≤ X ≤ 6) = (0.6915 - 0.5) + (0.7734- 0.5)

P (0 ≤ X ≤ 6) = 0.1915+ 0.2734

P (0 ≤ X ≤ 6) = 0.4649

The value for P (0 ≤ X ≤ 6) is 0.4649.

Fundamental Theorem of Calculus Proof

The fundamental theorem of calculus determines the association the two basic operations of calculus called as the differentiation and integration. The first fundamental theorem of integration deals with the indefinite integration and the second theorem of integration deal with the definite integral of the function.

In this article we are going to see about the proof of the fundamental theorem for calculus.

Proof of fundamental theorem for calculus:

Proof of first fundamental theorem:

Let us take a real valued function f which is given by

x
F(x) = int f(t) dt, here f is also a real valued function
a

Then F is said to be continuous on [a,b] and can be differentiated on the open interval (a,b) which is given by

F′(x) = f(x), for all the values of x in the interval (a,b)

Let us consider two numbers x1 and x1+∆x , in the closed interval (a, b), then we have

x1                                                                        x1+∆x
F(x1) = int f(t) dt   -------> (1)     and       F(x1+∆x) = int f(t) dt ----------> (2)
a                                                            a

When we subtract the first equation and the second equation we get

x1+∆x                x1
F(x1+∆x) - F(x1) =  int   f(t) dt  -  int   f(t) dt ------------> (a)
a                    a

a              x1+∆x
=  int f(t) dt + int f(t) dt
x1                 a

a              x1+∆x           x1+∆x
int f(t) dt + int f(t) dt =  int f(t) dt
x1                 a                x1

When we substitute in equation (a) we get,

x1+∆x
F(x1+∆x) - F(x1) =  int   f(t) dt  -------------> (b)
x1

According to the theorem of integration

x1+∆x
int   f(t) dt  = f(c) ∆x  ,  given that there exists c in [x1 and x1+∆x]
x1

When we substitute this value in equation (b) we get,

F(x1+∆x) - F(x1) = f(c) ∆x

When we divide by ∆x on both sides we get,

F(x1+∆x) - F(x1) = f(c)
∆x

Take the limit ∆x  ->0, on both sides we get,

lim      F(x1+∆x) - F(x1) = lim  f(c)
∆x ->0            ∆x                        ∆x ->0

This left side of the equation is the derivative of F at x1

F′(x1)  = lim  f(c) -----> (3)
∆x ->0

We know that

lim    x1 = x1  and lim    x1+∆x  = x1
∆x ->0                 x1+∆x->0

Then according to the squeeze theorem, we get,

lim    c  = x1
∆x ->0

Substituting this in (c), we get

F′(x1) = lim  f(c)
c -> x1

The function f is continuous and real valued and thus we get,

F′(x1) = f(x1)

Hence the proof of the fundamental theorem for calculus.

Corollary of fundamental theorem proof:

The fundamental theorem is used to calculate the definite integral of a given function f if f is a real-valued continuous function on the interval [a, b], then

b
int f(x) dx = F(b) – F(a)

Learn Analytic Geometry Online

Another name of Analytical  geometry is co-ordinate geometry and it describes as a graph of quadratic equations in the co-ordinate plane. Analytical geometry grew out of need for establishing uniform techniques for solving geometrical problems, the aim being to apply them to study of curves, which are of particular importance in practical problems. In this article we shall discuss the learning analytic geometry ans some example problems

Example Problem 1 to learn analytic geometry online:

Find the equation of the parabola if the curve is open upward, vertex is (− 1, − 2) and the length of the latus rectum is 4.

Solution:

Since it is open upward, the equation is of the form

(x − h)2 = 4a(y − k)

Length of the latus rectum = 4a = 4 and this gives a = 1

The vertex V (h, k) is (− 1, − 2)

The equation of parabola is [x-(-1)]2=4*1 [y-(-2)]

[x+1]2=4[y+2]

Example Problem 2 - Learn analytic geometry online  :

Find the equation of the parabola if

(i) the vertex is (0, 0) and the focus is (− a, 0), a > 0

Solution: (i) From the given data of the parabola is open leftward

The equation of the parabola is of the form

(y − k)2 = − 4a(x − h)

Here, the vertex (h, k) is (0, 0) and VF = a

The required equation is

(y − 0)2 = − 4a (x − 0)

y2 = − 4ax.

Example Problem 3 - learn analytic geometry online :

Find the equation of the parabola if the curve is open rightward, vertex is (2, 1) and passing through point (6,5).

Solution: Since it is open rightward, the equation of the parabola is of the form

(y − k)2 = 4a(x − h)

The vertex V(h, k) is (2, 1)

∴ (y − 1)2 = 4a (x − 2)

But it passes through (6, 5)

(5-1)2 = 4a (6 − 2) -- > 16= 4a * 16

4a= 1 ---- > a = 1/4

The required equation is (y − 1)2 = 1/4 (x − 2)

Example Problem 4 - learn analytic geometry online :

Find the equation of the parabola if the curve is open leftward, vertex is (2, 0) and the distance between the latus rectum and directrix is 2.

Solution: Since it is open leftward, the equation is of the form

(y − k)2 = − 4a(x − h)

The vertex V(h, k) is (2, 0)

The distance between latus rectum and directrix = 2a = 2 giving a = 1 and the equation of the parabola is

(y − 0)2 = − 4(1) (x − 2)

or y2 = − 4(x − 2)

Practice problems- learn analytic geometry online:

1.Find the equation of the parabola whose vertex are (1, 2) and the equation of the directrix x = 3.

The required equation is

(y − 2)2 = 4(2) (x − 1)

(y − 2)2 = 8(x − 1)

2.The separate equations of the asymptotes of the hyperbola 4x2-25y2=100

Answer   :x/5-y/2=0 and x/5+y/2=0

Definition to Population Variance

Population variance is defined as the square of the mean deviation value by the total number of data. Population variance is calculated to find the variance in the probability of data.

Formula for finding the

sigma^2 =( sum_(k=1)^n(x_k - mu)^2) / N

Where as

sigma^2 - symbol for population variance

mu - It is the mean of the given data.

N  - Total number of values given in data set.

For finding the population variance we have to find the sample mean. For the take the average for the given values.

 mu = (sum_(K=1) ^n (x_k))/N

This mean value is used in the population variance to find its value.

Population Variance Values - Example Problems:

Population Variance Values - Problem 1:

calculate the population variance for the given data set. 2, 4, 3, 3, 5, 6, 5

Solution:

Mean:  Calculate the mean for the given data

 mu = (sum_(K=1) ^n (x_k))/N

using the above formula find the average for the given values.

 mu = (2+4+3+3+5+6+5)/7

mu = 28/ 7

 mu = 4

Population Variance: Calculate the population variance value from the mean.

sigma^2 =( sum_(k=1)^n(x_k - mu)^2) / N

substitute the mean values to find the deviation values from the given values.

 sigma^2 = ((2-4)^2+(4-4)^2+(3-4)^2+(3-4)^2+(5-4)^2+(6-4)^2+(5-4)^2)/7

sigma^2 = 12/7

sigma^2 = 1.7142857142857

Hence the population variance value is calculated using the mean value.

Algebra is widely used in day to day activities watch out for my forthcoming posts on Differentiation Math and What are Composite Numbers. I am sure they will be helpful.

Population Variance Values - Problem 2:

Find the value for the population variance of the given data set. 367, 378, 365, 366.

Solution:

Mean: Calculate the mean for the given data set using the formula,

 mu = (sum_(K=1) ^n (x_k))/N

Find the mean value that is average of the giave data set

mu = (367+378+365+366)/4

mu = 1476 / 4

mu = 369

Population variance: Calculate the population variance value from the mean.

sigma^2 =( sum_(k=1)^n(x_k - mu)^2) / N

substitute the mean values to find the deviation values from the given values.

sigma^2 =((367-369)^2 + (378-369)^2 + (365 - 369)^2+(366-369)^2) / 4

sigma^2 = 110/4

sigma^2 = 27.5

Hence the population variance value is calculated using the mean value.

Population Variance Values - Problem 3:

Find the value for the population variance of the given data set. 36, 37, 36, 39.

Solution:

Mean: Calculate the mean for the given data set using the formula,

 mu = (sum_(K=1) ^n (x_k))/N

Find the mean value that is average of the giave data set

mu = (36+37+36+39) / 4

mu = 148 / 4

mu = 37

My previous blog post was on Natural Numbers please express your views on the post by commenting.

Population variance: Calculate the population variance value from the mean.

sigma^2 =( sum_(k=1)^n(x_k - mu)^2) / N

substitute the mean values to find the deviation values from the given values.

"sigma^2 = ((36-37)^2 +(37-37)^2+(36-37)^2+(39-37)^2) / 4

sigma^2 = 6/4

sigma^2 = 1.5

Hence the population variance value is calculated using the mean value.

Population Variance Values - Practice Problems:

Find the value for the population variance of the given data set. 45, 47, 49, 43.

Answer: Population Variance = 5

Find the value for the population variance of the given data set. 87, 89, 78, 82, 89.

Answer: Population Variance = 18.8

Polynomial Matrices

A polynomial matrices or matrix polynomial is a  matrix whose elements are univariate or multivariate polynomials.A univariate polynomial matrix P of degree p is define like:sum_(n=0)^pA(n)x^(n)=A(0)+A(1)x+A(2)x^(2)+....+A(p)x^(p) where A(p) is non-zero and A(i) indicate a matrix of constant coefficients. Hence a polynomial matrix is the matrix-equivalent of a polynomial, by means of every one element of the matrix satisfying the classification of a polynomial of degree p.

Properties of polynomial matrices

A polynomial matrix in excess of a field with determinant equivalent to a non-zero constant is called unimodular, and have an inverse, which is also a polynomial matrix.
Note, that the simply scalar unimodular polynomials are polynomials of degree 0 - nonzero constants, for the reason that an inverse of an arbitrary polynomial of high degree is a rational function.
The roots of a polynomial matrix in excess of the complex numbers are the points in the complex plane wherever the matrix loses rank.

Characteristic polynomial of a product of two matrices

If A and B are two square n×n matrices then,attribute polynomials of AB and BA match:

PAB(t)=PBA(t).

If A is m×n-matrix and B is n×m matrices such that m
PAB(t)=tn-mPAB(t)

Polynomial in t and in the entry of A and B is a general polynomial identity. It consequently suffice to verify it on an open set ofparameter value in the complex numbers.

The tuples (A,B,t) wherever A is an invertible complex n by n matrix,

B is any complex n by n matrix,

and t is any complex number since an open set in complex space of dimension 2n2 + 1. When A is non-singular our result follow from the fact that AB and BA are similar:

BA=A-1(AB)A.

Example 1:

An example the 3x3 polynomial matrices

P=[[1,x^(2),x],[0,2x,2],[8x+2,x^2-1,0]]

=[[1,0,0],[0,0,2],[2,-1,0]]+[[0,0,1],[0,2,0],[3,0,0]]x+[[0,1,0],[0,0,0],[0,1,0]]xx^(2)

Example 2:

Find the eign value of given polynomial matrices

P=[[3,3],[0,6]]

The polynomial has the characteristic equation

0=det(P-λI)

=det[[3-lambda,3],[0,6-lambda]]

18-6lambda -3lambda + λ2

18-9lambda +18

λ2-3λ-6λ+18

λ(λ-3)-6(λ-3)

(λ-3)(λ-6)

λ=3,andλ=6

The eigenvalues of these matrices are 3,6

Example 3: Find the product of the given matrices M1=[[1,2],[3,4]] and M2=[[8,3],[2,7]]

The given polynomial is

M1=[[1,2],[3,4]]


M2=[[8,3],[2,7]]

The product of the given matrices M1 and M2  =M1xM2

M1xM2    =[[1,2],[3,4]]xx [[8,3],[2,7]]

The product of the given  matrices is=[[12,17],[32,37]]

Least Common Denominator

If the denominators are unlike, then we can find LCD (Least common Denominator) of the given denominators.

Least Common Denominator is the smallest positive(least) integer which is common in multiples of the denominators.

For example, given fractions are 1/3 and 1/6. Find LCD.

List the multiples of 3:   3, 6, 9, 12, 15, 18, 21,...

Multiples of 6:   6, 12, 18, 24,...

Here, 6 is the lowest common term for both the multiples of 3 and multiples of 6.

The answer is 6, and that is the Least Common Denominator.

There are five examples for least common denominator. From these examples for least common denominator, we can get clear view about least common denominator. Let us see the examples for least common denominator in the following section.

Examples on examples for least common denominator:

We are going to explain examples for least common denominator.

Example 1:

Find the least common denominator of the fractions; 1/5,1/3

Solution:

Here, the denominators are 5 and 3.

The common denominator of 5 and 3 is 15.

Multiples of 5: 5,10,15,20,…..

Multiples of 3: 3,6,9,12,15,18,….

Here, 15 is the lowest common term for both the multiples of 5 and multiples of 3.

The answer is 15, and that is the Least Common Denominator.

Example 2:

If the given fractions are 3/4,1/3, find the least common denominator.

Solution:

Here, the denominators are 4 and 3.

The common denominator of 4 and 3 is 12.

Multiples of 4: 4,8,12,16,20….

Multiples of 3: 3,6,9,12,15….

Here, 12 is the lowest common term for both the multiples of 4 and multiples of 3.

So, the Least Common Denominator is 12.

Example 3:

Find the least common denominator of ; 5/6, 2/15

Solution:

Here, the denominators are 6 and 15.

The common denominator of 6 and 15 is 30.

Multiples of 6:  6,12,18,24,30…

Multiples of 15: 15,30,45,60….

Here, 30 is the lowest common term for both the multiples of 6 and multiples of 15.

So, the Least common denominator is 30.

Example 4 on examples for least common denominator:

What is the least common denominator of the fractions;  5/12, 11/18

Solution:

Here, the denominators are 12 and 18.

The common denominator of 12 and 18 is 36.

Multiples of 12:  12,24,36,48,….

Multiples of 18: 18,36,54,….

Here, 36 is the lowest common term for both the multiples of 12 and multiples of 18.

So, the Least common denominator is 36.

Example 5:

1/5 + 1/6 + 1/15  What is the LCD?

Solution:

First we list the multiples of each denominator.

Multiples of 5 are 10, 15, 20, 25, 30, 35, 40,...

Multiples of 6 are 12, 18, 24, 30, 36, 42, 48,...

Multiples of 15 are 30, 45, 60, 75, 90,....

Here, 30 is the lowest common term for both the multiples of 5 and multiples of 6 and multiples of 15 .

So, the Least common denominator is 30.

Therefore, Examples for least common denominator are explained.