Average Deviation Calculator

The average deviation in which it is defined to calculate the average for the deviations for the given data. It is calculated by taking sum for all values of the deviation divided by the total number  of values in the given data set.

For finding the average deviation value, find the mean for the given data is using the formula,

`barx = (sum_(k = 1)^n (x_n))/ (n)`

Calculate the deviation value by using the calculated mean value, by using the formula given below,

`(x - barx)^2`

Average Deviation is calculated by taking average for the founded deviation value of the data set.

Average Deviation = `"(sum_(K=1)^n (x-barx^2)) / N`

` T It is nothing but the above shown formulas the population of the variance formula in which it is also called as the average deviation.`

Steps to Calculate the Average Deviation:

1. Give the average for all the given dimensions of the data set .
2. Give the difference of the initial value of the data and the average value we have found which is called as mean difference.
3. Take the all absolute value from this mean difference of the given data.
4. Repeat the steps 2 and 3 for all the other given values and find the mean difference to the data set.

Understanding example of algebraic expression is always challenging for me but thanks to all math help websites to help me out. 

Screen Shot for the Calculator to Find the Average Deviation:

For finding the average for the squared mean difference which is known as the average deviation of the given data set.



Select the formula to find the result and put the values in input field  then press calculate button.



Average Deviation Calculator - Example Problems:

Average deviation calculator - Problem 1:

Calculate the average deviation for the given data set. 35, 36, 37, 38.

Solution:

Mean: Formula for finding mean,

`barx = (sum_(k = 1)^n (x_n))/ (n)`

`barx = (35+36+37+38) / 4`

` barx = 146 / 4`

` barx =` 36.5

Calculate the deviation for the given data set from the mean,

Deviation  =  `(x - barx)^2`

= `((35-36.5)^2+(36-36.5)^2+(37-36.5)^2+(38-36.5)^2)`

Deviation = 5

Calculate the average deviation for the given data,

Average Deviation = `(sum (x-barx)^2) / (n)`

= `5 / 4`

Average Deviation   = 1.25

Average deviation calculator - Problem 2:

Calculate the average deviation for the given data set. 33, 35, 36, 38.

Solution:

Mean: Formula for finding mean,

`barx = (sum_(k = 1)^n (x_n))/ (n)`

`barx = (33+35+36+38) / 4`

` barx = 142 / 4`

` barx =` 35.5

Calculate the deviation for the given data set from the mean,

Deviation  =  `(x - barx)^2`

=`((33-35.5)^2+(35-35.5)^2+(36-35.5)^2+(38-35.5)^2)`

Deviation = 13

Calculate the average deviation for the given data,

Average Deviation = `(sum (x-barx)^2) / (n)`

= `13 / 4`

Average Deviation   =  3.25

Average deviation calculator - Problem 3:

Calculate the average deviation for the given data set. 2, 5, 6, 5, 4, 7, 8, 5.

Solution:

Mean: Formula for finding mean,

`barx = (sum_(k = 1)^n (x_n))/ (n)`

`barx = (2+5+6+5+4+7+8+5) / 8`

` barx = 42/ 8`

` barx =` 5.25

Calculate the deviation for the given data set from the mean,

Deviation  =  `(x - barx)^2`

= `((2-5.25)^2+(5-5.25)^2+(6-5.25)^2+(5-5.25)^2+(4-5.25)^2+(7-5.25)^2+(8-7.25)^2+(5-5.25)^2)`

Deviation = 16.5

Calculate the average deviation for the given data,

Average Deviation = `(sum (x-barx)^2) / (n)`

= `16.5 / 8`

Average Deviation   = 2.0625

Average Deviation Calculator - Practice Problems:

1 Calculate the average deviation for the following 55.3, 56.6, 50.9 and 54.0.

Answer: Average Deviation = 4.47500

2. Calculate the average deviation for the following data

Answer: Average Deviation = 2

Algebra is widely used in day to day activities watch out for my forthcoming posts on Standard Deviation Calculator and relative standard deviation. I am sure they will be helpful.

Statistical Power Table

Statistical power is important in math. In math statistical power means probability of reject a false null hypothesis. In statistics to test hypotheses and also test the null hypothesis. The power is equal to 1-beta. In odd position we want to reject our null hypothesis in favor of the alternative. It is more helpful for exam preparation.

Basic Statistical Power of Negative Z-score Table:

In the following statistical power of negative z score table to explain the how to calculate the probability value




For example,

P(0 > Z > 2.31) = P( 0> Z > `oo` ) − P(-1.13 > Z >0)

= 0.5 − 0.1292 (Use statistical power of negative z score table to calculate the probability value)

= 0.3708

Select the first column value -1.13 then choose the right side eight column value 0.03 in the same direction we got an answer as 0.1292

Example Problems for Statistical Power Table:-

Problem 1:-

Calculate the standard deviation range of P(0 > Z > 1.9) by using statistical power table

Solution:

P(0 > Z > 1.9) = P(-1.9< Z<0)

= 0.0287 (Use statistical power of negative z score table to calculate the probability value)

Select the first column value -1.9 then choose the right side eleventh column value 0.0 in the same direction we got an answer as 0.0287

I like to share this algebra problems with you all through my article.

Problem 2:-

Calculate the standard deviation range P(-2.2 > Z >3.1) by using statistical power table

Solution:

P(-1.2 > Z >2.1) = P(− 2.2 > Z > 0) + P(-3.1 > Z > 0)

= 0.0139+ 0.0010 (Use statistical power of negative z score table to calculate the probability value)

= 0.133

Select the first column value -2.2 then choose the right side eleventh column value 0.0 in the same direction we got an answer as 0.0139

Select the first column value -3.1 then choose the right side eleventh column value 0.0 in the same direction we got an answer as 0.0010

Algebra is widely used in day to day activities watch out for my forthcoming posts on statistical graphs and free algebra help. I am sure they will be helpful.

Adding and Subtracting Vectors

The addition may be represented graphically by placing the start of the arrow b at the tip of the arrow a, and then drawing an arrow from the start of a to the tip of b. The new arrow drawn represents the vector a + b.



To subtract b from a, place the end points of a and b at the same point, and then draw an arrow from the tip of b to the tip of a. That arrow represents the vector a − b.
Source: Wikipedia



Adding and subtracting of vectors operation and example problems are given below.

Operations for on Vectors:

1. Addition of vectors:

Let `vec(OA)` = `veca` , `vec(AB)` =` vecb` . Join OB.

Then `vec(OB)` represents the addition (sum) of the vectors veca and vecb.



This is written as `vec(OA) ` +` vec(AB)` = `vec(OB)` Thus `vec(OB)` = `vec(OA) ` + `vec(AB) ` = `veca ` +` vecb`

This is known as the triangle law of addition of vectors which states that, if two vectors are represented in magnitude and direction by the two sides of a triangle taken in the same order, then their sum is represented by the third side taken in the reverse order.

Applying the triangle law of addition of vectors in




ΔABC, we have BC + CA = BA ⇒ BC+ CA = − AB

⇒ AB + BC + CA = 0

Thus the sum of the vectors representing the sides of a triangle taken in order is the null vector.

2. Subtraction of vectors:

If `veca` and `vecb` are given two vectors, then the subtraction of `vecb` from `veca` is defined as the sum of `veca` and − `vecb` and  denoted by `veca` − `vecb` .



`veca ` − `vecb` = `veca` + ( − `vecb` )

Let `vec(OA)` = `veca` and `vec(AB)` =` vecb `

Then `vec(OB)` = `vec(OA)` + `vec(AB)` = `veca` + `vecb`

To subtract `vecb` from `veca` , produce BA toAB' such that AB = AB'.

∴ `vec(AB')` = − `vec(AB)` = −` vecb`

Now by the triangle law of addition

` vec(OB')` = `vec(OA)` +` vec(AB')` = `veca` + ( `-vecb` ) =` veca ` − `vecb`

Example Problems for Adding and Subtracting of Vectors:

Example problem 1:

The position vectors of the points A, B, C, D are `veca` , `vecb` , `2veca` + `3vecb` ,`veca` − `2vecb` respectively. Find `vec(DB) ` and `vec(AC)`

Solution:

Given that

` vec(OA)` = `veca` ; `vec(OB)` =` vecb` ; `vecOC` =` vec2 ` + `3vecb` ; `vec(OD)` = `veca ` − `2vecb `

`vec(DB)` = `vec(OB)` − `vec(OD)` = `vecb ` − (`veca` − `2vecb` ) = `vecb ` − `veca` + `2vecb` = `3vecb` − `veca`

`vec(AC)` = `vec(OC)` − `vec(OA)`

= (`2veca` + `3vecb` ) − `veca` = `veca` + `3vecb `

Having problem with cbse maths sample papers for class 10 keep reading my upcoming posts, i will try to help you.

Example problem 2:

` vec(OA)` = `2veca` + `3vecb` - `vecc` , `vec(OB)` = `4veca` + `2vecb` + `2vecc` , Find adding and subtracting of vectors

Solution:

Subtracting the two vector,

`vec(AB) ` = `vec(OB)` - `vec(OA)`

`vec(AB)` = `4veca` + `2vecb ` +` 2vecc` - `2veca` - `3vecb` +` vecc`

`vec(AB)` = `2veca` - `vecb ` + `3vecc`

Adding the two vector,

` vec(AB)` = `2veca` + `3vecb` - `vecc` + `4veca` + `2vecb` + `2vecc`

` vec(AB)` = `6veca` +` 5vecb` + `vecc`

I like to share this Adding Fraction with you all through my Blog.

Accurate to Two Decimal Places

Decimal numbers contains decimal places. Here it is very important to know about the place value in decimal numbers.

So what is Decimal Place Value?

It can  be explained by taking example:



let us discuss it using whole number which does not have decimal point in it.

As we go from  right to left the value, the position or the place value gets 10 times bigger.

Similarly as we move from left to right the place value gets 10 times smaller.

What do we get as we move beyond unit place?

Here is where the decimal point comes into picture. If we move beyond unit place the value becomes 1/10, 1/100, 1/1000 and so on. These number which contains decimal point is called DECIMAL NUMBER.

Below is the example for decimal number:



here we can see that the place value of number goes on decreasing as we move beyond decimal points. The value of one in the above example is 1/10 , the place of 2 become 1/100

I like to share this class xii cbse sample papers with you all through my Blog.

Conversion of Decimal Number to Whole Number (accurate to Two Decimal Places)

HOW TO CONVERT  DECIMAL NUMBER TO WHOLE NUMBER?

This can be explained using eg given below:

consider 125.43 and also 125.56

First check the number immediately after the decimal  place. If it is greater than or equal to 5 then add 1 to the unit place and write it as whole number otherwise discard the numbers after decimal point and write the whole number as it is.

In the above example 1)  the number 4 appears immediately after decimal point which is less than 5 .So there is no change and we have to discard all the number proceeding decimal point i,e 4 and 3 and write the number as it is.

whereas in case of example 2) the number which appears immediately is 5 so add one to unit place which becomes 126


Accurate Writing of the Decimal Numbers to 2 Decimal Places

How to write numbers accurate to 2 decimal places?

Even here we have to follow the same rules mentioned above.

examples:

123.4612 ----------------------->123.46

123.4689 ------------------------> 123.47


In the above example 1) check the third number after the decimal point. In this case the third number is 1 which is smaller than 5 so don't add 1 to the number which in 1/100's  place and discard the 3rd digit after the point and write as it is. Here  the final number accurate to 2 decimal point becomes  1 2 3 4 . 4 6

example 2) here the 3rd digit after decimal point is 8 which is greater than 5 so add 1 to the second digit after decimal. Here  the final number accurate to 2 decimal point becomes  1 2 3 4 . 4 7

Algebra is widely used in day to day activities watch out for my forthcoming posts on  list of irrational numbers. I am sure they will be helpful.

Intersection of Two Lines Calculator

Straight line :-

A straight line is generally termed as line. The  curvature of a straight line is 0 . The orientation of a straight line is given by its slope.General representation of straight line is given by  AB.

Explanation to Intersection of Two Lines Calculator

Intersecting lines:

Two lines are said to be intersecting if and only if the have a common root  or  solution.The general form of equation of a line is given by Y=mX +c Where   m= slope, c= y intercept of line .

Features of straight line  of form Y= mX +c :-

i) The straight line is parallel to X axis of m = 0 ie slope of line is 0.

ii) The straight line passes through orign when constant "c" =0.

iii)The straight line makes an angle 45owith "X" axis when m=1 and c=0.

iv) The straight line makes an angle 135o with "X" axis when m=-1 and c=0.

v) The straight line parallel to "y" axis has slope infinity .

Please express your views on Algebra by commenting on blog.

Condition for two line to  intersect :-

For two lines to intersect, their slopes must not be equal .Lines having same slopes are called as parallel lines and intersection point of two parallel lines is 0 or we can say that they meet at infinity. There can be  only one point of intersection for two distinct lines  if they  are not parallel.Multiple intersection of two lines is not possible.

Finding point of intersection of two lines :-

The point of intersection of two lines-can be found by two methods :-

i) By solving the equations

ii) By graphing the equation

Both of these approaches lead to same answer

lets us take two lines  y = 2x+3 ; y= -0.5x + 7 and find the point of intersection

here  primary examination of slopes is to be done.

Slopes are 2& -0.5 so these lines are not parallel lines

So we go by first method solving equations

y = 2x+3 ; ------ equation 1

y= -0.5x + 7 --------equation 2

=> 2x+3  = -0.5x + 7

=> 2.5x = 7-3

=> x= 4/2.5

=> x=1.6

substituting in equation 1 or 2 we get

 y= 2*(1.6) +3

=> y= 6.2

so (1.6 , 6.2) is the point of intersection of  y = 2x+3 ; y= -0.5x + 7

on graphing the equations and plotting them



we can observe that the point of intersection is (1.6, 6.2).

Hence both the approaches give the same result .

Having problem with 12th cbse sample paper keep reading my upcoming posts, i will try to help you.

Examples to Intersection of Two Lines Calculator

 EX 1:-

find point of intersection of  lines 3y= 6x +3 and y= 2x+3

Solution:-

Given equations 2y= 4x +2 and y= 2x+3

comparing with standard form of equation y=mx +c

slope of 3y= 6x +3 => y= 2x+1 is 2

slope of line y= 2x+3 is 2

here slopes of lines are equal so they are parallel

so there will be no point of intersection for lines  3y= 6x +6 and y= 2x+3.

 EX 2 :-

FInd the point of intersection of lines y= 2x+1  & x=2

solution :-

Here both lines are not parallel as on comparing with standard equation y=mx+c

slope of  y= 2x+1  is 2 and   x=1 is infinite as its parallel to y axis

substituting x=2 in y= 2x+1

=> y= 2*2+1

=> y= 5

so point of intersection of  y= 2x+1  & x=1 is (1,5)

EX 3:

Find the point of intersection of "X" axis and "Y" axis

solution:

equation of  x axis is  y=0

equation of y axis is x=0

So point of intersection of x, y axis is (0,0) ie origin.

Solve Real Life Algebra Problems

Introduction:

In this branch of mathematics, we use the alphabetical letters in problems like a, b, x and y to denote numbers. In real life algebra, the various operations are addition, subtraction, multiplication, division or extraction of roots on these symbols and real numbers are called algebraic expressions. In the real life a Greek mathematician Diaphanous has developed and solve this subject to a great extent and hence we call him as the father of algebra.

Solve Real Life Algebra Problems-solved Problems :

Example1:

A woman on tour travels first 160 km at 60 km/hr and the next 160 km at 80 km/hr. The average speed for the first 310 km of the hour. Find the total time?

Solution:

The given data’s which are taken and can be solve as follows,

Total time taken = (160 / 60+160 / 80) hr.

=12/3 hr.

Example 2:

The ratio between the speeds of two trains is 5 : 6. If the second train runs 300 km in 3 hours, then what is the speed of the first train?

Solution:

This can be solve as below,

Speed of two trains 5 and 6 km/hr

6x = (300 / 3)

=100

X = (100 / 6) =16

Speed of first train = 16 × 5km/hr

= 80km/hr

Average speed= (340 ×12 / 3)km/hr

= 87.62km/hr.

I like to share this 10 class cbse syllabus with you all through my blog.

Solved Problems for Real Life Algebra:

Example1:

A man traveled a distance of 30 km in 6 hours. He traveled partly on foot at 2 km/hr and partly on bicycle at 6 km/hr. What is the distance traveled on foot?

Solution:

The problem which can be solve as,

Distance traveled on bicycle = (30 - x) km

x / 2 + (30-x) / 6 = 6

6x + 2 (30-x) = 6 × 12

x = 12 km

Example 2:

A bus crosses a 600 m long street in 2 minutes. What is his speed in km per hour?

Solution:

This is a real life problem in algebra which can be solve as,

Speed = (600 / 2 × 60)m/sec

= 5 m/sec

We have to convert  m/sec to km/hr

= (5 × 18/2)

= 45 km/hr

Example 3:

The ratio between the speeds of two trains is 3 : 5. If the second train runs 300 kms in 3 hours and average speed is 50 km/hr.what is the speed of the first train?

Solution:

The problem in the real life algebra which can be solve as follows,

Speed of two trains 3 and 5 km/hr

5x = (300 / 3)

=100

X = (100 / 5) = 20

Speed of first train = 20 × 5 km/hr

= 100 km/hr

Average speed = (50 ×9 /3 ) km/hrs

= 150 km/hr.

These are all the solve problems which are in the real life algebra.

Envision Math California

Envision MATH California is Based on problem-based, interactive knowledge and theoretical understanding. What does that mean? Students with theoretical thoughtful know more than inaccessible details and techniques, they are capable to study fresh thoughts by linking to the thoughts they previously be acquainted with. They study ideas by interact with problems; this communication takes place on a daily basis and is exact to the idea being trained.lets see some problems on envision MATH California.

Envision Math California Problems:-

Problem 1:-

Find the perimeter of the following figure:-  

                                                                                                         
Solution:-

We need to find the perimeter of the following figure

Perimeter can be found by adding lengths of the given figure.

Here the length of each sides are the following

4.8 yards, 4.3 yards, 1.1 yards, 4.5 yards, 4.6 yards, 1.6 yards.

Perimeter = sum of all given sides

= 4.8 yards + 4.3 yards + 1.1 yards + 4.5 yards  + 4.6 yards + 1.6 yards.

= 20.09  yards.

So the perimeter of the given figure is 20.09 yards.

Having problem with 10 class cbse syllabus keep reading my upcoming posts, i will try to help you.

Problem 2:-

Find the Perimeter of the given figure.

Solution:-

We need to find the perimeter of the following figure

Perimeter can be found by adding lengths of the given figure.

Here the length of each sides are the following

4.7 yards, 2.3 yards, 1.1 yards, 1.1 yards,  4.3 yards, 2.4 yards, 3 yards.

Perimeter = sum of all given sides

= 4.7 yards+ 2.3 yards+1.1 yards+ 1.1 yards+  4.3 yards+ 2.4 yards+ 3 yards.

= 18.9 yards.

So the perimeter of the given figure is 18.90 yards.

Problem 3:-

Find the Perimeter of the given figure.

Solution:-

We need to find the perimeter of the following figure

Perimeter can be found by adding lengths of the given figure.

Here the length of each sides are the following.

2 feet, 2feet, 5 feet, 5 feet, 2 feet.

Perimeter = sum of all given sides

= 2 feet+ 2feet+ 5 feet+ 5 feet+ 2 feet.

= 16 feet

So the perimeter of the given figure is 16 feet.

Envision Math California Practice Problems:-

Find the area of the given parallogram:-

Answer:- 84 cm2


Between, if you have problem on these topics Business Math, please browse expert math related websites for more help on area of hexagon formula.