Inverse Variation

To understand the inverse variation,let is correlate it with our real life of inverse variation examples such as the total time taken with the speed traveled in a trip. Assuming distance of a trip would be fixed, let us say the distance of a trip would be 250 mile.

I like to share this Inverse Sine Function with you all through my article. 

Here  inverse variation equation will be used as T=250/r , according to the time will be  T be time of the trip will be equal to 250  divided by the Rate in miles per hour which will called r .so if rate is 50 miles per hour it would take,  5 hours  .

The next of the inverse example is total time taken to spread the soil and the number of people working on a part of land. so let us say if the total time of the job will take 12 hours the total time T to complete the job would be  12 hours divided by n, ( n is the number of workers).

Here the equation will be T= 12/n.  so if have 1 worker , we take 12 hours, 3 workers we  take 4 hours and so on.

Total amount of cost required by per person for petrol and the number of people in a cab. So if we know that it will cost 45 dollars’ worth of petrol to go for a picnic trip. The cost represents C= 45 dollars divided by number of people in the cab.

That is C= 45/no of people. So if n will increases the total cost decreases.
Here in inverse variation definition is as the ones quantityvalue gets bigger and the others value gets smaller in such that there product remains the same or proportional. Let us understand with one more example, Y varies inversely as X. where Y= 4 and when X=2.

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Find the value of the inverse-variation equation. Then determine Y when X= 16.  By performing substitution we can solve this question.  Y= k/x.  by doing cross multiplication , we get 4= k/2 which is k=8.  So our inverse-variation will be  Y=K/x that is 8/x. Now we can answer the second query on determining the y when x is 16. So, y= 8/x , substituting the value of x as 16. So y= 8/16. And we get  1/2 . y= ½ and x= 16.

Solve Laplace Equation

In mathematics, Laplace equation is one of the most interesting topics in second order partial differential equations. The two dimensional heat equations in Cartesian form in unsteady state is

`(delu)/(delt)` = `alpha^2 [(del^2u)/(delx^2) + (del^2u)/(dely^2)]`

I like to share this Laplace Transform Pairs with you all through my article.

For the steady state `(delu)/(delt)` = 0, then the above equation becomes

`[(del^2u)/(delx^2) + (del^2u)/(dely^2)]` = 0

(or)

`grad^2 u` = 0

This is the two dimensional heat equation in unsteady state (or) Laplace equation. In this article, we shall discuss about how to solve the Laplace equation. The following are the example problem in solve Laplace equation.

Solve Laplace equation - Laplace equation:

Finite difference solution for the two dimensional heat equation in steady state (or) Laplace equation is

`grad^2 u` = 0

(or)

`(del^2u)/(delx^2)` + `(del^2u)/(dely^2)` = 0

Here we know the values of `(del^2u)/(delx^2)` , `(del^2u/dely^2)`

`(del^2u)/(delx^2)` = `(u_(i+1, j) + u_(i-1, j)-2u_(ij))/(h^2)`

`(del^2u)/(dely^2)` = `(u_(i, j+1) + u_(i, j-1)-2u_(ij))/(h^2)`

substitute the values of `(del^2u)/(delx^2)` , `(del^2u)/(dely^2)` in the Laplace equation, then we get

`4u_(ij)` = `(u_(i+1, j) + u_(i-1, j) + u_(i, j+1) + u_(i, j-1))`

`u_(ij)` = (`1/4` ) `(u_(i+1, j) + u_(i-1, j) + u_(i, j+1) + u_(i, j-1))`

This is the finite difference solution for the Laplace equation, which is also called as the standard five-point formula.

Solve Laplace equation

Solve Laplace equation - Example problem:

Example 1:

Solve the laplace equation for the given region

Solve Laplace equation - Example

Solution:

The finite difference sheme for the Laplace equation is

Solve Laplace equation - Example

u(ij) = (`1/4` ) [`(u_(i+1, j) + u_(i-1, j) + u_(i, j+1) + u_(i, j-1))` ]

Using this we can solve this,

u1 = `1/4` [200 + 50 + u2 + u4]

= `1/4` [250 + u2 + u4]                                → (1)

u2 = `1/4` [u1 + 0 + 0 + u3]

= `1/4` [u1 + u3]                                         → (2)

u3 = `1/4` [u4 + u2 + 100 + 50]

= `1/4` [150 + u2 + u4]                                → (3)

u4 = `1/4` [100 + 200 + u1 +  u3]

= `1/4` [300 + u1 + u3]                                → (4)

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Solve the equation  by using the gauss seidal method

u1 = `1/4` [250 + u2 + u4]

u2 = `1/4` [u1 + u3]

u3 = `1/4` [150 + u2 + u4]

u4 = `1/4` [300 + u1 + u3]

Let the intial approximation is

Learn Slope Form

The ratio of rise value of the line equation to the run value of the line equation is called as slope. Slope form learning is one of the important part of algebra. In other words, it can be defined as the ratio of the change of  Y values to the change of the X value. Generally, the slope of the equation is denoted as "m". It is necessary to learn the slope form. learn slope form is mainly used in linear equation.

learn slope form of the linear equation

The slope form of the linear equation can be written as,

y = mx + b

Where, y is the line equation,

m is the slope of the equation and

b is the Y intercept value.

In point form, the line equation can be written as,

Y - Y1 = m (X - X1)

Where, X1 and y1 are the points of the line

m is the slope of the equation.

I like to share this Math Slope Formula with you all through my article.

Given two points, we could find the slope using the formula

m = (Y2 - Y1) / (X2 - X1)

Where, (X1, Y1 ) and  (X2, Y2 ) are any two points in the line.

Example problems for learn slope form

Ex:1 Find the slope of the given equation Y = 7X + 6.

Sol: In general form, the line equation is given as,

y = mx + b

where, m is the slope of the equation.

In the above equation, the slope is coefficient of x,

slope = 7

Ex:2 Find the slope of the given points (1, 3) and (3, 6).

Sol:

The given points are (1, 3) and (3, 6).

Here, X1 = 1, Y1 =3

X2 = 3, Y2 = 6

The formula for finding the slope is given as,

m = (Y2 - Y1) / (X2 - X1)

Substitute all  values in the above equation,

m = (6 – 3) / (3 -1)

= 3 / 2

Ans:

The slope of the given equation is m= 3/2

Ex:3 Find the slope of the given points (2, 6) and (6, 14).

Sol: The given points are (2, 6) and (6, 16)

Here, X1 = 2, Y1 = 6

X2 = 6, Y2 = 16

The formula for finding the slope is given as,

m = (Y2 - Y1) / (X2 - X1)

Substitute all  values in the above equation,

m = (16 – 6) / (6 -2)

= 5 / 2

Ans:

The slope of the given equation is m = 5 / 2

Expressions With Rational Exponents

A rational number is a number represented as p/q where p and q are integers and q != 0.Exponent means  power.

The large numbers are very difficult to read and write and understand. To make them simpler we can use exponents, with the help of this many of the large numbers are converted to simpler forms. The short notation 74 stands for the product 7 x 7 x 7x 7. Here ‘7’ is called the base and ‘4’ the exponents. The number 74 is read as 7 raised to the power of 4 or simply as fourth power of 4. 74 are called the exponential form of 2401.

The above example is of a natural number as exponents.

Ex 5 ^(3/2) here 3/2 is a rational number so this is an example of rational exponents.

Rules invovled for solving rational exponents:

The Rational exponents rules are:

p,q are any real numbers except zero and m,n are positive integers.

Rule I:The base value is same under multiplication so we can directly add the power values. Then the exponent take the following form

pm  × pn = pm+n

Solving the expression 73  × 75 = 73+5 =78

Rule II:The base value is same under division so we can directly subtract the power values. Then the exponent take the following form such that m > n then

(p^m)/(p^n) = pm-n

Solving the expression  (4^4)/(4^2)= 44-2=42

Rule III:If m < n, (p^m)/(p^n) =(1)/(p^(n-m))

Solving the expression 8^3/8^2 =1/8

Rule IV:A real number to the power of power

(pm )n  = pm*n

Solving (82 )3  = 82*3=86

Rule V:The power of 0 is

p0 = 1. Anything power zero is equals to 1.

p^m/p^m   = pm-m =p0 = 1

Rule VI:p to the power of negative number is

p-m =1/p^m
Solving the expression 2-5 =  1/2^5

Rule VII:Two numbers to the same power can be written as

pm  × qm = (p*q)m

Solving the expression 32 × 82 = (3*8)2=242

Rule VIII A number to rational power is written as

p^(m/n) = root(n)(p^m)

Examples of expressions having rational exponents:

Ex : 1Solve the exponent    2x^(1/3)xx8^(5/3)xxx^(2/3)

Sol:Step 1:Given

2x^(1/3)xx8^(5/3) xx x^(2/3)

Step 2:Exponents are added

=2x^(1/3+2/3) (2^3)^(5/3)

Step 3: 8 is written as 2 power 3

Step 4 : Simplify=2x2^5

=2xx32 x

=64 x

Ex 2: Solve x^(-1/3)8^(-2/3)

x^(-1/3)8^(-2/3)

Solu: Step 1  x^(-1/3) 8^(-2/3)

Step2: 8^(-2/3) = (2^3)^(-2/3) = 2^(3 xx-2/3) = 2^-2 = 1/2^2 = 1/4

My forthcoming post is on  Define Arithmetic Mean will give you more understanding about Algebra.

Step3 Simplify the variable with negative exponent

=1/(x^(1/3)) 1/4

Step 4 Simplify:=1/(4x^(1/3))

=1/(4root(3)(x))

Power Coefficient Equation

A coefficient is a multiplicative factor in a few phrase of an expression. It is typically a number, but in any case does not occupy some variables of the equation. For example in equation 7x2 − 3xy + 1.5 + y the first three stipulations correspondingly have coefficients 7, −3, and 1.5. In the third expression there are no variables, so the coefficient is the idiom itself called the power constant term or constant power coefficient of this expression

I like to share this Sample Correlation Coefficient with you all through my article.

Example Problem for the Coefficient:

Find the coefficient of the given equation:

X2 + 5y3 +2xy2 + 5yx2

Solution:

Given that X2 + 5Y3+ 2XY2 + 5YX2

Here X2+ 5Y3+ 2XY2+ 5YX2 the expression with the coefficient

The coefficient of x2 is 1

The coefficient of  5Y3 is 5

The coefficient of 2xy2 is 2

The coefficient of 5yx2 is 5.

Coefficient Of Varience

Coefficient of variation is a relative determine for standard deviation.

It is defined as 100 times the coefficient of spreading base in the lead standard difference is called coefficient power of variation.

Less C.V. designate the fewer variability or additional power.

More C.V. indicate the more variability or less power.

According to Karl Pearson who suggested this compute the equation, C.V. is the percentage dissimilarity in the mean, standard deviation being considered as the total variation in the mean.

With the help of C.V. we can find which salesman is more consistent in making sales, which batsman is more consistent in scoring runs, which student is more consistent in scoring marks, which worker is more consistent in production.

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Coefficients Of Dispersion

Whenever we want to compare the variability of two equations which differ widely in their averages or which are measured in different units.

We do not merely compute the measures of dispersions in the equations but we calculate the coefficients of dispersion which are pure numbers autonomous of the units of measurement.

Quotient Rule for Integration

The quotient rule is a method of finding the integration of a function that is the quotient of two other functions for which derivatives exist.

By the Product Rule,

if f (x) and g(x) are differentiable functions, then
d/dx[f (x)g(x)]= f (x)g'(x) + g(x) f' (x).
Integrating on both sides of this equation,

∫[f (x)g'(x) + g(x) f '(x)]dx = f (x)g(x),
which may be rearranged to obtain

∫f (x)g'(x) dx = f (x)g(x) −∫g(x) f' (x) dx.     (A)
Letting U = f (x) and V = g(x)

then differentiating it  we get

dU = f '(x) dx and dV =g'(x) dx,

pluging these values in (A), we get

∫U dV = UV −∫V dU.                  (1).


By the Quotient Rule,

if f (x) and g(x) are differentiable functions, then

d/dx[f (x)/g(x)]= g(x) f '(x) − f (x)g'(x)/[g(x)]2 .
Integrating both sides of this equation, we get
[f (x)/g(x)]=∫g(x) f '(x) − f (x)g'(x)/[g(x)]2 dx.
That is,

f (x)/g(x)=∫f '(x)/g(x)dx -∫f (x)g'(x)/[g(x)]2 dx,
which may be rearranged to obtain

∫f '(x)g(x)dx = f (x)g(x)+∫f (x)g'(x)/[g(x)]2 dx.      (B)

Letting u = g(x) and v = f (x) and then differentiating it , we get

du = g'(x) dx and

dv = f '(x) dx,
we obtain a Quotient Rule Integration by Parts formula:

plugging values of u , v , du and dv in B we get
∫dv/u= v/u+∫(v/u²) du.                                  (2)

quotient rule for integration-Application

∫[sin(x−½)/x²] dx.
Let
u = x½, du=1/2(x-½)

v=2cos(x-½),dv = sin(x−1/2)/x³/²

Then

∫sin(x−½)/x² dx = 2 cos(x-½)/x½+∫2 cos(x-½)/x• 1/2(x-½) dx

= 2 cos(x-½)/x½− ∫2cos(x-½) ·•[−(x-³/²)/2]dx

= 2 cos(x-½)/x½− 2 sin(x-½) + C,
which may be easily verified as correct.

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quotient rule for integration -Illustration


The Quotient Rule Integration by Parts formula (2) results from applying the
standard Integration by Parts formula (1) to the integral

Let ∫dv/u

with

U = 1/u,V = v,then differentiating ,we get

dU = − 1/u² du,dV=dv
plugging these values

∫dv/u=∫U dV

= UV −∫V dU

= 1/u·( v) −∫v (− 1/u²)du
= v/u+∫v/u² du

Terminating Decimals are Rational Numbers

Rational numbers are contrast among irrational numbers like Pi and square roots and sins and logarithms of information. On rational numbers and the ending of the article you be able to click on a linkage to maintain studying about irrational numbers. A number is rational but you can note down it in a form a/b where a and b are integers, b not zero. Visibly all fractions are of that form.

Terminating decimals are rational numbers:

A rational number is several number that be able to be expressed as the section a/b of two integers, among the denominator b not equivalent to zero. While b could be equal to 1, every one integer is a rational number. The set of every rational in sequence is frequently denoted by a boldface Q stands for quotient.

The terminating decimal extension of a rational numbers forever whichever terminating follows finitely several digits or begins to replicate the similar sequence of digits over and over. Moreover, any repeating or terminating decimal represent a rational number. These statements hold true not now meant for base 10, but as well for binary, hexadecimal, or any further integer base. Terminate decimal numbers be able to simply be written in that form: for example 0.67 is 67/100, 3.40938 = 340938/100000.

A real number so as to be not rational is called irrational. Irrational numbers contain square root 2, pi, and e. The decimal extension of an irrational number continues evermore without repeating. Because locate of rational numbers is countable, and the position of real numbers is uncountable, just about each real number is irrational.

In terminating conceptual algebra, the rational numbers shape a field. This is the representative field of characteristic zero, with is the field of fractions for the ring of integers. Finite extensions of Q are called algebraic number fields, and the algebra finality of Q is the field of algebraic number.

In mathematical study, the rational numbers shape a dense separation of the real numbers. The real numbers is able to be constructed from the rational numbers by achievement, using either Cauchv sequences or Dedekind cuts.

Zero separated by any other integer equals zero, consequently zero is a rational number even though division by zero itself is indeterminate.

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Solving Positive Direction Coordinates

A coordinate is a number that determine the position of a point the length of a number of line or arc. A list of two, three, otherwise additional coordinates can be use to establish the position of a point on a surface, volume, otherwise higher-dimensional area.

The Positive direction represents the positive value for both the x axis and y axis. The x axis value and the y axis value should be positive to move towards the positive direction. Any of the negative change in the axis will not lead to positive direction.

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Solving positive direction examples:

Solve the following equation:

Y = x2 +2

Solution :

Now we have to substitute values for the variable x in the given equation.

When we substitute any positive value for the given equation we get only positive output values.

That is when  x = 1

Y = (1)2 +2

Y = 1+2

Y = 3

The co ordinates will (1,3) which lies on the positive direction.

when we substitute x =2

We get

Y = (2)2 + 2

Y = 4 + 2

Y = 6

The co ordinates are (2,6) which lies on the positive direction.

Similarly when we substitute any positive value for the given equation the output will be positive.

Even though we give negative values, the output will be (that is y) positive. But we need to give positive values to move in the positive direction.

Solving positive direction example:

Solve the following equation:

Y = 3x +4

Solution:

Now we have to substitute values for the variable x in the given equation.

When we substitute any positive value for the given equation we get only positive output values.

That is when x = 4

Y = 3(4) +4

Y = 12+4

Y = 16

The co ordinates will (4,16) which lies on the positive direction.

when we substitute x =6

We get

Y = 3(6) + 4

Y = 18 + 4

Y = 22

The co ordinates are (6,22) which lies on the positive direction.

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Similarly when we substitute any positive value for the given equation the output will be positive.

Here when we substitute negative values for the variable x, we get negative values as output also. So we need to substitute positive values for the equation.

Poisson Distribution

If n is large, the evaluation of the binomial distribution can involve considerable computation. In such a case a simple approximation to the binomial probability could be considerable use. The approximation of binomial when n is large and p is close to zero is called the Poisson Distribution Mean.

Definition:

A random variable is said to follows Poisson distribution if it assumes only non-negative values and its probability mass function is given by

P(x,lambda ) =P(X=x) = elambda lambda x  /  X!    

x=0, 1, 2…

O            

otherwise

lambda is known as parameter of Poisson distribution .X~p() denotes it.

Characteristic function of Poisson distribution

phi x ( t )  =E[ei t x ]

=sum_(x=0)^oo ei t x   e-lambda lambda x / x!

=  e-lambda sum_(x=0)^oo ei t x lambda x  / x!

=  e-lambda [1+(lambda eit ) + (lambda eit)2 / 2! +(lambda ei t)3  /3!  + ............. ]

=e-lambda elambda eit

=  elambda (ei t -1)

Additive property of Poisson distribution:

Independent Poisson variate is also a Poisson variate xi (i=1,2,......n). xi follows Poisson with parameter lambda i.

xi ~ P(lambda i )    (i=1,2.....n)

then sum_(i=1)^n    xi ~ P(sum_(i=1)^n lambda i )

Proof:

Mxi(t) = elambda i(e^t -1)

Mx1+x2+x3+.....Xn(t) = Mx1 (t) Mx2 (t). . . Mxn (t)

= elambda 1(e^t-1). elambda 2(e^t-1) ..........elambda n(e^t-1)

=e(lambda 1 +lambda 2 +lambda 3+.......lambda n )(et-1)

=esum_(i=1)^n lambda i(et-1)

M.G.F of Poisson distribution:

Mx ( t ) = E [ etx ]

= sum_(x=0)^oo etx   e-lambda lambda x  / x!

= e-lambda sum_(x=0)^oo etxlambda x / x!

= e-lambda [ 1+ et lambda +  (lambda et )2 /2! + (  (lambda et )3  / 3! +...... ]

=e-lambda elambda

= elambda (et -1)

Applications of Poisson distribution

The following are some instance where the distribution is applicable

Number of deaths from a disease
Number of suicide reported in a particular city.
Number of defective materials in packing manufactured by good concern.
Number of printing mistakes at each page of the book.
Number of air accidents in some unit of the time.

Needs Assessment Definition

Assessment is an important and essential part of teaching. If teachers are to ask themselves whether what they are teaching and how they are teaching has the desired outcomes, they will needs to assess what children are able to do according to a set of criteria.

The criteria, which indicate what children, should be able to do and think in mathematics by the end of the foundation phase.

Teachers need to ask themselves with assessment:

The following are two very important questions that teachers have to ask when teaching.

Is what I am doing helping children to develop a desire to learn mathematics?

Is what I am doing teaching children to become numerate?

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Why we needs assessment?

Some reasons for assessing learners are so that we can measure whether a lesson has been effective for each learner (whether we have taught the lesson effectively enough). We assess learners to see what each one can do, for example, which learners are able to calculate change, or which learners are able to solve relevant word problem. We assess learners to see which of the children are ready for a new challenge and which must still practice what has already been taught. We assess learners so that we can plan further lessons that suit the needs of the children.

Types of assessment:

There are two main ways in which to assess children

Formative assessment

Summative assessment.

Formative assessment is assessing a learner while the learner is forming the new knowledge.

Example for formative assessment:

An example of formative assessment would be sitting with a learner while he or she is doing a task (say using a number line to count in groups), watching how the child goes about the task and asking the child to explain how and what he or she is doing. In this way, you find out what strategies the child is using and developing and what strategies you should be helping the child with; you are getting direct and instant feedback on hoe the child is coping and you are able to respond to the situation immediately through re-teaching and explaining again, asking anther learner to help, or planning another lesson on that needs for the next day.

Summative assessment is assessing a learner at the end of the lesson, section, topic, quarter or year as a summing up of what the learner knows. Therefore, tests and exams are summative versions of assessment.

When both formative and summative assessments are used, that is continuous assessment. In an outcomes-based education system, continuous assessment is used. The teacher studies the learning outcomes required of the learners and then plans lessons to teach to achieve these outcomes. During the lessons, the teacher observers what children are doing and saying and how children are doing a task. The teacher asks for explanations from the children as to what and how they are doing a take. The teacher helps those learners who are confused and continually monitors which learners are gaining control of the skills and concepts. Once a child can do something independently within the number range for that learner, a teacher can say that that child has learnt what was intended by the lesson and so can record that performance as a desired learning outcome for that child.

Learn Horizontal Line Test

Definition of function:

The mathematical concept of a function expresses the intuitive idea that one quantity (the argument of the function, also known as the input) completely determines another quantity (the value, or the output). A function assigns a unique value to each input of a specified type. The argument and the value may be real numbers, but they can also be elements from any given sets: the domain and the codomain of the function. An example of a function with the real numbers as both its domain and codomain is the function f(x) = 2x, which assigns to every real number the real number with twice its value. In this case, it is written that f(5) = 10.

A relation f:A`->B ` is said to be a function if every element in domain(A) has an unique image in codomain(B).

Definition of One-to-One Function:

A function is said to be One-to-One if no two different elements in domain have same image in codomain.The definition of one-to-one function can be written algebraically as follows:

Let a  and b any elements of domain.

A function f(x) is said to be one-to-one

1.if a is not equal to b then f(a) is not equal to f(b)

OR contra positive of the above

2.if f(a)=f(b) then a=b

Horizontal Line Test::

The horizontal line test is used to determine if a function is one-to-one.The lines used for this test are parallel to x axis.

If the function is one-to-one, then it can be visualized as one whose graph is never intersected by any horizontal line more than once.



If and only if  f is onto, any horizontal line will intersect the graph at least at one point (when the horizontal line is in the codomain).



If f is bijective, any line horizontal or vertical will intersect the graph at exactly one point.


Graph of one-to-one function:

If a line is drawn parallel to x axis (horizontal line) to this curve  then it will cut the curve at only one point so it is an one-to-one function.

Graph of a function which is not one-to-one:

If a line is drawn parallel to x axis (horizontal line) to this curve  then it will cut the curve at more than  one point so it is not an one-to-one function.

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Meaning of Correlation Coefficient

Two variables are related in such a way that:

(i) if there is an increase in one accompanied by an accompanied by a decrease in the other, Then the variables are said to be correlation The value of  the correlation will be in the interval [1, -1].

If the value of the correlation is positive then it is direct and if it is negative, then it is inverse.

If  the value of correlation is 1, it is said to be perfect positive correlation. If it is -1, it is said to be perfect negative correlation. If  the correlation is zero, then there is no correlation.

The formula to find the correlation is [sum dx dy] /[ sqrt [sum d_x^2 . sum d_y^ 2]]

where dx = x – barx ,  dy = y – bary

Now let us see few problems on correlation.

Example problems on meaning of correlation coefficient:

Ex 1: Find the correlation between two following set of data:

Cor_Tab1

Soln: barX = 180 / 9 = 20,

barY = 360 / 9 = 40.

Cor_SolTab1

Therefore the correlation =  [sum dx dy] /[ sqrt [sum d_x^2 . sum d_y^ 2]] = 193 / [sqrt [120 xx346]] = 0.94

This value shows that there is a very high relationship between x and y.

More example problem on meaning of correlation coefficient:

Ex 2: Find the correlation between the following set of data.

Cor_Tab2

Soln: barX = 36 / 6 = 6,  barY = 60 / 6 = 10

Cor_SolTab2

Therefore the correlation is   [sum dx dy] /[ sqrt [sum d_x^2 . sum d_y^ 2]]  = -67 / [sqrt [50 xx 106]]

= -0.92

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This value shows that there is a very low relationship between x and y.

By now the meaning of correlation coefficient will be more clearer. I believe that these examples would have helped you to do problems on correlation coefficient.

Average Velocity

Before going to the concept of average velocity,you should know about the concept of velocity.So, here is the introduction to velocity.

Velocity - the rate of change of position of an object. Velocity is a vector physical quantity because it requires both magnitude and direction to define. Speed is an absolute scalar value (magnitude) of velocity. Velocity is measured in meter/second (m/s).

In mathematics, velocity is the ratio of  displacement and  during the time taken to the interval. As it is associated to how fast waves travel between layers of the earth, it also tells us of how compact they are in between.

For example, 10 meter per second is a scalar value of velocity, but 10 m/s east is a vector. The rate of change in the velocity is called as acceleration.

Average velocity:

The Average velocity topic is dealt under mathematics. Students get to learn how to calculate the average velocity when the rest of the values are given. Students get to learn to practice various word problems on the concept of average velocity and also they get ample support from the online tutors regarding their homework problems and examination preparation.They can get all the answers for homework problems.

The “Rate of change(derivative) of position at which an object ” is known as velocity . It is a vector physical quantity; both magnitude and direction are essential to define it. The scalar complete value (magnitude) of velocity is speed, a quantity that is measured in meters per second (m/s or ms-1) when using the SI (metric) system.

The average velocity v of an object affecting through a displacement  (?x) during a time interval (?t) is described by the formula.Here is the Formula for average velocity:

V = Final displacement / total time taken

=  ?x / ?t

Students can get more detailed explanation on the topic on the Physics help page.

Finding average velocity

Example 1: If an object is thrown from the ground at an initial velocity of 10 meter per second, after t seconds the height of the object “h” in meter is given by,   h = t2 + 10t.  Find the time taken to the object to reach a height of 200 meter.

Solution:

Given, Height, h = 200 meter

h = t2 + 10t --> (1)

t =? (At 200 meter)
Plug h = 200 in equation (1)

200 = t2 + 10t

t2 + 10t – 200 = 0

By solving the quadratic equation,

t2 + 20t -10t -200 = 0

t(t+20) – 10(t+20) = 0

(t+20) (t -10) = 0

t = -20 or t = 10

We have two values for time “t”. Since time cannot be negative, any problem that gives a negative answer for one of its answers is always false, so we just go for the positive value.
So the final answer is 10 seconds.

So, the object reaches the height (200 meters) in 10 seconds.

Students can get more solved examples and problems to practice upon on the mastering physics page.

Calculate average velocity

Here is one more example of how to Calculate average velocity.

Example :  A particle moving along the x axis is located at 17.2 m at 1.79 s and at 4.48 m at 4.28 s. what is the average velocity in the particular time interval?

Solution:

Distance traveled by particle beside x-axis = Final location  - Initial location

= 17.20 - 4.48

= 12.72 meters

Time taken to cover up this distance = Final time - Initial time

= 4.28 - 1.79

= 2.49 seconds

Average velocity    = (Distance traveled)/(Time taken)

= 12.72 / 2.48

= 5.1084 (approximately)

Average velocity of the particle is    5.1084 m/s

Bijective Function Examples

In mathematics, a bijection, or a bijective function, is a function f from a set X to a set Y with the property that, for every y in Y, there is exactly one x in X such that f(x) = y and no unmapped element exists in either X or Y.

A one - one onto function is said to be bijective or a one-to-one correspondence.

A few examples of bijective function is given below which helps you for learning bijective function.

(Source: Wikipedia)

Examples of bijective function:

Example 1:

Show that the function f : R → R : f(x) = 3 - 4x  is one-one onto and hence bijective .

Solution:

We have

f(x1) = f(x2)

3 - 4x1 = 3 - 4x2

x1 = x2

Therefore, the function f is one-one.

Now, let y = 3 - 4x. Then, x = (3 - y)/4

Thus, for each y ε R (codomain of f), there exists x = (3 - y)/4 ε R

such that f(x) = f((3 - y)/4)

= {3 - 4 (3 - y)/4 }

= 3 - (3 - y)

= y

This shows that every element in codomain of f has its pre-image in dom(f).

Therefore, the function f is onto.

Hence, the given function is bijective.

Example 2:

Let A = R - {3} and B = R - {1}. Let f : A → B : f(x) = (x - 2)/(x - 3) for all values of x ε A.

Show that f is one-one and onto.

Solution:

f is one-one, since

f(x1) = f(x2)

(x1 - 2)/(x1 - 3) = (x2 - 2)/(x2 - 3)

(x1 - 2)(x2 - 3) = (x2 - 2)(x1 - 3)

x1x2 - 3x1 - 2x2 + 6 = x1x2 - 2x1 - 3x2 + 6

x1 = x2

Let y ε B such that y =  (x - 2)/(x - 3) .

Then, (x - 3)y = (x - 2)

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

Clearly, x is defined when y ≠ 1.

Also, x = 3 will give us 1 = 0, which is false.

Therefore,

x ≠ 3.

And, f(x) = ((3y - 2)/(y - 1) - 2)/((3y - 2)/(y - 1)- 3) = y

Thus, for each y ε B, there exists x ε A such that f(x) = y.

Therefore, f is onto.

Hence, the given function is one-one onto.

These examples of bijective function help you to solve the following practice problems.

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Practice problems of bijective function:

Following examples of bijective function is given for your practice which helps you to learn more about bijective function.

1) Show that the function f : R → R : f(x) = x3  is one-one and onto.

2) Let R0 be the set of all non zero real numbers. Show that f : R0 → R0 : f(x) = 1/x is a one-one onto function.

Probability Distribution

In probability theory and the statistics, a probability distribution identifies either the probability of each value of a random variable (when the variable is discrete), or the probability of the value falling within a particular interval (when the variable is continuous). The probability distribution describes the range of the possible values that a random variable can get and the probability that the value of the random variable is within any subset of that range.

I like to share this Non Central T Distribution with you all through my article.

The Normal distribution is often called as the "bell curve", when the random variable takes the values in the set of real numbers. Let us see some sample problems on probability distribution statistics.

Examples

Given below are some of the examples on Probability Distribution Statistics.

Example 1:

A continuous random variable X has probability distribution function f(x) = 3x2, 0 ≤ x ≤ 1, Find a and b such that. (i) P(X ≤ a) = P(X > a) and (ii) P(X > b) = 0.05. Calculate probability distribution for this function.

Solution:

(i) Since the total probability is 1, [Given that P(X ≤ a) = P (X > a)

P(X ≤ a) + P(X > a) = 1

i.e., P(X ≤ a) + P(X ≤ a) = 1

⇒ P(X ≤ a) =1/2⇒ `int`3x2dx = 12

i.e.,[3x3/3]a0=1/2 ⇒ a3 =1/2

i.e., a = 1/213

(ii) P(X > b) = 0.05

∴ `int` f(x) dx = 0.05 ∴ `int` b1 3x2  dx = 0.05

[3x3]31b= 0.05 ⇒ 1 − b3 = 0.05

b3 = 1 − 0.05 = 0.95 =95

100 ⇒ b = 19/2013

Example 2:  A random variables X has probability mass function as in the probability distribution tables given below
X
0 1 2 3 4 5 6
P(X = x) k 3k 5k 7k 9k 11k 13k



(1) Find k.

(2) Evaluate P(X < 4), P(X ≥ 5) and P(3< X ≤ 6)

Solution:

(1) Since P(X = x) is a probability mass function `sum_(n=0)^6` P(X = x) = 1

ie.,P(X=0) + P(X = 1) +P(X = 2) +P(X = 3) +P(X = 4) +P(X = 5)+P(X = 6) = 1.

⇒ k + 3k + 5k + 7k + 9k + 11k + 13k = 1 ⇒ 49 k = 1 ⇒ k =1/49

(2) P(X < 4) = P(X = 0) + P(X = 1 ) + P(X = 2) + P(X = 3) =1/49 +3/49 +5/49 +7/49 =16/49

P(X ≥ 5) = P(X = 5) + P(X = 6) =11/49 +13/49 =24/49

P(3 < X ≤ 6) = P(X = 4) + P(X = 5) + P(X = 6) =9/49 +11/49 +13/49 =33/49

∴ The smallest value of x for which P(X ≤ x) > 1/2 is 4.

Example 3:

In a class, the average number of marks obtained by student in Physics is 0.52, chemistry is 0.48 and in both Physics and chemistry they obtained 0.37. Find the total average obtained in either Physics or chemistry.

Solution:

P(A) = Average number of marks in Physics = 0.52

P(B) = Average number of marks in Chemistry = 0.48

P(A and B) = Average number of marks in both physics and chemistry = 0.37

P(A or B)    = P(A) + P(B) – P(A-B)

= 0.52 + 0.48 – 0.37

= 0.63

Calculating Interquartile Range

Calculating interquartile range is referred as a measure of variability, spread or dispersion, H-spread. It is the difference sandwiched between the 75th percentile (often called Q3 or third quartile) and the 25th percentile (often-called Q1 or first quartile). The primary step is to find the interquartile range is to arrange the given set of numbers in ascending order. The standard formula to find the interquartile range is given as

Interquartile ranges = Quart3 – Quart1.    ----- Standard formula

Whereas Q1 = first quartile.

Q3 = third quartile.

Example Problems for calculating interquartile range:

Ex : Determine the interquartile range of following set of numbers

34, 15, 8, 26, 22, 9, 19

Sol :  The following are the steps to find the interquartile range of a set of numbers.

Step 1:  Arranging of numbers

The initial step is to modify the given data in order, from smallest to biggest.

9, 8, 15, 19, 22, 26, 34

Step 2:    Calculating 1st quartile Q1.

The next step is to find the lower median (1st quartile Q!).

This is the middle of the lower three numbers.

1st quartile Q1 is 8.

Step 3:    Calculating 3rd quartile Q3.

Now find the upper median (The 3rd quartile Q2).

This is the middle of the upper three numbers.

The 3rd quartile Q3 is 26.

Step 4:     Calculating interquartile range.

The formula used to find the interquartile range is

Interquartile range = Q3 – Q1.

Q1 = first quartile.

Q3 = third quartile.

Here

Q1 the first quartile = 8

Q3 the third quartile = 26

Interquartile range = Q3 – Q1. ----- Standard formula

Plug in the Q1 and Q3 values in the standard formula Q3 – Q1.

Interquartile range = 26 – 8.

Interquartile ranges = 18.

Practice Problems for calculating interquartile:

Pro 1:  Find the interquartile range of following set of numbers

42, 81,56,21,63,12,5

Ans : Interquartile range = 51.

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Pro 2:  Find the interquartile range of following set of numbers

56, 23, 14, 25, 86, 45, 63, 15, 49, 18, 16

Ans:   Interquartile range = 40.

Pro 3:   Find the interquartile range of following set of numbers

12, 8, 5, 22, 15, 45, 2

Ans :   Interquartile range = 17

Permutations in Math

In mathematics, the notion of permutation is used with several slightly different meanings, all related to the act of permuting (rearranging in an ordered fashion) objects or values. Informally, a permutation of a set of values is an arrangement of those values into a particular order. Thus there are six permutations of the set {1,2,3}, namely [1,2,3], [1,3,2], [2,1,3], [2,3,1], [3,1,2], and [3,2,1]. (Source – Wikipedia.)

Example problems for permutations in math:

Example 1

1) How many permutations of 5 apparatus are probable from 2 apparatus. what is a permutation in given math problem?

Solution:

By means of the formula we can evaluate the answer,

P (n, r) = n! / (n – r)!

Here, n = 5, r = 2

P (5, 2) = 5! / (5–2)!

P (5, 2) = 5! / 3!

=   `(5xx4xx3xx2xx1)/ (3xx2xx1)`

Now 3, 2, 1 gets crossed out

= 5 x 4

= 20

Answer is: 20

Example 2

How many permutations of 9 files are possible from 5 books. what is a permutation in given math problem?

Solution:

Formula for calculate permutations,

P (n, r) =n! / (n – r)!

Now, n = 9

r = 5

P (9, 5) = 9! / (9 – 5)!

P (9, 5) = 9! / 4!

= `(9 *8 * 7 * 6 *5 * 4 *3 *2 *1)/ (5 * 4 *3 *2 *1)`

Here 5, 4, 3, 2, 1 gets crossed out

= 9x 8 x 7x 6

= 3,024

The answer is: 3, 024

Example 3

How many ways can 6 graduates from group of 15 are lined up for a function. what is a permutation in given math problem?

Solution:

There are 15P6 possible permutations from a group of 15.

15P6 = (15!)/(15-6!)

= (15!)/ (9!)

= 15x 14x 13x 12x 11x 10

= 3, 603, 600 different lineups.

The answer is: 3, 603, 600 different lineups.

Example 4

How many ways can 3 kids from group of 13 are lined up for a photograph. what is a permutation in given math problem?

Solution:

There are 13P3 possible permutations of 3 students from a group of 13.

13P3 = (13!)/ (13- 3!)

= (13!)/ (10!)

= 13 x 12 x 11

= 1716 different lineups.

The answer is: 1716 different lineups.

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Practice problems for what is a permutations in math:

Practice problem -1

1) How many permutations of 9 components are possible from 2 elements. what is a permutation in given math problem?

Ans: 72

Practice problem -2

2) How many ways can 3 girls from group of 14 are lined up for a photograph?

Ans: 2184

Practice problem -3

3) How many permutations of 17 books are possible from 4 books?

Ans: 57, 120

Practice problem -4

4) How many 4-digit numbers can be ordered from the digits 1, 2, 3, 4, and 5, if each digit is unique. what is a permutation in given math problem?

Ans: 120.

Non linear Differential Equations

An ordinary differential equation is an differential equation in For instance (i)dy/dx = x + 5 (ii) (y′)2 + (y′)3 + 3y = x2 which  is a single independent variable enters either explicitly or implicitly are all ordinary non linear differential equations And now we see about the non linear differential equations.

Non linear differential equations in First order

Order and degree of a non linear differential equations:

Definition of non linear differential equations :

The order of an differential equation is the order of the highest order derivative occurring in it. The degree of the differential equation is the highest order derivative which occurs in it, after the differential equation has been made free from radicals and fractions as far as the derivatives are concerned.The degree of a non linear differential equation does not require variables like r, s, t … to be free from radicals and fractions.

Find the order and degree of the following differential equations:

ii) y = 4dy/dx + 3xdx/dy

y = 4dy/dx + 3xdx/dy  y = 4(dy/dx) + 3x 1/(dy/dx )

Making the above equation free from fractions involving

dy/dx we get y .dy/dx = 4(dy/dx)2 + 3x 2Highest order = 1

Degree of Highest order = 2

(order, degree) = (1, 2)

Problem for non linear differential equations:

Formation of non linear differential equations :

Form the differential equation from the following equations.

y = e2x (A + Bx)

ye−2x = A+ Bx … (1)

Since the above equation contains two arbitrary constants, differentiating

twice, we get y′e−2x − 2y e−2x = B

{y′′e−2x − 2y′ e−2x} − 2{y′e−2x − 2y e−2x} = 0

e−2x {y′′ − 4y′ + 4y} = 0 [‡ e−2x ≠ 0]

y′′ − 4y′ + 4y = 0 is the differential equation.

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Non linear Differential equations of first order and also first degree :

Solve: 3ex tan y dx + (1 + ex) sec2y dy = 0

`=>` 3 log (1 + ex) + log tan y = log c

`=>` log [tan y (1 + ex)3] = log c

`=>`   (1 + ex)3

required solution to the problem is tan y = c

Learning Sample Space

Percentage is a number expressed  as a fraction of  100. The word percent is short form of the Latin word "Percentum" meaning out of hundred.  It means ratio of some number to hundred. We use the sign" %" to denote percentage.  It is basically a fraction, with 100 as the denominator and the number as the numerator.

For example 25% :

25% = (25)/(100)

In everyday life, we relate percentage to profit, loss, discount,rate of interest in banks, sales tax, income tax.

(i) Convert a ratio into percent , we write it as a fraction and multiply it with 100.

Example:  To express 20 : 50 as percent , we write (20)/(50) x 100 = 40%

(ii) To convert decimal into percent  we also multiply with 100

Example:   0.175 to percent   0.175 x 100 = 17.5%

(iii) To convert percent to fraction or decimal.  We drop the sign % and divide the remaining number by 100

Example:   45% = (45)/(100)

(iv) To find the certain percent of a given quantity, we multiply the given percent with the given quantity

Example:  30 % of 120   = (30)/(100) x 120 = 36

Some examples on solve finding percentages

1) what is 25 % of 200 ?

Solution:   (25)/(100)  x 200   = 0.25 x 200 = 50

2) What is 45% of 20 ?

Solution:   (45)/(100) x 20 = 0.45 x 20 = 9

3) We have to pay sales tax of 2% on product for $6.25. What is the amount we pay in all for that product.

Solution:  2% = (2)/(100) = 0.02

0.02 x 6.25 = 0.125

Total amount paid is $ ( 6.75 +0.125) = $  6.875

4) A shoe shop marks 40% discount on each pair of shoes , what is the cost price of a pair of shoes that cost $ 67.

Solution:     40 % = (40)/(100) = 0.4

0.4 x 67 = 26.8,

Selling price = cost price - discount = 67 - 26.8 = $ 40.2

5) 43 % of a number is 53.75, find the number .

Solution:   Let the number be x

43% = (43)/(100) = 0.43

0.43 x  = 53.75   divide by 0.43

x = (53.75)/(0.43) = 125

Word problems on solve finding percentages

1) Mary had 24 pages to write .  By the evening, she completed 25% of her work .  How many pages did she have left ?

Solution:    Mary completed 25% means  (25)/(100) x 24  = 6 pages
Number of pages left is 24 - 6 = 18 pages

2) John's income is 20% more than that of Tom .  How much percent is the income of Tom less than that of Jim.

Solution:   Let Tom's income is  $ 100

John's income is 20% more means 20 + 100 = $ 120

Tom's income is 100

Jim's income is 120

Tom's income is $20

% is  (20)/(120) x 100 = (100)/(6) % = (50)/(3) % = 16 (2)/(3) %

3)A 90 kg solution has 10% salt .  How much water must be evaporated to leave the solution  with 20% salt.

Solution:    Let amount of water evaporated is x
90 kg solution has 10  % salt means   (10)/(100) x 90  = 9 kg
After evaporation total quantity left is ( 90 -x )
In this solution salt is 20% means   (20)/(100)  (90-x) 20/(100) = (1)/(5) (90-x)
Quantity of salt remains same in both of these , as only water evaporated
(1)/(5) (90 - x) = 9
90 - x = 9 . 5,   90 -x = 45 ,  90 -45 = x ,  x = 45
45 Kg water evaporated.

Logistic Growth

Logistic growth is always explained with the help of the logistic curve. Logistic curve can be shown in form of sigmoid curve. Logistic growth models the S-shaped curve of growth of  set P, where P might be referred as  population. At the initial stage of the growth the graph is in exponential then saturation begins and the growth slows and finally at maturity the growth stops.

I like to share this Logistic Regression Model with you all through my article.

Learn the definition of Logistic growth:

learn the Logistic Growth formula:

P (t) = 1/ (1+e –t)

where P is the population and t is considered as a time.

The S-curve is obtained if the range of the time over the real numbers from −∞ to +∞. In practice, due to the nature of the exponential function e−t, it is then enough to calculate time t over a small range of real numbers like [−7, +7].

Learn the derivative which is most commonly used:

d/dt P(t) = P(t).(1-P(t))

The computation of function P is

1-P(t) = P(-t)


Learn the logistic differential equation:

learn the logistic growth function, it can be used to calculate the first order nonlinear differential equation.

d/dt P(t) =P(t)(1- P(t))

Here P is a variable with respect to time t and by applying the condition P(0) we can get ½ .

One may readily find the solution to be

P(t) = et / (e t + e c )

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Decide the constant of integration e c = 1 gives the other well-known form of the definition of the logistic curve

P (t) = e t / (e t + 1) = 1 / (1 + e –t)

The logistic curve demonstrates the exponential increase for negative t, which can slow down the curve to linear growth. It is in the slope of 1/4 at t = 0, then it approach exponentially decaying gap at y=1

The relationship among the logistic sigmoid function and the hyperbolic tangent is given by,

2P(t) = 1 + tanh (t/2)

Laws of Exponents

The laws of exponents are used for combining exponents of numbers. Exponents is a number raised to another number, it is denoted as,   a n,  here, n is known as the exponent of the nth power of a.

Laws of exponent are as follows:

x1 = x

x0 = 1

Negative exponent

x - n = (1)/(x^n)

Multiplication law of exponent

x a x b =  (x) a+b

Division law of exponent

(x^a)/(x^b)   =  (x) a-b

Power of power law of exponent

(x a) b  =  x ab

(xy)a   =  xa y a(x/y)^a = (x^a)/(y^a)

Fractional law of exponent x^(a/b)     =    (x^a)^(1/b)  =   root(b)(x^a)


Examples on laws of exponents

1)  Solve  the exponent (32) 5 =  (3) 2x5 = (3) 10      ( using Multiplication law)

2) Solve the exponent  (5^4)/(5)  =  (5) 4-1 =  5 3 = 125    (Using division law)

3) Simplify the exponent 2 (- 5)  = (1)/(2^5)  = (1)/(32)

4) Simplify the exponent  (1/4)^(-3) = (1)/[(1/4)^3] = (1)/(1^3/4^3)  = (4^3)/(1^3) = 4 3 = 64   (using Division law)

5)Simplify using the law of exponents  (sqrt(4) ) -3 = (4^(1/2))^(-3)        (using fractional law)

=  (4)^[(1/2)*(-3)]   (using power of power  law)

= 4^(-3/2)   (using multiplication law)

=  (1)/(4^(3/2))           (using negative law)

= (1)/((4^3)^(1/2)) = (1)/((64)^(1/2))

= (1)/((8^2)^(1/2)) = (1)/(8^(2*(1/2))) = (1)/(8)

Solved examples

Below are the solved examples on laws of exponents:

1)Solve the exponents  3 7 * 3 2 = 3 (7+2) = 3 9       (using Multiplication law)

2) Solve the exponents 2 (-3) * (-7) (-3) =  (2 * (-7)) (-3) =  (-14) (-3)    (using Power of power law)

3) root(3)((343)^-2)  = (343^(-2))^(1/3)      (using Fractional law)

=  (343^(1/3))^(-2) =  (1)/((343^(1/3))^2)   ( using negativel law)

=  (1)/(7^3^(1/3))^2   = (1)/(7^(3*1/3))^2       (using power of power law)

=  (1)/(7^2)  =   (1)/(49)

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4) Simplify using the law of exponents [{(1/5)^(-2)}^2]^(-1) =  {(1/5)^(-2)}^(2*(-1))

=  {(1/5)^(-2)}^(-2)

= (1/5)^((-2)*(-2))

= (1/5)^4

= (1^4)/(5^4) = (1)/(625)