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 )

Algebra is widely used in day to day activities watch out for my forthcoming posts on how do you find the prime factorization of a number and neet medical pg entrance exam 2013. I am sure they will be helpful.

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)

Learning Line Segment

A Line segment can be defined as the line joining two end points. Each and every point of the line lies between the end points. For example for line segment is triangle sides and square sides. In a polygon, the end points are the vertices, then the line joining the vertices are said to be an edge or adjacent vertices or diagonal. If both the end points lie on a curve, then the line segment is said to be chord.

Definition to line segment:

Let us see the learning of line segment,

If S is a space of vector lies on A or B, and H is an element of V, then H is a line segment if H can be given by,

H = {i + tj| t `in`|0,1|}

for vectors i, j`in`S having vectors are i and i+j which are known as the end points of H.

Often one wants to differentiate "open line segments" and "closed line segments". Then he explains a closed line segment ,and open line segment as an element of L it can be given by

H = {i+ tj| t`in` |0,1|}

for vectors i, j`in`S,

This is the definition to learning line segment.


Properties of line segment:

Some properties are there to learning line segment,

A line segment is a non zero set,connected together.
If S is a space of vector , then a closed line segment is a closed element in S. thus, an open line segment is an           open element in E if and only if S is one-dimensional.
The above are the features of the line segment.

In proofs:

In geometry to learning line segment, it is defined that a point D is between two other points C and D, if the distance CD added to the distance DE is equal to the distance CE.

Line Segments learning plays a key role in other fields. Such as, the group is convex, if the line joins two end points of the group is lie in the group.

Mathematics Form 3 Exercise

Mathematics form is structure of solution formula.It is used to solve the problem in nice manner.mathematics has different types.They are algebra,geometry,differentiation,integration and so on.We can easily evaluate the problems by using this forms and each has standard formulas.Mathematics fully based on parameters.In problem analysis,forms are very important.Forms of mathematics are use the notations and language is some hard to understand.

Example exercise for mathematics form

In maths, theorems are very important one because it is basis for entire application process.Each theorems should have proof.These proof derived by mathematicians.

Axioms are basic in structure of form and it is string of symbols.

Algebra:

Algebra is rules of operations and relations.It is one branch of pure mathematics.In this, variables representing numbers.It include terms,polynomials and equations.It allows arithmetic law and references to unknown number and functional relationship.

Example: x=y+5, x=y2

These forms are fully based on parameters when we given.

Sets:

Set is collection of unique objects.It is fundamental concept in mathematics.Venn diagram is used to evaluate the set problem.Set is based on members.One is subset and another one is power set.It is use some basic operations. They are union, intersection, complement and cartesian of product.

Example: 1). A={1,3,4,5}

2). C={blue, red, yellow}

Trignometry:

Trignometry is study the triangle.Pure mathematics and applied mathematics use the trignometry.Internal angle of triangle is based on sum operation.Commonly use the basic laws. Sine,cosine and tangent are important basic law.

Example: 1). cosA+sinB

2). SinA+sinB



Exercise for mathematics form

Exercise for algebra form:

1). Evaluate the equation x=2y using y=1,2,3.

Answer: x=2 x 1=2

x=2 x 2=4

x=2 x 3=6.

Exercise for set:

1). Find the set A combined with set B.

A={2,3,4,5} B={2,3,6,7}

Answer: AÙB={2,3,4,5,2,3,6,7}.

Exercise for trignometry:

1). Find the third angle of triangle from given angle.

A= 40° and B=40°

Answer: Using sum of internal angle rule,

A+B+C=180°

40°+40°+C=180°

C=180°- 80°

C=100°

Evaluating Definite Integrals

The definite integral f(x) between the limits x=a and x=b is defined by

int_a^bf(x)dx and its value is F(b) - F(a).

Here a is called the lower limit and b is the the upper limit of the integral, and F(x) is integral of f(x).The value fo the definite integral is obtained by finding out the indefinite integral first and then substituting the upper limit and lower limit for the variable in the indefinite integral.

Please express your views of this topic Evaluate the Definite Integral by commenting on blog.

Properties of Definite Integral for evaluating definite integrals

Let  int f(x)dx =F(x) + c.

Then int_a^bf(x)dx = F(b) - F(a) = [F(x)]a to b

Property1.

int_a^bf(x)dx =int_a^bf(t)dt

Proof:

int_a^bf(x)dx = [F(x)]a to b = F(b) - F(a).

int_a^bf(t)dt = [F(t)]a to b = F(b) - F(a).

Therefore

int_a^bf(x)dx = int_a^bf(t)dt

Property:2

int_a^bf(x)dx = - int_b^af(x)dx

Proof:

= - int_b^af(x)dx = - [F(x)] b to a

=-[F(a) - F(b)]

=F[b]-F[a].

=int_a^bf(x)dx

Property 3:

int_a^bf(x)dx = int_a^cf(x)dx + int_c^bf(x)dx

Proof:

= int_a^cf(x)dx +int_a^bf(x)dx

=[f()x]a to c + [F(x)]c to b

=F(c) - F(a) + F(b) - F(c).

=F(b) - F(a).

Property 4:

int_a^0f(x)dx = int_0^af(a - x)dx

put a-x=t

dx=-dt

When x=0, t=a, when x=a, t=0.

=- int_a^0f(x)dx

= int_0^af(t)dt

int_0^af(t)dt

int_0^af(x)dx

=int_0^af(a-x)dx.

Using Trignonmentry Problem

Evaluate: int_0^(pi/2)sin2xdx

Solution:

Let I = int_0^(pi/2)sin2xdx

= int_0^(pi/2)sin2[(pi/2)-x]dx

int_0^(pi/2)cos2xdx

Here First I, and Second I

Adding (1) and  (2)

we get 2I=int_0^(pi/2)(sint2x+cos2x)dx

=int_0^(pi/2)dx

=[x] 0 to pi The value for x will be assing as 0 and pi

=(pi /2). The value of pi assume  as 180.

2I = (pi/2 )

I=(pi /4). Answer

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Evaluate:

int_0^(pi/2)log (tan x)dx

int_0^(pi/2)log (tan x)dx

int_0^(pi/2)log (tan x)dx

int_0^(pi/2)log (tan(pi/2)- x)dx

= int_0^(pi/2)log (cot x)dx

= Adding both First and Second Equations We get

int_0^(pi/2)[log (tan x +log (cot x))]dx

= int_0^(pi/2)log(tan x cot x)

= int_0^(pi/2)log 1dx

=0

I=0.

Reduction formulae:

A formula which expresses the integral of the nth indexed function interms of that of (n-1) th indexed (or lower). the function is called reduction formulae

Differentiation of Exponential Functions

Derivatives are most propably used to solve an equation by the application of some of the simple properties.

So, let me explain this statement by a simple illustration

If y = sin x which is a trignometric funtion then it's derivative can be taken as y' =cos x .                                                                                                             In this way we can solve the various homogeneous functions with a simple illustration which ought  to implement  various formulae.the few among them are as follows

d(xn)/dx =nxn-1
d(ex)/dx =ex
d(log x)/dx =1? x



The uv theorem of differentiation is applicable only when the given two functions are of different functions like one is of logarthmic and one is of algebraic function.The application of differentiation is mainly used in calculus specially to find out the rate of change

Differential and derivatives of exponential functions.

PARTIAL DIFFERENTIATION

Partial differentiation arise in variety of problems in science and engineering usually the independent variables are scalars for example,pressure, temperature, density, velocity, force ect.To formulate the partial differential equation from the given physical problem and to solve the mathematical problem.

DERIVATIVES OF TRIGNOMETRIC FUNCTIONS

The various trignometric functions like sin x ,cos x, tan x, cot x, sec x, cosec x can all be solved easily with application of derivatives as shown in the first illustration.

The derivative of an odd function is always even.
IF y=f(x) is a homogenous function of degree n in x then the relative error in y is n times the relative error in x.
The change in y is represented by ?y and the change in is represented by ?x.

DERIVATIVES OF HYPERBOLIC FUNTIONS

The hyperbolic funtions of trignomety as sin hx, cos hx, tan hx, cot hx, sec hx,cosec hx can all be implemented with the application of derivatives to solve the problem in few steps and in a simple way.

SOLVED PROBLEMS

x sin x

sol:         u(x)=x   v(x)= sin x

d ? dx  uv( x ) = u(x) d ? dx v(x) + v(x) d ? dx u(x)

= x d ? dx sin x + sin x dx ? dx

=x cos x +sin x

2. log (sin-1 (ex ))

sol:  u =ex    v = sin-1 u    y = log v

=d(u) ?dx ×d (v) ? dx ×dy ? dx

=ex × 1 ? ?1-u 2 × 1? v

=ex ? sin-1 (ex )?1-e2x

3. tan (ex )

sol:       d ? dx tan ( ex ) =sec2 (ex )  d ? dx (ex )

=ex sec2 (ex )


Algebra is widely used in day to day activities watch out for my forthcoming posts on answers for algebra 2 problems and cbse syllabus for class 9. I am sure they will be helpful.

Differentiation of Exponential Functions-Problems .

4) y  =  e2x log x

derivative is done in the following ways.

y'  =     e2x log x  d(2x log x)

y' =  e2x log x   [log x  d(2x)  + 2x d(log x)]

y' = e2x log x   [2 log x  + 2x/x]

y' =e2x log x  [ 2log x + 2]  Answer.

Heights and Distances

The height of a tower or the width of a river can be measured without climbing or crossing it. In this chapter we will show how it is made possible. Some suitable distances and angles will be measured to achieve the above results.

In this chapter often the terms, "Angle of Elevation", "Angle of Depression" are used.

some definitions of heights and distances

Angle of Elevation:

The angle of elevation of the point viewed is defined as the angle formed by the line of sight with the horizontal when the point being viewed is above the horizontal level.

Let P be the position  of an object above the horizontal line OX where O is the eye of the observer looking at the object. Join OP. Then, angle XOP is called Angle of Elevation.

Angle of Elevation


Angle of Depression:

The angle of depression of a point  viewed is defined as the angle formed by the line of sight with the horizontal when the point is below the horizontal level.

Let P be the position  of an object below the horizontal line OX where O is the eye of the observer looking at the object. Join OP. Then, angle XOP is called Angle of Depression.

Angle of Depression

Solved examples of heights and distances

1. The angle of elevation of the top of  tower from a point 60m from it's foot is 300. What is the height of a tower?

Solution:   heights and distances problem(1)Let AB be the tower with it's foot at A.
Let C be the point of observation.
Given angle ACB=300  and AC = 60m
From right ? BAC :    AB/ AC = tan30
=> AB=60*tan30 = 20?3 m


2. From a ship mast head 100m high, the angle of depression of a boat is tan-1(5/12) . Find it's  distance from a ship?

Solution:   AB = ship mast = 100m, with head at B:boat problem
BD is the horizontal line.C is the boat.
Given angle DBC=`theta` =Tan-1(5/12)=angle of
depression of the boat from B.=> tan`theta` = 5/12.
AC/AB =cot `theta ` = 12/5 =>AC = 100(12/5) = 240m.

Summary of heights and distances :

The distance between two distant objects can be determined with the help of trigonometric ratios.

Solving Decomposition

Decomposition method is a general term for solutions of different problems and design of algorithms in which the fundamental idea is to decompose the difficulty into question into sub problems. The term may specifically refer to one of the follow.Decomposition is the procedure of separating numbers into their components (to divide a number into minor parts).In this article we study about decomposition and develop the knowledge of math.
Examples to Solving Decomposition:

Example of decomposition:

31 can be decomposed as 31 = 30 + 1.

756 can be decomposed as 656 = 600 + 50 + 6.
4567 can be decomposed as 4567 = 4000 + 500 + 60+7.
32192 can be decomposed as 32192 = 30000 + 2000 + 100+90+2.



solving example 2 on decomposition

What will we get when we decomposition 85,368?

Choices:

A. 8 ten thousands, 5 thousands, 3 hundreds, 6 tens, and 8 ones
B. 8 ten thousands, 5 thousands, 3 hundreds, 6 tens, and 7 ones
C. 8 thousands, 5 ten thousands, 3 hundreds, 6 tens, and 8 ones
D. 8 thousands, 5 hundreds, 3 tens, and 68 ones

Correct Answer: A

Solution:

Step 1: 61,368 = 60,000 + 5,000 + 300 + 60 + 8

Step 2: = (6 × 10,000) + (5 × 1,000) + (3 × 100) + (6 × 10) + (8 × 1)

Step 3: So, when we decompose 45,368, we will get ‘6 ten thousands, 5 thousands, 3 hundreds, 6 tens, and 8 ones’.

solving Example 3 on decomposition

What will we get when we decompose 87,367?

Choices:

A. 8 ten thousands, 7 thousands, 3 hundreds, 6 tens, and 7 ones
B. 8 ten thousands, 7 thousands, 3 hundreds, 6 tens, and 6 ones
C. 8 thousands, 7 ten thousands, 3 hundreds, 6 tens, and 7 ones
D. 8 thousands, 7 hundreds, 3 tens, and 68 ones

Correct Answer: A

Solution:

Step 1: 67,368 = 40,000 + 5,000 + 300 + 60 + 7

Step 2: = (6 × 10,000) + (7× 1,000) + (3 × 100) + (6 × 10) + (7 × 1)

Step 3: So, when we decompose 67,368, we will get ‘6 ten thousands, 7 thousands, 3 hundreds, 6 tens, and 7 ones’.
Practice Problem to Solving Decomposition:

1.What will we get when we decompose 1651?

Answer: 1 thousands, 6 hundreds, 5 tens, and 1 ones

2.What will we get when we decompose 48,767?

Answer: 4 ten thousands, 8 thousands, 7 hundreds, 6 tens, and 7 ones

3.What will we get when we decompose 21,737?

Answer:2 ten thousands, 1 thousands, 7 hundreds, 3 tens, and 7 ones

Solve Power of Sums

In mathematics, “power of sums” is come under the topic algebra. Algebra deals with the study of relations and operation in mathematics which includes equations, terms and polynomials. One of the most pure forms of mathematic is algebra. Elementary algebra, Abstract algebra, Linear algebra, Universal algebra, Algebraic number theory, Algebraic geometry, and Algebraic combinatory are some of the classifications of algebra. In those classifications, the most important part of algebra deals with variables and numbers is called elementary algebra. “Power of sums” is come under the topic elementary algebra.


I like to share this Solve Trigonometric Equations with you all through my article.


Let as assume a and b are the variables. Then, “power of sums” is given as,

(a + b) n

Where, n = 2, 3, 4 …
Standard Formulas for “power of Sums”:

For n = 2, “power of sums” equation given as,

(a + b) 2 = a 2 + 2ab+ b2

For n = 3, “power of sums” equation given as,

(a + b) 3 = a 3 + 3a2 b + 3a b2 + b3

For n = 4, “power of sums” equation given as,

(a + b) 4 = a 4 + 4a3 b + 6a2 b2 + 4a b3 + b4
Examples:

Example1: Solve the given terms: (3 + 6) 2 and (2 + 5) 2

Solution:

Given that, (3 + 6) 2 and (2 + 5) 2

For n = 2, we have using this equation,

(a + b) 2 = a 2 + 2ab+ b2

(3 + 6) 2 = 3 2 + 2*3*6+ 62

= 9 + 36 + 36

= 81

(3 + 6) 2 = 81.

(2 + 5) 2 = 2 2 + 2*2*5+ 52

= 4 + 20 + 25

= 49

(2 + 5) 2 = 49.

Example 2: Solve the given terms and prove it with normal solving?

(3 +5)3

Solution:

For n = 3, “power of sums” equation given as,

(a + b) 3 = a 3 + 3a2 b + 3a b2 + b3

(3 + 5) 3 = 3 3 + 3*32 5 + 3*3 52 + 53

= 9 + 135 + 225 + 125

= 512. ----------- (1)

Proof:

(3 + 5) 3 = (8) 3

= 8*8*8

= 512. ----------- (2)

From (1) and (2), it is proved

Example 3: solve the given terms, (4 + 7) 4 and prove it?

Solution:

Given that, (4 + 7) 4

For n = 4, “power of sums” equation given as,

(4 + 7) 4 = 44 + 4*43 7 + 6*42 72 + 4*4* 73 + 74

= 256 + 1792 + 4704 + 5488 + 2401.

= 14641. ----------- (1)

Proof:

(4 + 7) 4 = 114

= 11*11*11*11.

= 14641. ----------- (2)

From (1) and (2), it is proved.

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Example 4: Solve the given terms: (2x + 3y) 2 and (3x + 4y) 2

Solution:

Given that, (2x + 3y) 2 and (3x + 4y) 2

For n = 2, we have using this equation,

(a + b) 2 = a 2 + 2ab+ b2

(2x + 3y) 2 = 2x 2 + 2*2x*3y+ 3y2

= 4 x 2+ 12xy + 9 y2

(2x + 3y) 2 = 4x 2 + 12xy + 9y2.

(3x + 4y) 2 = 3x 2 + 2*3x*4y+ 4y2

= 9x 2+ 24xy + 16y2

(3x + 4y) 2 = 9x 2+ 24xy + 16y2

Practice problem:

Problem 1: Solve the given terms: (14 + 5) 2 and (9 + 3) 2

Answer is 361 and 144.

Problem 2: Solve the given terms: (5x + 2y) 2 and (9x + 6y) 2

Answer is 25x 2 + 20xy + 4y2.

81x 2 + 144xy + 36y2.

Problem 3: solve the given terms, (4 + 7) 4 and prove it?

Sets and its types

Sets are major concepts in mathematics that is taught in the middle school. In general terms, sets mean collection. A set is a collection of things, letters, words and more. For example: building block , puzzle, construction toys etc. can be termed as a set of kids’ educational toys. Let’ have a closer observation of sets in mathematics.

Sets: Any collection of things that can be grouped under one category is called as a set. For example: {Lego building block , Megabloks building blocks, Fisher Price building blocks, Peacock building blocks}. This is a collection of kids' building blocks and therefore a set. Sets can be classified into different types. Let’s have a look at the same in this post.

Finite Sets: A finite set is a type of set that consist a finite number of elements. For example: {kids laptop online, kids’ mobile online, kids’ electronic toys online}. This is a finite set of electronics for kids that include elements like kids’ laptop online and so on.

Infinite Sets: An infinite set is a type of set that consist infinite number of elements. For example: {1, 2, 3, 4, 5…..}. Here, the set consist infinite numbers and therefore is an infinite set.
Null Sets: A null set is a type of set that consists nothing. Null sets are also referred as empty sets or void sets.  A null set is denoted by {} or Ø.

Unit Sets: A unit set is a type of set that consist only one element. Unit sets are also referred as singleton sets or one point sets. For example: { baby food appliances }. Here, the set is a unit set as it consist only one element that is baby food appliances.
Find the type of sets for the below sets:
• {1, 2, 3, 4, 5} = Finite Set
• {} = Null Set
• {Barbie doll} = Unit Set
• {0, 1, 2, 3, 4, 5….} = Infinite Set
These are the basics about sets in mathematics.

Definition Value Proposition

Definition of proposition

Proposition is a statement which is either true or false. There are some statements which appear to be true and false at the same time. "The area of a circle is `pi`r2   ". , is a statement or proposition whose value is true. " 6 is an odd integer" is another proposition whose value is false. Consider the question " how are you? ". This is not a proposition since it does not possess a truth value. " What time is it now ?" is another example which is not  a proposition. Propositions are usually denoted by lower case letters like p, q, r, s etc.
Truth Value of a Proposition:

Proposition are statements which are either true or false. The statements which appear to be both at the same time are called paradoxes. For example consider the statement " I am a liar ". If this statement is true, then the speaker cannot be a liar. So the statement is false. If the statement is false then what the speaker says is false.Therefore the speaker is not a liar!

If a proposition is true, we say that the truth value of the proposition is True, denoted by T.

If a proposition is false, then we say that the truth value of the proposition is False, denoted by F.
Negation of a Proposition(definition Value Proposition)

" Mathematics is easy " is a  proposition.  Now, consider the proposition " Mathematics is not easy ". If the former is true, then the latter is false and vice-versa. Here the second proposition is called the negation of the first proposition. If p is a proposition, then the negation is denoted by the symbol ~p. The truth values of p and ~p are as follows :

p           ~p

T             F

F             T
Compound Propositions and Connectives

A combination of two or more propositions is called a compound proposition. There are four connectives used to make compound propositions. They are summarised below.

Compound proposition            connective                 symbol

Conjunction                                 and                               `^^`

Disjunction                                   or                                  `vv`

Conditional                                  if...then                          `|->`

Biconditional                               if and only if                   `harr`

High Line Construction

The line is a geometrical object in math and it is used for other shape construction. We can define the line by its properties. The single point is basis for high visible line construction. We can state the direction by line and it is straight. Now we are going to see about high line construction.
Explanation for High Line Construction

The high line in math:

The line is high symmetric and it is differentiated from other shapes by properties. The properties are straight, infinitely long, infinitely thin, zero width and the line is indicating the distance of two points.

High line construction:

We can construct the line easily and the parameters for line is simple one. Three types of geometry tools are used in high line construction. The tools are,

Pencil
Ruler
Protractor

What are all the steps followed in high line construction?

We should draw the line in white paper.
Take the pencil and sharpen the tip.
Start the line construction by dot and the line length is measured.
We are taking the line measurement in centimeter or millimeter.
The length is starting with dot till the length end point.
The ruler is used for measurement.
And joining the points with ruler.
Finally, we got the parallel line.
We can draw the perpendicular line by protractor and the angle mark and the starting point is joined.
The direction of line is represented with arrow mark in graph.
The arrow mark also indicated as the line is infinite length.
The construction of other shapes also done by line as basic tool.

More about High Line Construction

How to draw a line?

high line construction

The high visible line is drawn with ruler measurements like cm and mm. The mm value is represented as decimal values that is 6. 8 cm. Here the 8 is a mm value.

Pendulum Length

The length of the pendulum in the clock can be found by knowing the angle between the maximum points of the pendulum and the distance covered by the pendulum. The length of the pendulum can be calculated using the arc length formula where the known values are the central angle and the distance covered by the pendulum between the maximum points. In the following article we will see in detail about the topic pendulum length.

pendulum
More about Pendulum Length:

The pendulum oscuillstes from a single point and the pendulum covers a maximum position on the each side and the angle between the maximum positions can be measured and the distance covered by the pendulum between the maximum positions is also measured. Using all the values and substituting these values in the arc length formula we can calculate the length of the pendulum from the formula.

The formula for the arc length L = (theta/360)*2*pi*r

Here the θ is the angle between the maximum positions and r is the length of the pendulum and L is the distance covered by the pendulum between the maximum positions.
Example Problems on Pendulum Length:

1. The angle covered between the maximum positions of pendulum and the distance covered between the maximum positions are 110 degrees and 20 cm. Find the length of the pendulum.

Solution:

The length of the pendulum = (360*L)/(2*pi*theta)

= (360*20)/(2*pi*110)

= 360/11*pi

= 10.4 cm

2. The angle covered between the maximum positions of pendulum and the distance covered between the maximum positions are 120 degrees and 25 cm. Find the length of the pendulum.

Solution:

The length of the pendulum = (360*L)/(2*pi*theta)

= (360*25)/(2*pi*120)

= (3*25)/(2*pi)

= "12 cm"
Practice problems on pendulum length:

1. The angle covered between the maximum positions of pendulum and the distance covered between the maximum positions are 100 degrees and 18 cm. Find the length of the pendulum.

Answer: 10.3 cm.

Quadrilaterals Kite Tutor

The students are learn mathematics by using tutoring in online.  The tutor and students are communicated and the tutor share the information about related topics in online. The kites are quadrilaterals because it is drawn in geometry with four sides. The quadrilaterals have convex property in all types. So the kites also has convex property. Now we are going to learn about quadrilaterals kite by tutor.

Explanation for Quadrilaterals Kite Tutor

Tutor description for kite quadrilateral:

The quadrilateral kite has two pair of equal sides. So it is also known as deltoids. The adjacent sides are present next side. The parallelogram is a regular polygon. The kite also regular polygon.

The kite quadrilaterals property:

The right angle measurement is based on diagonals intersection.

The adjacent sides angles are equal.

The area of kite is find out by half of the diagonals product.

The perimeter of kite is find out by sum of length.

Quadrilaterals kite formula:

Area based on diagonals method – `(d_(1)d_(2))/2` .
Area based on trigonometry method – ab sin C.
Perimeter based on side’s sum – 2a + 2b.

More about Quadrilaterals Kite Tutor

Example problems for quadrilaterals kite tutor:

Problem 1: Calculating the kite area with diagonals 5.1 cm and 6.3 cm.

Tutor solution:

The diagonals of quadrilateral kite are d1 = 5.1 cm and d2 = 6.3 cm.

The area of quadrilaterals kite is `(d_(1)d_(2))/2` = `(5.1 * 6.3)/2` = 16 cm2.

Problem 2: Calculating the quadrilateral kite perimeter with  length of sides 21 cm and 10.8 cm.

Tutor solution:

The length of sides are a = 21 cm and b = 10.8 cm.

The quadrilateral kite perimeter is 2a + 2b = 2 x 21 + 2 x 10.8 = 63.6 cm.

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Exercise problems for quadrilaterals kite tutor:

1. The quadrilaterals kite with diagonals 14 cm and 12.5 cm. Calculating the kite area.

Tutor solution: The quadrilateral kite area is 87.5 cm2.

2. Calculating the kite perimeter with sides 18.3 cm and 13.8 cm.

Tutor solution: The kite perimeter is 64.2 cm.

Fixed Rate Calculator

A loan in which the interest rate do not alter through the whole expression of the loan reverse of adjustable rate A loan in which the interest rate does not modify through the whole term of the loan for an entity taking out a loan when rates are low, the fixed rate loan would permit him or her to "lock in" the short rates and not be worried with fluctuations. On the another hand, if interest rates were in the past elevated at the time of the loan, he or she would profit from a floating rate loan, since as the prime rate cut down to historically normal levels, the rate on the loan would reduce. Reverse of adjustable rate. A calculator is a small electronic device; it is used to achieve the fundamental operations of arithmetic. Modern calculators are more convenient than computers, while most PDAs are similar in size to handheld calculators.

Fixed rate calculator
Examples for the Fixed Rate Calculator:
Fixed rate calculator – Example 1:

Determine the payments and interest for a fixed rate loan, using monthly interest compounding and monthly payments. The purchase price $150, no of monthly payments is 2 months, and interest rate 5.000%, and the payment calculator computes the payment amount for you.
Solution:

Fixed rate calculator

The down payment amount is $11.00

The Loan amount is $ 139.00

Payment amount is $ 69.93

Interest rate is      5.000 %

Interest compounding: Monthly

Total amount financed: $139.00

Total payments: $139.87

Total finance charge: $0.87

Payment schedule:

Date           Payment InterestPrincipal Balance

Loan     08-26-2010                            139.00

1           09-26-2010   69.93       0.58    69.35    69.65

2           10-26-2010   69.94       0.29    69.65

2010 Total                139.87      0.87

Grand Total              139.87      0.87
More Examples for the Fixed Rate Calculator:
Fixed rate calculator – Example 1:

Determine the payments and interest for a fixed rate loan, using monthly interest compounding and monthly payments. The purchase price $100, no of monthly payments is 5 months, and interest rate 5.000%, and the payment calculator computes the payment amount for you.
Solution:

Fixed rate calculator

The down payment amount is $18.02

The Loan amount is $ 89.00

Payment amount is $ 18.02

Interest rate is      5.000 %

Interest compounding: Monthly

Total amount financed: $89.00

Total payments: $90.11

Total finance charge: $1.11

Payment schedule:

Event      Date             Payment Int    PrincipalBalance

Loan       08-26-2010                          89.00

1             09-26-2010  18.02    0.37    17.65  71.35

2             10-26-2010  18.02    0.30    17.72  53.63

3             11-26-2010  18.02    0.22    17.80  35.83

4             12-26-2010  18.02    0.15    17.87  17.96

2010 Total                  72.08    1.04    71.04

5             01-26-2011  18.03    0.07    17.96

2011 Total             18.03    0.07    17.96

Grand Total            90.11    1.11

Score as in Mathematical Terms

“Score” means 20! .The term "score" was came from the old English word “scoru”. It is the Middle age English word. Also, in turn, it came from the Norse skor. In mathematical terms score carried a value 20. Abraham Lincoln's celebrated Gettysburg consists of the phrase, "Four score and 7 years ago". In bible the term mentioned as “three score years and ten”.

Orgin of score:

In mathematical terms "Scores" way "groups of 20", just as "dozens" way "groups of 12". "Score" is eventually from the Proto-Indo-European root (s)ker-, sense "to cut" . Another English offspring is "shear". English still uses the word "score" to submit to an indentation or line made by a sharp instrument. Scores cut in wooden tally firewood were used in including, a fact we still make the unwitting suggestion to today when we talk regarding "keeping score". For a while in the history of the Germanic languages, the word for the cut used to record the number 20 became a word for the number itself.
Examples

4 score

Score mean 20 in mathematical terms

`4xx20=80`

Therefore four score is 80

3 score and 10

Score mean 20 in mathematical terms

`3 x 20` and 10

60+10=70.

Therefore 3 score and 10 is 70.

In Abraham Lincoln's celebrated Gettysburg consists of the phrase, "Four score and 7 years ago".

The phrase denotes, in mathematical terms

`4xx20` and 7

80+7

87

The phrase means 87 years ago.

Three score difference 10

Three score

`3xx20`

60 difference 10

60-10

50

Some mistakes arises that in some cases score is taken as 10. It wii lead to wrong answer

We cant assume any value to score. It means to 20 alone.

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Related examples like score:

Like score there are a lot of examples are in mathematical terms.

Foursquare

It means symmetrical in all dimensions or equal in all dimensions like cubical.

Like the words numbers also holds some meaning, Let take an example

223

It bibbiblically mean things involving to an order of survival beyond the visible, demonstrate-able space; that which is above and away from the laws of nature; and/or matters concerning the stroke or influence of the ghostly or psychic powers and the top-secret knowledge of them.

General Probability

General probability is the part of mathematics that learns the feasible outcomes of given events together with the outcomes' relative possibilities and distributions. In general usage, the word "probability" is used to mean the opportunity that a particular event will occur expressed on a linear scale from 0 to 1, also expressed as a percentage between 0 and 100.Now we will see the examples of probability.

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Examples- General Probability

Example 1

In a class there are 8 students got top eight marks in English. The marks are 78,83,86,88,92,93,95,98.  What is the probability for the following outcomes?

i) Select the marks are below 85.

ii) Select the marks between 80 and 90.

Solution:

i) Take P(A) is the probability for the marks are below 85.

Given marks are 78,83,86,88,92,93,95,98.

Total numbers n(S)=8

Here the following marks are below 85 n(A)={78,83}=2

So P(A)=(n(A))/(n(S))

=2/8

=1/4 .

ii) Take P(B) is the probability for the marks between 80 and 90.

The marks 83,86,88  are available between the 80 and 90.So n(B)=3

Total outcomes n(S)=8

So P(B)=(n(B))/(n(S))

= 3/8 .

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Example 2

What is the probability for select the letter ‘G’ from the word ‘GENERAL KNOWLEDGE’?

Solution:

Given word is GENERAL KNOWLEDGE.

Total letters n(S)=16

Number of ‘G’ letter n(A)=2

So the probability=2/16

=1/8 .

Example 3

Paul has 20 caps and Alex has 15 caps. What is the probability for select the Alex’s caps?

Solution:

Paul’s caps n(A)=20

Alex’s caps n(B)=15

Total number of caps n(S)=20+15

=35

Probability for select the Paul’s caps= 15/35

= 3/7 .
Practice Problems- General Probability

1) What is the probability for select the letter ‘E’ from the word ‘ELEMENT’?

Answer:

Probability=3/7 .

2) Tom has 5 white balls, Joseph has 7 yellow balls. What is the probability for getting the yellow balls?

Answer:

Probability=7/12 .

These examples and practice problems are used to study the general probability.

Solving Summation Notation

A summation notation (Σ) is used for sum the numbers or quantities indicated. We can sum all types of numbers such as real, integer, complex, rational numbers using sigma notation. Summation notation is also called as sigma notation. Using summation notation we can solve the problems easily. Summation a concise and convenient way of writing long sums.


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Solving Summation Notation – Solving Example Problems

Example 1: Find the sum for the given expression and explain the terms of summation notation;

7

∑ (n + 1)3 =?

n=1

Solution:

7

∑ (n + 1)3 =?

n=1

Lower Bound: ‘n’ is the lower bound of the summation notation.

Upper Bound:  ‘7’ is the upper bound of the summation notation.

Formula: (n + 1)3 is the formula to describe the each term of the summation notation.

Start: n = 1, here 1 is indicating the first number of the terms of summation notation. Usually value of start is zero (0) or one (1).

End: 7 is indicating the last umber of the terms of the summation notation.

Index Variable: Index variable ‘n’ is used to labeled each term of summation notation.

7

∑ (n + 1)3 = 8 + 27 + 64 + 125 + 216 + 343 + 512 = 1295

n=1

7

Therefore, ∑ i3 = 1295

i=1

Example 2: Solve and find the sum for the given expression;

6

∑ 3(n + n2) =?

n=1

Solution:

6

∑ 3(n + n2) =?

n=1

Formula: 3(n + n2), starting term: 1, end term: 6, and index variable: n

6

∑ 3(n + n2) / n = 3(1 + 12) + 3(2 + 22) + 3(3 + 32) + 3(4 + 42) + 3(5 + 52) + 3(6 + 62)

i=1


= 3(1 + 1) + 3(2 + 4) + 3(3 + 9) + 3(4 + 16) + 3(5 + 25) + 3(6 + 36)

= 3(2) + 3(6) + 3(12) + 3(20) + 3(30) + 3(42)

= 6 + 18 + 36 + 60 + 90 + 126

= 336

6

Therefore, ∑ 3(n + n2) = 336

i=1

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Example 3: Solve and find the sum for the given expression;

6

∑ n(2n + 1) / 2 =?

n=1

Solution:

6

∑ n(2n + 1) / 2 =?

n=1

Formula: n(2n + 1) / 2, starting term: 1, end term: 6, and index variable: n

6

∑ n(2n + 1) / 2 = (1(2(1) + 1) / 2) + (2(2(2) + 1) / 2) + (3(2(3) + 1) / 2) + (4(2(4) + 1) / 2)

n=1  + (5(2(5) + 1) / 2) + (6(2(6) + 1) / 2)


= (1(3) / 2) + (2(5) / 2) + (3(7) / 2) + (4(9) / 2) + (5(11) / 2) + (6(13) / 2)

= (3 / 2) + (10 / 2) + (21 / 2) + (36 / 2) + (55 / 2) + (78 / 2)

= 1.5 + 5 + 10.5 + 18 + 27.5 + 39

= 101.5

6

Therefore, ∑ n(2n + 1) / 2 = 101.5

n=1
Solving Summation Notation – Solving Practice Problems

Problem 1: Solve and calculate the sum for the given expression;

5

∑ n2(n) =?

n=1

Answer: 225

Problem 2: Solve and calculate the sum for the given expression;

5

∑ (2n + 2) / n =?

n=1

Answer: 14.57

Define Expanded Form

Expanded Form is a method to break up a number to illustrate how much each digit in the number represents and in other words, expanded form is the method of pulling a number separately and expressing it as a sum of the values of every digit. In this article we study about expand form define of and develop the knowldge of the math .

Example on Define Expanded Form

Choose the expanded form of the number 15,793.

Solution:

The expanded form of the number 15,793 can be found using a place value chart.

Now, let’s aim to write the number in words and define how it helps us identify the expanded form.

The number 15,793 is read as: 15 thousands 7 hundreds 93 which is the same as 1 ten thousand  5 thousands 7 hundreds 9 tens and 3 ones.

So, the expanded form of 15,793 is 10,000 + 5,000 + 700 + 90 + 3.

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Example to Define Expand Form Math:

Example 1:

57  can be expand as 57 = 50 + 7.

226 can be expand as 296 = 200 + 20 + 6.

5467 can be expand as 5467 = 5000 + 400 + 60+7.

89192 can be expand as 89192 = 80000 + 9000 + 100+90+2

Example 2:

What will we get when we expand form 23,368?

Choices:
A. 2 ten thousands, 3 thousands, 3 hundreds, 6 tens, and 8 ones
B. 2 ten thousands, 3 thousands, 6 hundreds, 6 tens, and 7 ones
C. 5 thousands, 5 ten thousands, 6 hundreds, 6 tens, and 8 ones
D. 5 thousands, 8 hundreds, 3 tens, and 68 ones

Correct Answer: A

Solution:

Step 1: 23,368 = 20,000 + 3,000 + 300 + 60 + 8
Step 2: = (2 × 10,000) + (3 × 1,000) + (3 × 100) + (6 × 10) + (8 × 1)
Step 3: So, when we expand 23,368, we will get ‘2 ten thousands, 3 thousands, 3 hundreds, 6 tens, and 8 ones’.

Example 3:

What will we get when we expand form 13,369?

Choices:
A. 1 ten thousands, 3 thousands, 3 hundreds, 6 tens, and 9 ones
B. 2 ten thousands, 3 thousands, 3 hundreds, 6 tens, and 5 ones
C. 3 thousands, 7 ten thousands, 9 hundreds, 6 tens, and 9 ones
D. 4 thousands, 7 hundreds, 9 tens, and 68 ones

Correct Answer: A

Solution:

Step 1: 13,369 = 10,000 + 3,000 + 300 + 60 + 9
Step 2: = (1 × 10,000) + (3× 1,000) + (3 × 100) + (6 × 10) + (9 × 1)
Step 3: So, when we expand 13,368, we will get ‘1 ten thousands, 3 thousands, 3 hundreds, 6 tens, and 9 ones’.

Systematic Sampling Method

In this article, let us study what is sampling and the types of sampling.

Preliminary concepts of sampling:

In testing the quality of bulbs produced, by a company it is impossible to test every bub manufactured by a company; it is quite sufficient if a sample is taken for testing and based on this test it is possible to draw conclusion about the population. If each and every item is tested, in some cases it may not be possible to send them to the market. thus, process of sampling is to get information about the population from the sample. To the extent to which we can do this with any accuracy depends on the choice of our sample or samples.

Samples are classified as follows:

non-probability samples

probability samples

Some non -probability samples  are 1. deliberate sampling , 2. Quota sampling 3. Block sampling.

Some probability samples  are1. Random sampling , 2. Stratified sampling , 3. Systematic sampling , 4. Multi stage sampling
What is Systematic Sampling

Systematic sampling with a random beginning is a form of restricted random sampling which is highly useful in surveys concerning enumerable population. In this method, every member of the population is numbered in a serial order and every ith element, starting from any of the first i items is chosen.

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Example for Systematic Sampling:

Suppose we require 5% sample of students from a college where there are 2000 students and where everyone is given a departmental number from 1 to 2000, we should first select one number at random from 1 to 20. Suppose the number chosen is 12; then our sample consists of students with departmental numbers 12, 32, 52, 72,..1992

If, however, the students are arranged in groups of 20 such that the first student belongs to the first of 20 specified communities, the second to a second community and so on, then our sample may consist of students of all belonging to the same community and consequently we cannot consider the sample to be random

Integral of ln u

Integral calculus is a branch of calculus that deals with integration. If f is a function of real variable x, then the definite integral is given by

int_a^bf(x)dx

Where a and b are intervals of a real line.

Integral of ln u means integration of natural logarithmic (ln) function with respect to the variable ' u '. In this article, we are going to see the list of natural logarithmic integral (ln) rules with few example problems.
List of Natural Logarithmic (ln) Integral Rules:

int 1/u dx = ln |u| + C

int  ln u du = uln u  - u + C

int uln u du = (u^2)/2 ln u  - 1/4 u2 + C

int u2 ln u du = (u^3)/3 ln u - 1/9 u3 + C

int (ln u)2 du = u(ln u)2 - 2uln u + 2u + C

int ((ln u)^n)/u dx = 1/(n + 1) (ln u)n+1 + C

int  (du)/(uln u) = ln (ln u) + C

Learning Natural Logarithmic (ln) Integral Rules with Example Problems:

Example problem 1:

Integrate the function f(u) = 5/u

Solution:

Step 1: Given function

f(u) = 5/u

int f(u) du = int5/u du

Step 2: Integrate the given function f(u) = 5/u with respect to ' u',

int5/u du = 5ln (u)+ C

Example problem 2:

Integrate the function f(u) = log (9u)

Solution:

Step 1: Given function

f(u) = log (9u)

int f(u) du = intlog (9u) du

Step 2: Integrate the given function  with respect to ' u' using integration by parts,

Let x = log (9u)    dv = du

dx = 1/u  du         v = u

int x dv = xv - int v dx

int log (9u) du = ulog (9u) - int u 1/u du

= ulog (9u) - int du

= ulog(9u) - u + C

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Example problem 3:

Integrate the function f(u) = 10log (3u)

Solution:

Step 1: Given function

f(u) = 10log (3u)

int f(u) du = int10log (3u) du

The above function can be written as

int f(u) du = 10intlog (3u) du

Step 2: Integrate the given function  with respect to ' u' using integration by parts,

Let u = log (3u)    dv = du

du = 1/u  du         v = u

int x dv = xv - int v dx

10 int log (3u) du = 10[ulog (3u) - int u 1/u du]

= 10ulog (3u) - 10int du

= 10ulog(3u) - 10u + C

Equilateral Pentagon Angles

In geometry an equilateral pentagon is a polygon with five sides of equal length. Its five internal angles, in turn, can take several values, thus permitting to form a family of pentagons. In contrast, the regular pentagon is unique, because is equilateral but at the same time its five angles are equal. Four intersecting equal circles disposed in a closed chain, are sufficient to describe an equilateral pentagon. ( Source Wikipedia )


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Examples

Problems 1

Find the missing angle on the pentagon

Is the four angle on the equilateral pentagon are 150, 100, 58 and 67

Solution

The sum of the 5 angles of pentagon is 540

The given angles are

Angle 1 a1=150

Angle 2 a2=100

Angle 3 a3=58

Angle 4 a4=67

Let the missing angle be x

Angle 5 a5=x

a1+a2+a3+a4+a5=540

150+100+58+67+x=540

250+125+x=540

375+x=540

Subtract 375 on both sides

375-375+x=540-375

x=165

The missing angle is 165

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Problems 2

Find the missing angle on the pentagon

Is the four angle on the equilateral pentagon are 150, 70, 158 and 67

Solution

The sum of the 5 angles of pentagon is 540

The given angles are

Angle 1 a1=150

Angle 2 a2=70

Angle 3 a3=158

Angle 4 a4=67

Let the missing angle be x

Angle 5 a5=x

a1+a2+a3+a4+a5=540

150+70+158+67+x=540

220+225+x=540

445+x=540

Subtract 375 on both sides

455-455+x=540-455

x=95

The missing angle is 95

Problems 3

Find the missing angle on the pentagon

Is the four angle on the equilateral pentagon are 100, 150, 90 and 40

Solution

The sum of the 5 angles of pentagon is 540

The given angles are

Angle 1 a1=100

Angle 2 a2=150

Angle 3 a3=90

Angle 4 a4=40

Let the missing angle be x

Angle 5 a5=x

a1+a2+a3+a4+a5=540

100+150+90+40+x=540

250+130+x=540

380+x=540

Subtract 375 on both sides

380-380+x=540-380

x=160

The missing angle is 160

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Problems 4

Find the missing angle on the pentagon

Is the four angle on the equilateral pentagon are 155, 120, 60 and 70

Solution

The sum of the 5 angles of pentagon is 540

The given angles are

Angle 1 a1=155

Angle 2 a2=120

Angle 3 a3=60

Angle 4 a4 =70

Let the missing angle be x

Angle 5 a5=x

a1+a2+a3+a4+a5=540

155+120+60+70+x=540

275+130+x=540

405+x=540

Subtract 375 on both sides

405-405+x=540-405

x=135

The missing angle is 135