Number Percentage Calculator

Percent means ‘for every 100 ’. So, when we say P% .it means P  out of 100 . Thus, P%=P/100 . It is often denoted by symbol “% ”. Any percentage can be expressed as a fraction. For example, 40%=40/100=2/5 .

Percentages are used to find whether one quantity is large or small compared with another quantity. The first term usually represents a part of, or a change in the second term, which should be greater than zero.

Percentages are usually used to express numbers between zero and one, any dimensionless proportionality can be expressed as a percentage.

Please express your views of this topic Percentage of Difference by commenting on blog.

Let us now express some percentages as fractions:

a.       5%=5/100=1/20 .

b.      10%=10/100=1/10 .

c.      25%=25/100=1/4 .

d.     75%=75/100=3/4 .

e.      125%=125/100=5/4 .

f.        175%=175/100=7/4 .

g.       (3 1/8)%=25/800=1/32 .

h.     (6 1/4)%=25/400=1/16 .

i.        (8 1/3)%=25/300=1/12 .

j.        (16 2/3)%=50/300=1/6 .

k.       (66 2/3)%=200/300=2/3 .

l.        (87 1/2)%=175/200=7/8 .



Calculation of Percentage:

Calculation of percentage:

The percent symbol can be treated as being equivalent to the pure number constant 1/100=0.01,  while performing calculations with percentage.

If a number is first changed byP%  and then changed by Q% , then the net change in the number =[P+Q+((PQ)/100)] . Remember that any decreasing value in the formula should be taken as ‘negative’ and increasing value should be taken as ‘positive’.

Similarly, if A’s salary is P%  less than B’s salary, then the percentage by which B’s salary is more than A’s salary is(100P)/(100-P) .

If expenditure also, then percentage change in expenditure or revenue=[P+Q+((PQ)/100)] . Where ‘P’ is the percentage change in price and ‘Q’ is the percentage change in consumption.

Problems on number percentages:

Ex1 : What percentage of 1600 is 40?

Sol:  Let 40  be "P% of 1600.

So, 40=P%   of 1600 =(P/100)(1600)=16P .

Thus, P=40/16=2.5% .

Ex2 : Calculate 40%  of 625 .

Sol: 40%  of a number =2/5  of the number =2/5  of 625=(2/5)(625)=250 .

Ex :3 A number is first increased by 30%  and then decreased by 20% . Find the net change in the number.

Sol:  Let the original number be 100 .

Increasing by 30%, " it becomes " 130 .

Now, if 130  is decreased by 20% , it becomes 104 .

Thus, the net change =(104-100)=4%  increase.

Algebra is widely used in day to day activities watch out for my forthcoming posts on Divide Fractions by Whole Numbers and sample paper of class 9 cbse sa2. I am sure they will be helpful.

Practice problems on number percentages:

Q:1  A’s salary is 25%  more than B’s salary. By what percent is B’s salary less than A’s?

Sol:  Let B’s salary be .

Since A’s salary is 25%  more than that of B, his salary will be Rs. 125 .

Thus, B’s salary is Rs. 25 less than the A’s salary.

So, in percentage: (25125)(100)=20% . Hence, B’s salary is 20%  less than A’s salary.

Multiplication Properties

In this page we are going to discuss about multiplication properties concept. Integer used to measuring and counting in general things. Positive integer and negative integer is the two types of integers. Integer is nothing but using notational symbol represents numbers. Multiplying integer’s mean scaling a number by another.

3 x 4 = 3 + 3 + 3 + 3 = 12

3 x 4 = 4 + 4 + 4 = 12

Properties of multiplication

To study multiplying integers, we have to know five basic properties such as associative, commutative, distributive, and multiplicative identity. These multiplication properties are described as below.

Commutative property
Associative property
Multiplicative identity property
Distributive property
Zero property



Commutative property:

This is the product the two number is same even we change the order. That is, a * b = b * a.

Example: 5 * 4 = 4 * 5

= 20

Associative Property:

This is the product the three more number is same even we change the order. That is, (a * b) * c = a * (b * c).

Example: 5 *( 4 * 3 ) = (5 * 4) * 3

= 60

Identity Property:

This is the multiplication of any number by 1 is that number. That is, a * 1 = a.

Example: 9 * 1 = 9.

Distributive property:

Multiplication of a number with the addition of two numbers is same as addition of two products. That is, a* (b + c)=ab + ac.

Example: 5 * (4 + 3) = 5*4 + 5*3

= 35

Zero property:

Any thing multiply with zero is zero. That is, a * 0 = 0.

Example: 7 * 0 = 0

Multiplication rules

Positive * Positive = Positive

Negative * Negative =Positive

Positive * Negative = Negative

Negative * Positive = Negative

The following examples are to study multiplying integers with multiplication rule

5 * 4 = 20

-5 * -4 = 20

5 * -4 = -20

-5 * 4 = -20

Multiplication grid

Multiplication grid

Methods to multiplying integers

Below are the methods to multiplying integers-

Method 1:

Multiplying integers of 345 * 6

Solution:

Here, 5 is at unit place, so we do    5 * 1       =    5  * 6 =    30

4 is at 10s place, so we do   4 * 10     =  40  * 6 =   240

3 is at 100s place, so we do 3   * 100 = 300 * 6 = 1800
--------
The product result  =  2070
--------

Method 2:

Multiplying integers of 345 * 6

Solution:

345

x 6
--------------
30
240
1800
--------------
2070
--------------

Sample Space

In statistics, learning sample space basically represents the presentation and interpretation of the events of the possible outcomes that occur in a planned learning or scientific investigation. The learning sample space helps to refer all recording of information's are numerical or categorical, as an observation. The learning sample space assists in the following three cases, designed experiments, observational studies, and retrospective studies, the end result was a set of data that of course is subject to uncertainty.


Definition for Sample Space

The set of all favorable combinations of outcomes of a statistical experiment is labeled the sample space and is denoted by the mathematical symbol S.

Each possible outcome of a sample space is labeled a member of the sample space, in other words it is labeled the sample point. The trials of a sample space has a finite number of elements are separated by commas (,) and enclosed in braces ({}).

Thus the: sample space of possible outcomes when a coin is tossed, may be written S = {H, T),

Where, H means that “heads" and T means that “tails,”.


Examples for Sample Space:

Example 1:

Determine the sample space for the event of rolling a die, using the learning sample space.

Solution:

In rolling a die the number that shows on the top face.

The required sample space S1 = {1, 2, 3, 4, 5, 6}.

Example 2:

Determine the sample space for the number is even or odd.

Solution:

The required sample space S2 = {even, odd}.


More than one sample space:

If we know which element in S1 occurs, we can tell which outcome in S2 occurs; however, knowledge of what happens in S2 is of little help in determining which element in S1 occurs. Provides more information than S

Ex:  A coin is tossed twice. What is the Sample space?

Sol: The sample space; for this experiment is

S= {HH, HT, TH, TT}.

My previous blog post was on Multiplication Rule please express your views on the post by commenting.

Binary Number Representation

The binary numeral system, or base-2 number system, represents numeric values using two symbols, 0 and 1. More specifically, the usual base-2 system is a positional notation with a radix of 2. Owing to its straightforward implementation in digital electronic circuitry using logic gates, the binary system is used internally by all modern computers.

Understanding Binary Logistic Regression is always challenging for me but thanks to all math help websites to help me out.

The representaion of binary numbers are uses in mathematics are defined as follow,

Binary numbers representation in math:

In the binary number representation consists of octal, decimal, and hexa decimal numbers in the column. We can be represent the binary numbers by use of its operation. In the binary number representation, the decimal is easiest method of understanding binary numbers.

For example we can represent: 4567,

4 is represent the 1000’s

5 is represent the 100’s

6 is represent the 10’s

7 is represent the 1’s

Which means the representaion of 4567 is given as follow,
4567 = 1x1000 + 2x100 + 3x10 + 4x1

Given binary number representation,

1000


= 103 = 10x10x10

100


= 102 = 10x10

10


= 101 = 10

1


= 100 (any number to the exponent zero is 1)

The table above can be represented as the binary numbers,

Such that,

4567 = 4x1000 + 5x100    + 6x10     + 7x1

= 4x103 + 5x102 + 6x101 + 7x100

Examples for binary number representation in math:

The examples of binary number representation in math is given as follow:

Example:1

To determine the decimal number in 10102?

Solution:

Step 1: 1 => 1×2×2×2 = 8

Step 2:  0 => 0×2×2 = 0

Step 3: 1 => 1×2 (=2)

Step 4: 0 => 0

Answer is: 1010 = 8+0+2+0 = 10.

Example 2:

To determine the decimal number in 10112?

Solution:

Step 1:  1=> 1×2×2×2 (=8)

Step 2: 0 => 0×2×2 (=0)

Step 3: 1 => 1×2 (=2)

Step 4: 1 => 1

Answer is: 1001 = 8+0+2+1 = 11.

My forthcoming post is on how to find the prime factorization of a number and neet entrance exam syllabus will give you more understanding about Algebra.

Example 3:

To determine the decimal number in 1.112?

Solution:

Step 1:  1 => 1

Step 2:  1 => 1×(1/2)

Step 3: 1 => 1×(1/4)

Answer is :1.75.

Example 4:

To determine the decimal number in 11.112?

Solution:

Step 1: 1 => 1×2 (=2)

Step 2: 1 => 1

Step 3:  1 => 1×(1/2)

Step 4: 1 =>  1×(1/4)

So, 11.11 is 2+1+1/2+1/4 = 3.75 in Decimal

Answer is: 3.75.

Examples of Poisson Distribution

Definition: A random variable X is  a Poisson distribution if the probability mass function of X is P(X = x) =e−λ  λx / x!,                                       x = 0,1,2, …for some λ > 0

The mean of  Poisson Distribution denoted by λ, and the variance is denoted by λ.
The parameter of Poisson distribution is λ.

The Poisson distribution is a restrictive case of Binomial distribution under the following conditions.
(i)   Number of trials(n) is indefinitely huge(large), that is n → ∞.
(ii)  The constant probability(p) of success in each trial is very less.
ie., p → 0.
(iii) np = λ is finite where λ is a positive real number. When an event occurs rarely, the distribution of such event may be assumed to follow a Poisson distribution. The example problems of poisson distribution is   given below

Examples of Poisson distribution:

Some Examples of poisson distributions are given below

(1) The number of gamma particles emitted by a radio active source in a given time interval.
(2) The number of phone calls received at a telephone exchange in a given time interval.
(3) The number of defective articles in a packet of 250, produced by a good industries limited.
(4) The number of printing errors at each page of a book by a good publication centre.
(5) The number of road accidents reported in a city at a particular time.

Example Problems for Poission distribution:

Example problem 1: If a publisher of  technical books takes a great pain to ensure that his books are free of typological errors, so that the probability of any given page containing atleast one such error is 0.005 and errors are independent from page to page

(i) what is the probability of its 400 page novels will contain exactly one page with error.

(ii) atmost three pages with errors.
[e−2 = 0.1353 ; e−0.2. = 0.819].


Solution :

n = 400 , p = 0.005
np = 2 = λ
(i)  P(one page with error) = P(X = 1) = e−λ λ1/1! = e-2 21/1!
= 0.1363 × 2 = 0.2726
(ii)  P(atmost 3 pages with error) = P(X ≤ 3)
= Σ e−λ λx / x!    [limits 0 to 3]
= `sum` e−2 (2)x  /  x!
= e2 [1 +2/1! + 22/2! + 23/3!]
= e−2 (19/3 )= 0.8569

Learning Simple and Compound Events

Probability of any event  is the ratio of several favorable outcomes of an event to the total number of outcomes of an event. Otherwise if we toss a single dice then it is called as a simple event. If we toss a two dice then it is called as a compound event. Inclusive means when two events are happen at a same time. Mutually exclusive means when two events cannot happen at a same time.

Learning Simple and Compound Events Formulas :

learning  simple and compound events formulas in the probability of an event can be expressed as

P(a) = number of a favorable outcomes / total number of a possible outcomes

learning  simple and compound events formulas in the probability of two independent events(A and B) are multiplied by the probability of the first event by the probability of the second event. Two events are said to be independent. if P(A and B) = P(A) P(B). P(A)and P(B) are are non zero

P(A and B) = P(A) . P(B)

learning  simple and compound events formulas in the probability of two dependent events (A and B ) are multiplied by the probability of A and the probability of B after A occurs.

P(A and B) = P(A) . P(B following A)

learning  simple and compound events formulas in the probability of one or other of two mutually exclusive events (A or B) are added to the probability of the first event to the probability of the second event.Two events are said to be disjoint. if and only if P(A and B) = 0

P(A or B) = P(A) + P(B)

learning  simple and compound events formulas in the probability of one or the other of two inclusive events(A or B) are added to the probability of the first event to the probability of the second event and subtract the probability of both events happening.

P(A or B) = P(A) + P(B) - P(A and B).

practice problems in learning simple and compound events

Example 1:Abraham is going to the Super market to pick a new sports bats. Today, the shopkeeper has 25 cricket bats and 50 Tennis bats are available for sales. If Abraham randomly picks the bats, then what is the probability that the bat would be a Cricket bat?

Solution:

Probability(randomly choosing a Cricket bat) = Number of Cricket Bats / Number of cricket bats + number of tennis bats

= 25 / (25 + 50 )

= 25 / 75

= 1 / 3

= 0.33

if Abraham randomly picks a bat (which is a Cricket bat) having a probability  0.33.

Example 2:Lenin is going to the Super market to pick a new sports bats. Today, the shopkeeper has 20 cricket bats and 40 Tennis bats are available for sales. If Lenin randomly picks the bats, then what is the probability that the bat would be a Cricket bat?

Solution:

Probability(randomly choosing a Cricket bat) = Number of Cricket Bats / Number of cricket bats + number of tennis bats

= 20 / (20 + 40 )

= 20 / 60

= 1/ 3

if Lenin randomly picks a bat (which is a Cricket bat) having a probability  = 0.33

Exponential Growth Formula

A function is said to be Exponential growth that including exponential decay when the growth rate of that mathematical function is proportional to the function's current value. In a discrete domain of definition with equal intervals of the function is called as geometric growth or geometric decay. The exponential growth model is also called as the Malthusian growth model.


Exponential growth formula:

Exponential formula defines the  X as exponentially on time t.

X(t) = a . b(t/r)

"a" denotes the initial value

a = x,

X(0) = a,

b= a

It denotes the positive growth of the factor, t = time required

Example for exponential growth formula:

If a power doubles every 5 minutes, initially there’s only one doubles, how many powers would be there after 2 hours?

Here, a= 1, b= 2, t = 5 minutes.

X(t) = a . b(t/r) = 1 . 2{(120 minutes)/(5 minutes)}

X(2 hour) = 1 . 2 24 = 16777216

After two hours, there would be 16777216 powers.


Exponential growth Problem:

A microbiologist is researching a newly-discovered species of fungi. At time t = 0 hours, he puts one hundred fungi into what he has determined to be a favorable growth medium. Six hours later, he measures 200 fungi. Assuming exponential growth, what is the growth constant "i" for the fungi? (Round i to two decimal places.)

For the given problem, the units on time t will be hours, because the growth is being measured in terms of hours. The starting amount P is the amount at time t = 0, so, for this problem, P = 100. The ending amount is A = 200 at t = 6. The only variable we don't have a value for is the growth constant i, which also happens to be what I'm looking for. So I'll plug in all the values we know, and then solve for the growth constant:
A = Peit
200 = 100e6i
2 = e6i
ln(2) = 6i
`ln(2)/6` = i = 0.11552453