Determining Sample Space

Introduction to determining sample space:

The probability separation be capable of be described regarding the sample space. The probability may be describing more requisites as, experiment, outcome, sample space, events. The probability is a one of dimension from the set of events. The many algorithms can be used to determining the sample spaces. The simply to determining the probable outcomes is straightforward represented as sample spaces. The probability triple is a further name of the sample space. The probability process can be classified three types like sample space, events and function of event.

Determining Sample Space-definition:

Sample space is defined as; the amount of probable outcomes is known as determining sample space. The determining sample space is helped to calculate the whole space, so it is normally represented as measure space.  The sample space can be specified as ?, and it is one of the chances of non- empty set.

Some of the results can be followed through the experiment so it is simply called as deterministic experiments. From the experiments, set of possible outcome is called as random experiment. The put of probable outcomes of a random research is known as determining sample space

Determining Sample Space-examples:

Problem 1:

Determining the sample space, when three coins are tossed randomly.

Solution:

Basically the coins contain two sample outcomes like head or tail. Here, we have to toss three coins randomly. The possible outcomes are,

The required sample space = {HHH,HHT,HTH,THH,THT,TTH,HTT,TTT}

Checking:

The number of coins (n) = 3.

Normal formula for sample space = 2number of entity.

The sample space       = 23.

So, the sample space = {HHH,HHT,HTH,THH,THT,TTH,HTT,TTT}.

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Problem 2:

Determining the sample space, where the two dice throw same time. Find the equal pair of dice.

Solution:

The each dice has been providing the six possible outcomes.

So, the two dice has been providing 36 possible outcomes.

The possible outcomes of two dice = { (1,1), (1,2),(1,3), (1,4),(1,5),(1,6),

(2,1),(2,2),(2,3) ,(2,4), (2,5),(2,6),

(3,1),(3,2),(3,3),(3,4), (3,5),(3,6),

(4,1),(4,2),(4,3),(4,4),(4,5),(4,6),

(5,1), (5,2),(5,3),(5,4),(5,5),(5,6),

(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)}

These are the possible outcomes by throwing two dice.

From the given problem, we have to find the pair of outcomes.

The pair of sample space = { (1,1),(2,2),(3,3),(4,4),(5,5),(6,6)}.

This is the sample spaces for throwing two dice at the same time.

Solve Calculus Help Integrals

Introduction for solve calculus help integrals:

Calculus is divisions in mathematics. It contains limits, functions, derivatives, integrals, and infinite series. Calculus is divided into two major parts, differential and integral calculus, which are associated by the basic theorem of calculus. Calculus is the learning of vary, in the equal way that geometry is the learning of shape and algebra is the learning of procedure and their application to solving equations.A math tutorial is one technique of shifting knowledge and may be used as a piece of learning. More interactive and detailed than a book; a tutorial look for to tutor by example and provide the information to complete a certain task.I like to share this Triple Integrals with you all through my article.

3 important topics involved in Calculus are,

Limits

Derivatives

Integrals

Solve Calculus Help Integrals - Examples

Solve calculus help integrals - Example 1:

Integrate the given equation with respect to x, we get

`int ` x7 dx = `(x^7 +1)/(7 + 1) ` + c

= `(x^8)/8 ` + c

Answer:

The final answer is  ` (x^8)/8` + C

Solve calculus help integrals - Example 2:

I am planning to write more post on Integration by Part, Fundamental Theorem of Calculus Examples. Keep checking my blog.

Evaluate:

`int` e(5x+3) dx

Solution:

Put (5x+3) =t so that 5 dx = dt or dx =`1/5` dt.

= `int` e(5x+3) dx =`1/5` et + C

= `1/5` e (5x+3) + C

Solve calculus help integrals - Example 3:

Integrate the given function int 2 sin 4x dx

Solution:

Integrate the given function with respect to x, we get

`int` 2 sin 4x dx = 2 `int` sin 4x dx

= 2 `xx`  - cos `(4x)/4`

= - cos `(4x)/4`

Answer:

The final answer is - cos` (4x)/4`

Solve Calculus Help Integrals - more Examples:

Solve calculus help integrals - Example 1:

Integrate int arcsin 2x dx

Solution:

U = arcsin 2x and du = dx = 1 dx

du =` 1/sqrt (1-(2x)^2) 2 dx = 2/sqrt(1-4x^2) dx ` and  v=x..

Therefore,

int arcsin 2xdx = x arcsin 2x – `int x 2/(sqrt(1-4x^2))` dx  = x arcsin 2x – 2` int x/(sqrt(1-4x^2))` dx.

Now apply u-substitution.

Let     U = 1-4x2

du = -6x dx,

`(-1/6)` du = x dx.

`int `  arcsin 2x dx = x arcsin 2x – 2` int` `x/sqrt(1-4x^2)` dx

= x arcsin 2x-2 `int 1/(sqrt(1-4x^2))` x dx

= x arcsin 2x-2 `int 1/sqrt(u) (-1/6)` du

= x arcsin 2x + `(1/3) int 1/sqrt(u)` du

= x arcsin 2x +` (1/3) int ` u-1/2 du

= x arcsin 2x+ `(1/3) (u^(1/2))/(1/2)` + C

= x arcsin 2x + `1/3(1-4x2)^(1/2)` + C

= x arcsin 2x +` (1/3) sqrt(1-4x^2)` + C.

Solve calculus help integrals - Example 2:

Evaluate:

1) `int` cos 2x dx

2) `int` e(4x+3) dx

Solutions:

1)    Put 2x =t so that 2 dx    = dt or dx =`1/2` dt.

= `int` cos 2x dx =`1/2` sin t + C

= `1/2` sin 2x + C

2)    Put (4x+3) =t so that 5 dx = dt or dx =`1/4` dt.

= `int` e(4x+3) dx =`1/4` et + C

= `1/4` e (4x+3) + C

Rules in Terminating Decimals

Introduction to rules in terminating decimals:
Since our numbering system decimals are fraction having their denominators as 10 or powers of 10, it is called a decimal system. Dec em in Latin means ten.

Here we will learn about numbers that are less than 1.

The numbers written in decimal form are called decimal numbers (or) simply decimals.
A decimal has two parts:
(1) whole number part and
(2) decimal part

Example:

In 473.619, the whole number part = 437 and the decimal part = .619

Rules in Terminating Decimals- Basic Definitions

Decimal point:

The dot that we use to separate the whole number part from the decimal part is called the decimal point.

Decimal places:

In a decimal, the places occupied by the digits after the decimal point are called decimal places.

For examples: 1) 7.37 has two decimal places.

2) 2.176 has three decimal places.

Decimal fractions:

A fraction whose denominator is 10 or a power of 10 is called decimal fractions

To round off a decimal:

Step 1: Look for the number of decimal point we would like to keep.

Step 2: Check the digit after the decimal point we would like to keep.

Step 3: Drop the digit if it is less than 5.

Step 4: If the digit is 5 or more, add 1 to the previous number before dropping

the extra digit

I am planning to write more post on What is an Irrational Number?, List of Prime Numbers to 100. Keep checking my blog.

Rules in Terminating Decimals-some Definitions

Definition of Terminating Decimal:

The word “terminate” means “end”. A decimal that ends is a terminating decimal. Inother words, if a fraction can be converted into a decimal completely, then such a decimal number is called a terminating decimal.

In the process of converting a fraction into a decimal by the division method, if we obtain a zero remainder after a certain number of steps, then the decimal obtained is a terminating decimal.

example for rules in terminating decimals:

Consider the following fractions namely 3/4 5/8, 1/8, 3/2,.by using convert fractions to decimals

Non-terminating decimals:

If a fraction cannot be converted into a decimal completely, then the division does not come to an end. Such a decimal number is called a Non-terminating decimal.

Example: ? = 22/7 = 3.412 . . . is a non-terminating decimal.

Recurring decimals:

Non-terminating decimals in which some digits are repeated continuously are called recurring decimals.

Example: 1.09090909. . . is a recurring decimal. It can be written as 1. 09 with ‘bar’ over the first and last digits of the repeated numbers.

Surface Integrals of Vector Fields

Introduction to Surface integrals of vector fields over parameterized surface:

The surface integral of vector field F on R3 over a surface S described in parametric form by r : D ? R3 is given by the number :

?? S F`*` dS = ?? DF(r(u,v)) `*` (ru `xx` rv) du dv

Surface integrals of vector fields for graphs:

Consider the vector field F = Pi + Qj + Rk and the surface S given by the graph z = f(x, y) on a domain D. Then the surface integral of vector field F over the surface S is :

`intint` S F`*` dS = ?? S F`*` ndS = ?? D(-Pfx - Qfy + R)dx dy

Example for Surface Integrals of Vector Fields

2?  ?

Let S be the unit sphere given by the parameterization:

x = cos ? sin f , y = sin ? sin f , z = cos f  for 0 = ? = 2p , 0 = f = p

Find the surface integral of vector field F(x, y, z) = xi + yj + zk  over the surface S.

Solution:

Here we have r(?, f) = ( cos ? sin f,  sin ? sin f,  cos f)

Hence,

r? = (-sin ? sin f, cos ? sin f, 0)

rf = (cos ? cos f,  sin ? cos f,  -sin f)

r? × rf = (-sin2 f cos ?, -sin2 f sin ?, -sin f cos f)

Next we evaluate,

F  · (r? × rf) = -sin f

Hence,

?? S F· dS = ?? DF(r(?, f)) · (r? × rf) d ? d ?

= $\int_{0}^{2\Pi }\int_{0}^{\Pi } -sin \varphi  d\varphi d\theta$

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Exercise on Surface Integrals of Vector Fields

Consider the vector field F  = x2i + y2j + zk. Let S be the surface given by the graph of the function z = x + y + 1 over the rectangle 0=x=1 , 0=y=1. Evaluate ?? S F`*` ndS

Hint: Here z = f(x, y) = x+y+1. Hence find fx and fy and plug in the formula given above for surface integrals of vector fields for graphs.

Conclusion on Surface Integrals of Vector Fields:

Surface integrals of vector fields are also referred to as flux across S. In other words, it is the volume of fluid moving with a velocity described by the vector field F(x,y,z) crossing the surface S per unit time.

Add Radical Numbers

Introduction to add radical numbers:

A square root of a number is said to be a radical number. Which is obtained by closing the numbers under add, subtract, multiply, and root extraction. A radical expression is containing a square root.  Radical is the (Sqrt) v symbol. That is used to denote square root, otherwise it is called as nth root. Here we have to go for add radical numbers, and how to work out addition of radical numbers.

Add Radical Numbers:

Add (Adding) radical expressions, which are containing radicals. Adding radicals are simple, when similar terms.

Similar terms:

If the radicals are similar terms means it has meet with following conditions.

The radical both have same index.
Both two radicals under the root sign is have identical quantities.
Any radicals, which are outside is same value.
Different terms:

If the radicals are different (not same) terms means it has meet with following conditions:

The radicals both are having same indexes.
The quantities which are under the radicals are different (not same), but both radicals need to simplify.
The outside variables are different (Not as same).

I am planning to write more post on Significant Figures Practice, Ratios and Proportions Word Problems. Keep checking my blog.

Example Problems for Add Radical Numbers:

Example 1:

Solve: v125 + v45. Add the radical numbers and find the answer.

Given: v125 + v45

First we have to simplify it. (125 = 25 * 5, and 45 = 9 * 5)

= v (25*5) + v (9*5)             (take common factor outside)

= 5v5 + 3v5

= 8v5.

Answer is: 8v5

Example 2:

Solve: v256 + v196. Add the radical numbers and find the answer.

Given: v256 + v196

Simplify the given terms, and we get (256 = 16* 16 and 196 = 14 * 14)

= v 16 * 16 + v 14 * 14                       (take common terms outside)

= 16 + 14

= 30

Answer is: 30

Example 3:

Solve: 2av3a3b3 + 3bv27a5b. Add the radical equation, and find the answer.

Solution:

Given: 2av3a3b3 + 3bv27a5b

Simplify it, and write it as separate,

= 2av3a2. a. b2.b + 3bv9.3a2. a2.a.b   (take common factors as outside) and write,

= 2a2bv3ab + 9a2bv3ab

Then we combine the co-efficient terms and we get 11a2bv3ab.

Answer is: 11a2bv3ab.

Double Integration by Parts

Introduction for double integration by parts:
The integration process represents the inverse process of the differentiation in progress. If `dy/dx` =h(x) here y is the function then the double integrals `int` `int`h(x) dx dy= y. There are two types of integration namely indefinite integral (without  limits) and definite integral (with limits). If we consider u = f(x), v = h(x), then the product rule in its simplest form is: `intint u dv/dx dxdy = int [uv - int v (du)/dx dx]dy`

In this article we shall discuss about the double integration by parts with the examples for the solution.

Examples to Explain "double Integration by Parts "

Double integration by parts on this function  `int int 5x^3 ln x dxdy` ,solve for the answer after calculating the integration on the first step by the integration by parts.
Solution:

First we take inner integral on the functions   `int 5x^3 ln x dx`

It is solved under the integration by parts as the first step for the prosecution of the steps.

The logarithmic form is exist in the given integration function.

Take ln x as u and `x^3 dx` as dv

u =5 `ln x`                   dv =`x^3 dx`

`(du)/dx = 5/x`                   v = `int x^3 dx`    

du = `5dx/x`                  v = `x^4/4 `      

We have  `int u (dv)/dx dx = uv - int v du/dx dx`

0r `int u dv = uv - int v du`

`int 5x^3 ln x dx`  = 5 ln x . `x^4/4 - ` `int x^4/4 5dx/x`             

=  5 ln x . `x^4/4 - ` `5/4int x^3 dx` 

=  5 ln x . `x^4/4 - `   `5/4 (1/4 x^4) `           

= `5 ln x (x^4)/4``-(5x^4)/16`          

This is the result obtained under the integration by parts for the inner integral.

Operate it with the outer integral functions by,

`int int 5x^3 ln x dxdy` =  `int [5 ln x (x^4)/4-(5x^4)/16 ] dy`         

=  `int` 5 ln x . `x^4/4 dy ` ` - int (5x^4)/16 dy` 

= `5y ln x (x^4)/4``-(5yx^4)/16 + c `          

This is the result obtained under the double integration by parts.

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Problems to Explain "double Integration by Parts "

Double integration by parts on this function  `int int x log x dx dy` ,solve for the answer after calculating the integration on the first step by the integration by parts.
Solution:

First we take inner integral on the functions   `int x log x dx`

It is solved under the integration by parts as the first step for the prosecution of the steps.

`int x log xdx` = `int ( log x) (x dx)`

The logarithmic function on the x function with another function x is not integrated directly ,so we take it as

u = logx       and        dv = x dx

? du = 1/x dx              v = `x^2/2`           

? `int x log x dx` = (logx) `x^2/2`  -  `int (x^2/2) (1/x dx)`                        

= `x^2/2`  (logx) `-1/2` `int x dx`       

? `int x log x dx` =  `x^2/2`  (logx) - `1/4 x^2 `

This is the result obtained under the integration by parts for the inner integral.

Operate it with the outer integral functions by,

`int int x log x dx dy`  =    `int [x^2/2(log x) - 1/4 x^2 ] dy`           

=    `int [x^2/2(log x)]dy - int [1/4 x^2 ] dy`

=       `x^2y/2(log x) - 1/4 x^2y dy + c`              

This is the result obtained under the double integration by parts.

Addition Fraction Calculator

Introduction to addition fraction calculator
A fraction is a number that can represent part of a whole. The earliest fractions were reciprocals of integers: ancient symbols representing one part of two, one part of three, one part of four, and so on. much later development still used today (½, ?, ¾, etc.) and which consist of a numerator and a denominator, the numerator representing a number of equal parts and the denominator telling how many of those parts make up a whole. Mathematical calculators are used to perform mathematical operations. With the help of fraction addition calculator we can easily adding the fraction.  In this article of addition fraction calculator we are going to discuss about adding the fraction using calculator.

Calculation of Addition of Fractions.

Addition Fractions with same denominators:

A fraction contains two numbers, the top number value is called the numerator, and the bottom number value is called the denominator.

numerator
denominator

To add two fractions the denominator have the same value, add the numerators that is top value and put that sum over the common denominator value.

Addition Fractions with different denominators:

Find the Least Common multiplication (LCM) of the fractions

Add the numerators values (top value) of the fractions

Simplifying the fraction value

Addition two mixed numbers the fractions have the same denominator:

Adding the numerators value (top value) of the two fractions

Place that sum of top value over the common denominator (top value).

If the fraction is improper fraction (numerator larger than or equal to the denominator) then we have to convert it to a mixed number

Adding the integer part of the two mixed numbers

I am planning to write more post on real numbers, solving inequalities. Keep checking my blog.

Fraction Addition Calculator Examples

Example for adding fraction with same denominator:

Find `(4/12)+(5/12)`

Adding top values

=`(9)/(12)`

Example for adding fraction with different denominator:

Find the sum of 2/3 and 4/6`(2/3)+(5/6)`

= `(2)/(3)` =`(5)/(6)`

=`(2*2)/(3*2)`  + `(5*1)/(6*1)`

`(4)/(6)`+  `(5)/(6)`

`(9)/(6)`

= 1 `(3)/(6)`

Example for adding fraction with different denominator:

Adding 2 `(3/4)+3(1/2)`

= 2 `3/4` = `11/4`

=3 `1/2` = `7/12`

=`11/4` +`14/4` =`25/4`

convert int to mixed fraction `25/4` = 6 `1/4`

Semi Circle Math Definitions

Introduction of Semi circle math definitions:

Another name of the semi circle is half of the  circle. The diameter of a circle is the distance from a one point on the sphere to an end point pi Radians, and is the distance from one point on a sphere to another.

If r is the semi circle radius, then d = 2r. d is the diameter of the semi circle.

Example for Semi Circle in Math Definitions:

Semi circle- Example 1 in math definitions:

Find the diameter of a semi circle given that its radius is 5

Solution:

The diameter is always multiply by 2 the length of the radius,

Diameter = 2 x 5 = 10

Semi circle- Example 2 in math definitions:

Find the diameter of a semi circle known that area of the semi circle is 153.86

Solution:

Step 1:

Find the radius of the semi circle

`( pi . r^2)/2` = 153.86

`(r^2)/2` = `153.86 / 3.14`

`(r^2)/2`   = 49

r2 = 49 x 2  = 9.899



Step 2:

Then calculate the diameter of the semi circle,

Diameter = 2 * r

=  2 * 9.899

= 19.79800

Therefore, the diameter of the semi circle is 19.798.

Semi circle- Example 3 in math definitions:

Find the diameter of a semi circle known that its radius is 6

Solution:

The diameter is always multiply by 2 the length of the radius,

Diameter = 2 x 6 = 12

Semi circle- Example 4 in math definitions:

Find the diameter of a semi circle known that area of the semi circle is 133.04

Solution:

Step 1:

Find the radius of the semi circle

` (pi . r^2)/2` = 113.04

`(r^2)/2` = `113.04 / 3.14`

r2  = 36 x 2 = 72

r = 8.485

Step 2:

Then calculate the diameter of the semi circle,

Diameter = 2 * r

=  2 *8.485

= 16.97

Therefore, the diameter of the semi circle is 16.97.

Semi circle- Example 5 in math definitions:

Find the diameter of a semi circle known that its radius is 8

Solution:

The diameter is always multiply by 2 the length of the radius,

Diameter = 2 x 8 = 16

Semi circle- Example 6 in math definitions:

Find the diameter of a semi circle known that its radius is 21

Solution:

The diameter is always multiply by 2 the length of the radius,

Diameter = 2 x 21 = 42

Semi circle- Example 7 in math definitions:

Find the diameter of a semi circle known that area of the circle is 1384.74

Solution:

Step 1:

Find the radius of the circle

`(pi . r^2)/2` = 1384.74

r2/2 = 1384.74 / 3.14

r2  = 441 x 2

r2 =  882

r   = 29.698



Step 2:

Then calculate the diameter of the circle,

Diameter = 2 * r

=  2 * 29.698

= 59.396

Therefore, the diameter of the semi circle is 59.396.

Semi circle- Example 8 in math definitions:

Find the diameter of a semi circle known that its radius is 32

Solution:

The diameter is always multiply by 2 the length of the radius,

Diameter = 2 x 32 = 64

Semi circle- Example 9 in math definitions:

Find the diameter of a semi circle known that its radius is42

Solution:

The diameter is always multiply by 2 the length of the radius,

Diameter = 2 x 42 = 84

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Practice Problems for Semi Circle in Math Definitions:

Practice Problem 1:

Find the diameter of a semi circle known that its radius is100

Answer:

200

Practice Problem 2:

Find the diameter of a semi circle known that area of the semi circle is 530.66

Answer:

36.768

Word Problems about Prime Factorization

Introduction :

In Mathematics, the basic theorem of arithmetic states that each composite number can be factorized uniquely into a product of prime factors. A number is a prime number if its only whole number factors are 1 and itself. 7 are primes because its only factors are 1 and 7. If a number is not primes, it is called a composite number.

Because 4 have factors of 2 and 2, 4 is a composite number. The number 1 is not considered a prime number. Therefore, it is not included in the following list of prime numbers less than 50

2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47

Word Problems about Prime Factorizations – Definition and Example Problems:
Word problems about prime factorizations – Definition:

A whole number greater than 1 with precisely two factors, itself and 1, is a prime number. A whole number greater than 1 with above 2 factors is a composite number. The numbers 0 and 1 are neither prime nor composite: 0 has infinite factor, and 1 only has one factor, itself.

The number 2 is the only even prime number. A number expressed as a produce of factors that are all prime is called the prime factorizations of the number. For example, the prime factorizations of 12 are 2 x 2 x 3. There are two main ways of finding the primes of a number: dividing and splitting.

Word problems about prime factorizations - Example problems:

Word problem 1:          

Jim purchased some oranges to distribute to her friends. If she gave 3 or 5 or 6 oranges to each of her friends she will be left with 100 oranges. What is the least number of oranges she purchased?

Solution:

The L.C.M of 3, 5, 6 = 30
100 + 30 = 130
The least number of oranges she purchased is 130.
Solution: 130 oranges.

Word problem 2:

George wished to buy some crackers. If he gave 2 crackers or 3 crackers or 5 crackers to each of his friends, he is left with no crackers. What is the least number of crackers he must purchase?

Solution:

George wished to buy some crackers.

The L.C.M of 2, 4, 5 = 20

The least number of crackers he purchased is 20.

Solution: 20 crackers.

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Word Problems about Prime Factorizations – Practice Problems:
Word problem 1:

Jane purchased a few oranges to distribute to her friends. If she gave 4 or 5 or 6 oranges to each of her friends she will be left with 15 oranges. What is the least number of oranges she purchased?

Answer: 75 oranges she purchased.

Word problem 2:

Three clocks ring once at the same time. After that, the first clock rings after every 3 hours, the second after every 7 hours, and third after every 5 hours. After how many hours will they again ring together?

Answer: 105 hours will they again ring together.

Word problem 3:

What is the smallest number of students required so that they may be uniformly arranged in rows of 12, 14 or 20?

Answer: 252 arranged students