Learning Geometric Probability

Numerical measure of the likelihood of an event to occur is called as Probability. The probability should be a range in between 0 and 1.In this case we say probability is 0. If the event is certain to occur, we say probability is likely 1. The probabilities involved in the geometric problem that also called as geometric probability. This geometry probability may be a circle or any polygon from the geometric. It involves the length, area and volume of any one of the geometric shapes.

The definition of probability of an event has shown in below that is depend on their outcomes from the possibilities



Number of successful outcomes

Probability  =        _____________________________

Total number of possible outcomes.

learning geometric probability example problem in triangle:

The figure shows a triangle divided into sectors of different colours. Find the probability of angle for blue sector and orange sector?

learning geometric probability

Solution:

We have to find the Probability for blue color:

Step 1:

Total angle of triangle is 180 degree

Step: 2

Probability finding blue color sector= The blue sector triangle angle/ The total angle of  entire triangle

The blue sector triangle angle=30 degree

=30/360

=1/12

Step 3:

Next we have to find Probability for orange color:

Probability finding orange color sector= The orange sector triangle angle/ The total angle of entire triangle

The orange sector triangle angle=45

=45/360

=9/72

learning geometric probability example in Rectangle:

Example 2:

A rectangle has four sides and that corner having each 4 ball. Find the probability of each corner having 4 balls.?

Solution:

Here the Rectangle has the five corners such as A, B, C, D

We have to find the probability here,

Probability = Number of successful outcomes / total number of possible outcome

Probability  of each corner having the balls=4/4=1

Example 3:

A triangle with area 15 cm2 is inscribed in a circle with radius 3 Cm.Find the probability that a ball thrown fall into the triangle?

Solution:

We have to find the area of the circle and area of the triangle is given. From this we can find the probability of a ball that fall into the triangle easily.

Area of the circle = `Pi` r2

Radius  = 3. So Area of the circle = 3.14 * 3 * 3 = 28.26 cm2.

Area of the triangle  = 15 cm2.

So the probability  = 15 / 28.26 = 0.53

Answer is 0.53.

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How to solve a percent problem

In mathematics, a percentage is a way of expressing a number as a fraction of 100 (per cent meaning "per hundred" in French). It is often denoted using the percent sign, "%", or the abbreviation "pct". For example, 45% (read as "forty-five percent") is equal to 45 / 100, or 0.45. Let us see how to solve percent problem.                                                                                                                                - Source from Wikipedia

Formula – How to solve percent problem:

To solve the percent problem convert the fraction else a decimal to a percent and then multiply it by 100:

To get the percent value multiply the fraction into hundred. For example

` 2 / 4` (100) = `2 / 4 ` * 100 = 0.5 * 100 = 50%

To convert the percent value to fraction divide it by hundred.

25% = `25 / 100 ` = `1 / 4`

Example: How to solve percent problem:

Problem 1:

What is the decimal value of 72%?

Solution:

To obtain the decimal value we have to divide the given number by 100. Which is `72 / 100` = 0.72.

Therefore the decimal value of 72% is 0.72.

Problem 2:

What is the decimal value of 86.3%?

Solution:

To obtain the decimal value we have to divide the given number by 100. Which is `(86.3) / 100` = 0.863.

Therefore the decimal value of 86.3% is 0.863.

Problem 3:

What is the decimal value 18% of 60?

Solution:

To find 18 percentages divide 18 by 100 and then multiply it by 60.

= `18 / 100` * 60 = 10.8

Hence 18% of 60 = 10.8

Problem 4:

What is the decimal value for 15.5% of 150?

Solution:

To find 15.5 percent divide 15.5 by 100 and then multiply it by 150.

=` (15.5) / 100` * 150 = 23.25

Hence 15.5% of 150 = 23.25

Problem 5:

The book cost is 5.00 dollars. Now it was 15% increased. What will be the new cost?

Solution:

Multiply the cost by 15%. That is `15 / 100` * 5.00.

= `15 / 100` * 5.00 = 0.75

Add this decimal value with original cost (5.00+0.75= 5.75)

Therefore the original cost of the book is 5.75 dollars.

Problem 6:

In a box there are 85 chocolates. Suppose if a boy took out 60% of chocolates, how many chocolates did he left in the box?

Solution:

The boy took out the chocolates from the box is 60% of 85 or else 60 / 100 × 85

`60 / 100` × 85 = 51%

So the box has 85 chocolates and the boy took out 51 chocolates. Hence the number of chocolates the boy left in the box is 85 – 51 = 34.
Therefore the boy left 34 chocolates in the box.

Problem 7:

A theatre contains 530 chairs. 470 of them are occupied. What percentages of the chairs not occupied?

Solution:

Number of chairs                        = 530

Number of chairs occupied       = 470

Number of chairs not occupied = 530 – 470 = 60

Number of chairs not occupied percentage = `60 / 530` × 100 = 11.32%
Practice problems – How to solve percent problem:

Problem 1:

What is the decimal value of 27%?

Solution:

27% = 0.27

Problem 2:

What is the decimal value 8% of 75?

Solution:

8% of 75 = 6

What is the Associative Property

The addition or multiplication of a collection of integer is the same separately from of how the numbers are grouped. A binary operation * is assumed to be associative if any three elements a, b, c of a set a *(b * c) = (a * b) * c. Multiplication of real integers are associative but the division of real integers are not assosiative. And addition is associative but subtraction is not.

What is the Associative Properties

What is the associative property?..The associative property is a property which always involve 3 or more numbers in calculations. The parenthesis indicates the terms that are consider one unit. The gathering (Associative Property) are within the parenthesis. Hence, the figures are 'associated' together. In multiplication, the result is always the same regardless of their combination.

There were two properties are involved in what is the associative property,

They were, Associative property in addition

Associative property in multiplication


Examples for what is the Associative Property (Addition and Multiplication)

Example for what is the Associative property in addition:

Let a, b, c be any real numbers, then

A + (B + C) = (A +B) + C

Let A = 5, B =9, C = 8

L.H.S = 5 + (9 + 8) = 5 + (17) = 22

R.H.S = (5 + 9) + 8 = (14) + 8 = 22

L.H.S = R.H.S

More Examples for addition:

When we change the groupings of addends, the sum does not change:

(8 + 5) + 9 = 22               (or)

8 + (5 + 9)  = 22

(9 + 1) +8 = 18                 (or)

9 + (1 + 8) = 18
Just consider that when the grouping of addends vary, the sum leftovers the same.

Example for what is the Associative property in multiplication:

Let A, B, C be any real numbers, then

A * (B * C) = (A *B) * C

Let, A = 5, B =9, C = 8

L.H.S = 5 * (9 * 8) = 5 * (72) = 360

R.H.S = (5 * 9) * 8 = (45) * 8 = 360

L.H.S = R.H.S

More Examples for Multiplication:

When we change the groupings of factors, the product does not change:

(1 x 9) x 5 = 45          or

1 x (9 x 5) = 45

(2 * 5) * 10 = 100      or

2 * (5 * 10) = 100

Just remember that when the grouping of factors changes, the product remains the same.

Simple Math Multiplication

In mathematics multiplication is used to multiple the numeric values or numbers. Multiplication can also be visualize as counting objects set in a rectangle or as result the area of a four-sided figure whose sides have given lengths. The area of a four-sided figure is not depend on which side is calculated first which illustrates that the order in which values are multiply as one does not matter

Multiplication examples:

Example 1:12 X 8

Step 1: Multiply the 1 of the 12 by the
number we are multiplying by, in this case 7. So 1 X 8 = 8.

Step 2: Multiply this 8 by 10 giving 80.

Step 3: Now multiply the 8 by the 2 of twelve giving 16. Add this to 80 giving 84.

Therefore 8 X 12 = 96

Example 2: 18 X 12

Step 1: Multiply the 1 of the 12 by the
number we are multiplying by, in this case 18. So 1 X 8 =1 8.

Step 2: Multiply this 18 by 10 giving 180.

Step 3: Now multiply the 18 by the 2 of twelve giving 36. Add this to 180 giving 216.

Therefore 18 X 12 = 216

Example 3 :12 X 9

Step 1: Multiply the 1 of the 12 by the
number we are multiplying by, in this case 7. So 1 X 9 = 9.

Step 2: Multiply this 9 by 10 giving 90.

Step 3: Now multiply the 9 by the 2 of twelve giving 18. Add this to 70 giving 108.

Therefore 9 X 12 = 108

Sample examples:

Example 4: 19 X 12

Step 1: Multiply the 1 of the 12 by the
number we are multiplying by, in this case 17. So 1 X 9 =1 9.

Step 2: Multiply this 19 by 10 giving 190.

Step 3: Now multiply the 19 by the 2 of twelve giving 34. Add this to 170 giving 228.

Therefore 19 X 12 = 228

Example 5 :23 X 7

Step 1: Multiply the 1 of the 12 by the
number we are multiplying by, in this case 7. So, 1 X 7 = 7.

Step 2: Multiply this 7 by 10 giving 70.

Step 3: Now multiply the 7 by the 2 of twelve giving 14. Add this to 70 giving 161.

Therefore 7 X 23 = 161

Example 6: 20 X 12

Step 1: Multiply the 1 of the 12 by the
number we are multiplying by, in this case 20. So 2 X 10 = 20.

Step 2: Multiply this 20 by 10 giving 170.

Step 3: Now multiply the 17 by the 2 of twelve giving 34. Add this to 170 giving 240.

Therefore 17 X 12 = 240

Example 7 :12 X 10

Step 1: Multiply the 1 of the 12 by the

number we are multiplying by, in this case 7. So 1 X 10 = 10.

Step 2: Multiply this 12 by 10 giving 120.

Step 3: Now multiply the 10 by the 2 of twelve giving 20. Add this to 70 giving 84.

Therefore 12 X 10 = 120.

Example 8: 30 X 12

Step 1: Multiply the 1 of the 12 by the
number we are multiplying by, in this case 17. So 1 X 30 =30.

Step 2: Multiply this 30 by 12 giving 360.

Step 3: Now multiply the 30 by the 2 of twelve giving 60. Add this to 170 giving 360.

Therefore 17 X 12 = 360.

Complex Analysis Problems

Complex analysis is the functions of the complex numbers. Complex numbers have the real and also the imaginary parts. The complex numbers can be represented as x +i y, where x denoted as the real part and y denoted as the imaginary part. The complex numbers comprises the addition of two complex numbers, subtraction of two complex numbers, multiplication of two complex numbers and also the division of two complex numbers. The conception of the complex numbers is the reflection of the fact. This article has the functions of the complex numbers.

Examples for learn complex analysis problems:

Example 1 to learn complex analysis problems:

Compute the value for the complex number f (x) = (75+42i) + (90+72i).

Solution:

The given complex number is f (x) = (75+42i) + (90+72i).

Step 1: (75+42i) + (90+72i) = (75+90) + (42i + 72i)

Step 2: (75+42i) + (90+72i) = 165 + (42i + 72i)

Step 3: (75+42i) + (90+72i) = 165 +114i

The value for the complex number f (x) = (75+42i) + (90+72i) is f (x) = 165 +114i.

Example 2 to learn complex analysis problems:

Resolve the value for the complex numbers f (x) = (13+i) x (11-i).

Solution:

The given complex numbers are f (x) = (13+i) x (11-i).

Step 1: (13+i) x (11-i) = 13(11 - i) + i (11 -i)

Step 2: (13+i) x (11-i) = (143- 13 i) + (11i - i2)

Step 3: (13+i) x (11-i) = 143 - 13i + 11i - (-1) (where i2= -1)

Step 4: (13+i) x (11-i) = (143+1) + (-13i +11i)

Step 5: (13+i) x (11-i) = 144 - 2i

The value for the complex number f (x) = (13+i) x (11-i) is f (x) = 144 - 2i.

Example 3 to learn complex analysis problems:

Calculate the value for the complex number f (x) = (180+53i) - (82+ 7i).

Solution:

The given complex number is f (x) = (180+53i) - (82+ 7i).

Step 1: (180+53i) - (82+ 7i) = (180-82) + (53i - 7i)

Step 2: (180+53i) - (82+ 7i) = 98 +46i

The value for the complex number f (x) = (180+53i) - (82+ 7i) is f (x) = 98 +46i.

Practice problem for learn complex analysis problems:

Compute the value for the complex numbers f (x) = (45+29i)-(2+7i).

Answer: f (x) = 43- 22i

Predict the value for the complex numbers f (x) = (145+88i) + (33+39i).

Answer: f (x) = 178+127i

Compute the value for the complex numbers f (x) = (5+3i) x (2+i).

Answer: f (x) = 7+11i

Calculate Ratio Math

In mathematics, a ratio expresses the magnitude of quantities relative to each other. Specifically, the ratio of two quantities indicates how many times the first quantity is contained in the second and may be expressed algebraically as their quotient.

Example:

For every Spoon of sugar, you need 2 spoons of flour (1:2)

Source: Wikipedia

Definition- calculate ratio math:

The ration of two numbers r and s(s≠0) is the section of the numbers. The numbers r and s are called the terms of the ratio.

Concept - calculate ratio math:

The numeric ratio of two numbers r and s(s≠0) is the section of the numbers. The numbers r and s are called the conditions of the numeric ratio.

Types of ratio- calculate ratio math:

Compounded ratio in math.

Duplicate ratio in math.

Triplicate ratio in math.

Define compounded ratio in math:

Regular format for compounded numeric ratio is

p/q *r/s=(pr)/(qs)

Example for compounded ratio in math:

How to calculate ratios: 7/5*3/3

Solution:

=7/5*3/3

=(7*3)/(5*3)

=21/15

=7/5 or 7:5

Define duplicate ratio in math:

Regular format for duplicate ratio is

(r/s)*(r/s)=(r^2)/(s^2)

Example for duplicate ratio in math:

How to calculate ratios: 6/5*6/5

Solution:

=6/5*6/5

=(6^2)/(5^2)

=36/25 or 36:25

Define triplicate ratio in math:

Regular format for triplicates ratio is

r/sr/sr/s=(r^3)/(s^3)

Example for triplicate ratios in math:

How to calculate ratios: 5/7 * 5/7 * 5/7

Solution:

=(5*5*5)/(7*7*7)

=(5^3)/(7^3) or 5^3:7^3

=125/343   or 125:343

More about ratios- calculate ratio math:

Define Inverse ratio in math:

It is often want to estimate the numbers of a ratio in the inverse order. To perform this, we simply switch over the numerator and the denominator. Therefore, the inverse of 21:7 is 7:21. When the terms of a ratio are switch over are called  the INVERSE NUMERIC RATIO results.

Example problem for inverse ratios in math:

In math, how to calculate inverse numeric ratios:  7/21*7/2

Solution:

=7/21*7/2    (7 Divided by both the numerator and denominator)

=(7*7)/(21*2)

=49/42

=7/6 or 7:6

Answer is 7/6 or 7:6

Function Examples Math

Let f be a function from X to Y.  Then it have to satisfy the following two conditions.

(i) No two different ordered pairs in f have the same first element.

(ii) All the elements of X will occur as the first elements in f.

Now let us discuss few examples that satisfies these conditions.


Example problems on functions:

Ex 1: Check whether the following ordered pair represents a function:

Here X = {1, 2, 3, 4}

Y = {a, b, c}

(i) {(1, a), (2, b), (3, c), (4, c)}.

(ii) {(1, a), (2, c), (3, b)}.

Sol: Given: (i) {(1, a), (2, b), (3, c), (4, c)}

Here all the first values are from x.  All the first values are different.  Therefore it is a function.

(ii) {(1, a), (2, c), (3, b)}.

Here 4 of X is not there in the ordered pair.  Therefore, this cannot be a function.

Ex 2: Let X = {7, 8} and Y = {1, 2}. Which one of the following is not a function of

f from X to Y.

(i) {(7, 1), (8, 2), (7, 2)}

(ii) {(7, 1), (8, 2)}

(iii) {(7, 1), (8, 1)}

Sol: (i) {(7, 1), (8, 2), (7, 2)}.  This is not a function from X to Y.  Because 7 is repeated in the first element.

(ii) {(7, 1), (8, 2)}.  This satisfies the definition of the function. Hence it is a function.

(iii) {(7, 1), (8, 1)}. This satisfies the definition of the function.  Hence it is a function.


More example problems on functions:

Ex 3: Let X = { 1,2,3,4,5,6,7,8,9} and  Y = {3,6,9}

R be a relation “is three times of”, find R on X.

Can this be a function {(3, 1), (6, 2), (9, 3)}?

Sol: R = {(3, 1), (6, 2), (9, 3)}.

Since here {(3, 1), (6, 2), (9, 3)} the first values are different, this can be a function from Y to X.

Continuous Probability Distribution

In statististics if we have the finite number of set and the probability of the set is called Discrete probability distribution. For an example for a cars on a road, deaths by cancer, tosses until a die shows a first 6. If we measure anything it is called as continuous probability distribution function. For an example we can to say the electric voltage, rainfall, and hardness of steel. In both cases we can determined the distribution by the following distribution function

F(x) = P(X ≤ x)

This is the probability that X will assume any value not exceeding x.

Continuous random variables and probability distribution:

Discrete random variables appear in experiments which is having finite set (defectives in a production, days of sun shines in Chicago, customers standing in a line etc.). Continuous random variables is appear in the experiments which we can't count but we can measure (length of screws, voltage in a power etc.). If we are find the the value probability for  the continuous  variable then it is called continuous probability distribution. By the definition of a random variable of X and its for the distribution are of continuous type will be defined as the following integral.

F(x) = `int_-oo^xf(v)dv`

it is called density of the distribution, is non negative, and is continuous perhaps except for finitely many x-values.

The continuous random variables are simpler than the discrete ones with respect to intervals. Indeed  in the continuous case the continuous probability distribution of the four probabilities corresponding to a < X ≤ b, a < X < b, a ≤ X < b, a ≤ X ≤ b with any fixed a and b (> a) are all same.

Sample problem for probability distribution:

Continuous probability distribution problem 1:

Let X have the density function f(x) = .75(1 - x2) if  -1 ≤ x ≤ 1 and zero otherwise. Find the value of distribution function. Find the continuous probability distribution of P(-1/2 ≤ X ≤ 1/2). Find x such that P(X ≤ x) = 0.95.

Solution:

F(x) = `int_-1^x`(1 - v2) dv= 0.5 + 0.75 x2 -.25x3              if  -1 ≤ x ≤ 1,

And F(x) = 1 if x > 1

P(-1/2 ≤ x ≤ 1/2) = F(1/2)  - F(-1/2) =  `int_(-1/2)^(1/2)`(0.5 + 0.75 x2 -.25x3) dx =  68.75%

Because for continuous probability distribution    P(-1/2 ≤ x ≤ 1/2) =    P(-1/2 < x ≤ 1/2)

Continuous probability distribution P(X = x) = F(x) = 0.5 + 0.75x -1.25x3  = 0.95

If we solve this we will get x = 0.73

Domain and Range of Trigonometric Functions

Domain of sine and cosine is set of all real numbers means it is defined for all values which is real

and domain of these function is [-1,1] means that whatever we  is put from domain ,we always get value with in it

 Domain of Tangent and Secant   Since tan ?=sin?/cos?  and sec?=1/cos?

,cos? being in denominator tan? and sec?  is not defined when cos? iszero

that is ? being odd multiple of ?/2 R- (2n + 1)/2| n`in`  Z

that is these are not defined at odd multiple of pi /2

and range of these functions is set of  all real numbers

Domain of cotangent and cosecant  since cot?= cos?/sin? and cosec?=1/sin?

sin? being in the denominator ,cot ? and cosec? is not defined when sin? is zero

that is  ? being multiple of ?

therefore for cot and cosec to be defined multiple of ? are dicarded

and range of these functions is all real numbers except interval  [-1,1]

Trignometric functions domain and range

Domain and Range of Trigonometric Functions




Explanation for domain and range of some trignometric functions

Domain of sine and cosine is set of all real numbers means it is defined for all values which is real

and domain of these function is [-1,1] means that whatever we  is put from domain ,we always get value with in it

 Domain of Tangent and Secant is R- {`pi` (2n + 1)/2| n `in`  Z } that is these are not defined at odd multiple of `pi`/2

and range of these functions is all values except (-1,1) interval

same can be explained  for cosecant and cotangent.

Learn Divisibility Tests

Divisibility tests means that to discover a number is divisible by a particular number (divisor) without perform the division operation. The rules shown below are used to translate a given number into a normally smaller number while preserve divisibility by the divisor. There are divisibility test for numbers in any radix, and they are all dissimilar, we represent rules simply for decimal numbers below. In this article we shall learn about divisibility tests rules with examples.

Divisibility tests rules

To learn a number is divisible by 2:

The number is end with even number (that is 0,2,4,6) the number are divisible by 2.

To learn a number is divisible by 3:

The sum of the digit is multiple of 3. The given number is divisible by 3.

To learn a number is divisible by 4:

The numbers are formed by the final pairs of digits is divisible by 4.

Example:

4672 => 72 ÷ 4 = 18

So 4672 ÷ 4 = 1168

A number is divisible by 8:

The numbers are formed by the final three digits are equally divisible by 8.

Example:

53,104 => 104 ÷ 8 = 13

So 53,104 ÷ 8 = 6638

To learn a number is divisible by 9:

The sum of the digits is a multiple of 9.

Example: 3,726

3,726 => 3 + 7 + 2 + 6 = 18

Since 18 = 9 × 2,

Then 3,726 ÷ 9 = 414

To learn a number is divisible by 6:

The numbers satisfy the rule for 2 and 3; that is, first it must be an even number, then a digit sum is a multiple of 3.

To learn a number is divisible by 12:

The numbers satisfy the rules for 3 and 4; that is, the digit sum is a multiple of 3, and its final digit pair is a multiple of 4.

Divisible tests practice problem

Problem1:

Check whether the given number 4652 is divisible by 4

Answer:

4652 => 52 ÷ 4 = 18

So 4672 ÷ 4 = 1163 therefore the given number is divisible by 4.

Problem2:

Check whether the given number 53112 is divisible by 8

Answer:

53,112 => 112 ÷ 8 = 14

So 53,112 ÷ 8 = 6639 therefore the given number is divisible by 8.

Vertical Stretch and Compression

Vertical stretch and compression mean transforming the graph based on the scale factors. Here we will see how we are performing the stretching and compression of graph using the scale factor. Stretching the graph is nothing but we are transforming the graph away from axis. Compression of the graph means squeezing the given graph towards the axis. We will see some example problems foe vertical stretch and compression of the graphs.
Vertical Stretch:

Vertical stretch is nothing but the stretching the graph away from x – axis. If the given function is f(x) then the vertical stretch of the given function is y = a f(x). Where 0 < a < 1

Example for vertical stretch:

Graph the following function and its vertical stretch. Where f(x) = x2 – 1.6x and the vertical stretch scale factor of the function f (x) is 0.4.

Solution:

Given function f(x) = x2 – 1.6x

We can write the given functions like y = x2 – 1.6x

If we want to graph the function we have to find the x and y intercept of the original function.

For x – intercept of the given function

We have to plug y = 0

So 0 = x2 – 1.6x

x (x – 1.6) = 0

x = 0 and x = 1.6

So the x intercept point is (0, 0) and (1.6, 0)

For y – intercept of the given function

We have to plug x = 0

So y = (x)2 – 1.6(0) = 0

So y intercept point is (0, 0)

Now we have to graph the vertical stretch function.

The vertical stretch function is

f(x) = a f (x)

So y = 0.4 (x2 – 1.6x)

If we graph both the function we will get the following graph

vertical stretch - stretch graph
Vertical Compression:

Vertical compression is nothing but the squeezing the graph towards the x – axis. If the given function is f(x) then the vertical compression of the given function is y = af(x). Where a > 1

Example for vertical compression:

Graph the following function and its vertical compression. Where f(x) = x2 – 3.5x and the vertical compression scale factor of the function f (x) is 1.25

Solution:

The given function f(x) = x2 – 3.5x

We can write the given functions like y = x2 – 3.5x

If we want to graph the function we have to find the x and y intercept of the original function.

For x – intercept of the original function

We have to plug y = 0

So 0 = x2 – 3.5x

x (x – 3.5) = 0

x = 0 and x = 3.5

So the x intercept point is (0, 0) and (3.5, 0)

For y – intercept of the original function

We have to plug x = 0

So y = (x)2 – 3.5(0) = 0

So y intercept point is (0, 0)

Now we have to graph the vertical compression function.

The vertical compression function is

f(x) = a f (x)

So y = 1.25 (x2 – 3.5x)

Solve Vector Spherical Coordinates

This article is to solve vector spherical coordinates. Solve vector spherical coordinates is nothing but solving problems under spherical coordinate with the help of the tutors. Tutor vista tutor have many highly qualified tutors in online. A quantity which has magnitude and direction is a vector. There are three methods  to describe a vector.  Specific lengths, direction, angles, projections or components are the methods. Below we can see about solve vector spherical coordinates.

Solve Vector Spherical Coordinates

From this system any point as the point of intersection of the spherical surface (radius r = constant) conical surface (theta phitheta and phi by dr, rd theta and dphi .The element sides are dr,rd theta and rsin theta dphi . = constant). A differential volume element is obtained in spherical coordinate. This happens by increasing r value,

The differential length dl = sqrt(dr^2+(rd theta)^2+(rsin theta dphi)^2)

The differential areas ds = dr.rd theta = rdrd theta

= dr.rsin theta dphi = r sin theta dphi dr

= rd theta. rsin theta dphi = r2sin2 theta d theta dphi

The differential volume dv = dr.rd theta.rsin theta dphi

dv = r2sin theta dphi dr.

So spherical co ordinates = (r, theta, phi)
Solve Vector Spherical Coordinates

A vector in cartesian co-ordinate system can be converted to spherical coordinate.

Conversion of  cartesian to spherical system

The Cartesian co-ordinates is (x,y,z). This cartesian coordinate is converted into spherical co-ordinates ( r, theta, phi )

Given                                  Transform

x                                     r = sqrt (x2+y2+z2)

y                                     theta = cos-1(z/(sqrt(x2+y2+z2))) = cos-1(z/r)

z                                    phi = tan-1(y/x)

Problem 1: Convert the cartesian coordinates x = 2, y = 1, z = 3 into spherical co-ordinates.

Given                        Transform

x = 2                           r = sqrt (x2+y2+z2)

= sqrt (4+1+9)

= sqrt (14) = 3.74

y = 1                            cos-1(z/r) =  cos-1(3/sqrt(14)) =  36.7°

z = 3                           phi = tan-1(y/x) tan-1(1/2)26.56°

Spherical co- ordinates are (3.74, 36.7°, 26.56°)

Information for Line Segment

Information for line segment, the division of a line with two end points is called a line segment. Line segment FG which we denoted by the symbol `bar(FG)` .

Note: We shell denote a line segment `bar(FG)` by FG only.

From the above figure, we call it a line segment FG. The points F and G are called end-points of the line segment FG.

We can also name it as line segment FG.

A line segments:

(a) A line segment has a definite length.

(b) A line segment has two end-points
Information for Drawing a Line Segment by Using Ruler:

Example:

Describe the information about to draw a line segment FG of length 11.5 cm?

Solution:

Given:

length of  line segment FG = 11.5 cm.

Steps to draw a line segment:

1. Place the ruler on a drawing paper.

2. Mark a point at the zero of the ruler. Here, we mark point F at the zero of the ruler.

3. Count the divisions in the ruler till we reach the required length. Here, we count the divisions in the ruler till we reach 11.5 cm.

4. Mark the required end point. Here, we mark the end point G.

5. Join points F and G.

information for line segment

Therefore, FG is the required line segment of length 11.5 cm. The required line segment is represented by `bar(FG)`
Information for Line Segment on a Polygon:
Describe the information to find the line segments of the given polygon. The polygon shown below figure,

information for line segment

Solution:

Given:

Polygon Triangle  EFG.

To find the line segments on a polygon:

We know that the line segments are consisting of two end points. Here, the triangle has three end points, such as E, F, and F. In the given polygon, the three end points to form the line segments on a triangle by connecting these end points consecutively, such line segments are EF, FG, and GE. These line segments are represented by `bar(EF)` , `bar(FG)` , and `bar(GE)` . Therefore, the given polygon triangle has three-line segments.

Simple Data Format

Any collection of information that makes a form to giving the required information is called data. Simple data format are used to compare the collection of data. Graphs are helping to analysis the various types of data. For example data analysis is statistics, bar graphs, histogram graphs, pie charts, and line graphs are used to simply analysis the data. In this article, we are going to discuss about simple data format with suitable example problems.
Example Problem for Simple Data Format:

1. Analyze the statistical data using the given data, find the mean, mode, median, and range of the given data.

14, 7, 13, 12, 10, 10, and 11

Solution:

Rearrange the data for ascending order.

7, 10, 10, 11, 12, 13, 14

Mean:

Mean is the sum of data divided by the number of data in the given data.

Mean = `("sum of data")/ ("number of data")`

Mean = `77/7`

= 11

Mode:

Mode is the most common value in given data.

Most common value is 10.

Therefore, mode is 10.

Median:

Median is a middle number else we have two middle value means; we find the average of their two values is median. We need to find it, we should make to arranged in ascending order from the given data.

Median:

In this problem we have a middle number.

Therefore, Median is the middle number 11.

Range:

Range is difference between maximum and least values in the given data.

Range = maximum value – minimum value

= 14 - 7.

Range = 7.
More Example Problem for Simple Data Format:

2. Analyze the data and create histogram the given simple data.

Solution:

histogram

The above histogram analyzed the given data that which one is high range and low range. This is the needed histogram.

Explanation:

Step 1:

First represented, the class intervals in the x-axis with the value 0 to 10, 10 to 20, 20 to 30, 30 to 40, and 40 to 50

Step 2:

Then represented the frequencies in the y-axis, which ranged as 250, 100, 50, 175, and 75

Given frequencies are represents to each area of rectangle in the histogram.

Step 3:

From the table values according to the class intervals and the frequencies are drawn. This is the required histogram for the simple data.