Addition of Whole Numbers

Introduction to addition of whole numbers:

One of the arithmetic operations is addition. In mathematics, addition represents to combining group of numbers together into a bigger collection. Add symbol denoted by the plus sign (+). For example, assume there are 2 + 2 objects meaning two objects and two other objects, add the form of object is four objects. Therefore, 2 + 2 = 2.

Whole numbers includes all integers. Addition of whole numbers is nothing but it is addition operation between whole numbers. For example, 3 + 2 + 5 = 10. The following are the arithmetic operation like addition, subtraction, multiplication and division performed on whole numbers.

Examples Problem for Addition of Whole Numbers:

1. Find the sum of value for given whole numbers, 2552, 5542, 7624, and 6751 using addition operation.

Solution:

Given whole numbers are 2552, 5542, 7624, and 6751,

Steps for addition of whole number,

First, arrange the numbers with placing values like ones under ones, tens under tens.
Addition operation starts with right side of the column.
Then add the each column in which sum is 9 or less than 9 means write the sum.
Else, sum is 10 or more than 10 means re-group the carry tens on the next column and write ones of that sum.Please express your views of this topic cbse class 12 question papers by commenting on blog.
2 1  --> carry

2,552

5,542

7,624

`+` 6,751

22,469

2  --> carry     1 --> carry

2 thousands +  5 hundreds +  5 tens +  2 ones

5 thousands +  5 hundreds +  4 tens +  2 ones

7 thousands +  6 hundreds +  2 tens +  4 ones

6 thousands +  7 hundreds +  5 tens +  1 ones

22 thousands +  4 hundreds +  6 tens +  9 ones

Answer for addition of whole numbers is 22469 or twenty two thousand four hundred and sixty nine.

2. Find the sum of value for given whole numbers, 6552, 9252, and 2751 using addition operation.

Solution:

Given whole numbers are 6552, 9252, and 2751,

Steps for addition of whole number,

First, arrange the numbers with placing values like ones under ones, tens under tens.
Addition operation starts with right side of the column.
Then add the each column in which sum is 9 or less than 9 means write the sum.
Else, sum is 10 or more than 10 means re-group the carry tens on the next column and write ones of that sum.
1 1 --> Carry

6,552

9,252

`+` 2,751

18,555

1 -> carry        1  --> carry

6 thousands +  5 hundreds +  5 tens +  2 ones

9 thousands +  2 hundreds +  5 tens +  2 ones

2 thousands +  7 hundreds +  5 tens +  1 ones

18 thousands + 5 hundreds +  5 tens +  5 ones

Answer for addition of whole numbers is 18555 or eighteen thousand five hundred and fifty-five.

Practice Problems for Addition of Whole Numbers:

1. Find the sum of value for given whole numbers, 52, 542, 624, and 651 using addition operation.

Answer for addition of whole numbers is 1869.

2. Find the sum of value for given whole numbers, 25, 54, 72, and 751 using addition operation.

Answer for addition of whole numbers is 902.

Frequency Tables and Histograms

Introduction of frequency tables and histogram:

Frequency distribution represent  a histogram is by means of the rectangles whose width represents class intervals and areas are directly proportional to the corresponding frequencies.” A frequency tables distribution applied for continuous quantitative data. The tabular collection of data is frequency distribution viewing how many interpretation lie above, or below the records. Histograms are helpful for interpreting tasks. The collections of groups of data called classes, and in the context of a histogram, they called as bins.

Constructing a Frequency Tables and Histogram:

Step 1 – Calculate the quantity of data
Step 2 - Review on a tally sheet
Step 3 – Calculate the range
Step 4 – Find the number of intervals
Step 5 - Calculate interval width
Step 6 - Identify interval of initial points
Step 7 - Calculate number of data in each interval
Step 8 - Plot the data
Step 9 - Add designation and tale


Explanation about Frequency Tables and Histogram:
A range of the Histogram equally separates all possible values in a data set of groups or classes. For each set of  a 4-sided figure constructed with a base length, which is identical to the collection of values in that definite group, and an area directly proportional to the number of observations decreasing into that group.

The height of a four-sided figure, which is the same to the frequency density of the interval, i.e. the frequency, should split by the breadth of the interval. The total region of the histogram that exactly is identical to the number of data.I like to share this 6th grade math homework with you all through my article.

Frequency Tables:

Marks           Frequency       Cumulative frequency

0-5                     3                 3

5-10                    9                12

10-15                  18                30

15-20                  26                56

20-25                  30                86

25-30                  38               124

i) Plot a points of coordinates such as (0,5),(5,10),(10,15),(15,20),(20,25),(25,30) and the related frequency can be identified and make the points.

ii) Take x -axis 1 cm = 5 unit, y -axis 1cm = 1 c.f.

iii) The number line represent by the horizontal axis,  which displays the data in equal interval. The vertical axis represents the frequency of each bar.

Trigonometric Functions Calculus

Introduction of Trigonometric Functions Calculus:

Trigonometry came from the Greek words ‘trigon-triangle’ and ‘metron-measure’. Trigonometric functions generally define the functions of angle. An equality that satisfies for any value of the variable is generally defined as an identity. An equation is an equality, which is true only for several values of the variable. An identity gives more clear and convenient form than the expression. Trigonometric function calculus are generally applied in modeling periodic phenomena and study of triangles.

Types of Identities in Trigonometric Functions Calculus:

There are several identities in trigonometry function calculus. They are as follows,

Reciprocal identity,
Tangent and Cotangent identity,
Pythagorean identity,
Co-function identity,
Even-Odd identity.

Example Problems for Trigonometric Function Calculus:

Ex 1:

Prove that, sec^2 a + csc^2 a = sec^2 a × csc^2 a, Using Trigonometric Functions calculus.

Proof:         L.H.S. = sec^2 a + csc^2 a

=> (1 / cos^2 a) + (1 / sin^2 a)                                    [Using Reciprocal Identity ]

=> (sin^2 a + cos^2 a) / sin^2 a × cos^2 a                   [Taking L.C.M. ]

=> 1 / sin^2 a × cos^2 a                                               [By Pythagorean Identity]

=> 1 / sin^2 a × 1 / cos^2 a                                          [Separating it Into Two Terms ]

=> csc^2 a × sec^2 a                                                   [Using Reciprocal Identity ]

=> sec^2 a × csc^2 a

=> R.H.S.

Hence proved that, sec^2 a + csc^2 a = sec^2 a × csc^2 a

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Ex 2:

Prove that, sin4 a - 2sin^2 a cos^2 a + cos4 a = cos^2 (2 a), Using Trigonometric Functions calculus.

Proof:

L.H.S. = sin4 a - 2 sin^2 a cos^2 a + cos^4 a

=> (sin^2 a)2 - 2sin^2 a cos^2 a + (cos^2 a)2

=> (sin^2 a - cos^2 a)2                                                [ Using ( a - b )2 = a2 - 2ab + b2 ]

=> ((1 - cos^2 a) - cos^2 a)2                                       [By Pythagorean identity ]

=> (1 - 2cos^2 a)2

=>( -(2cos^2 a - 1))2

=>( -cos 2a)2                                                          [ Using Formula: cos 2T = 2 cos^2T - 1 ]

=>cos^2 (2 a)

=> R.H.S.

Hence proved that, sin4 a - 2sin^2 a cos^2 a + cos^4 a = cos^2 (2 a).

Standard Deviation Dispersion

Introduction to standard deviation dispersion:

Standard Deviation can be defined as the measurement of describing the variability of given data set. Standard deviation is also used for measuring the exact value of the number in the given data set. Dispersion is defined as the measuring that’s also including the average deviation, variance, and then standard deviation. The standard deviation and the variance are most widely used for measuring the dispersion.Please express your views of this topic What is Dispersion by commenting on blog.

Formula for Standard Deviation Dispersion:

The formula for the standard deviation dispersion are given below,

`Variance = (sigma)^2 = (sum(x_r - mu)^2)/N`

The standard deviation, `(sigma)` , is the square root of the variance.

What the formula means:
(1)` x_r -mu` – take the mean value and subtract the mean value with each value given.
(2) `(x_r - mu)^2` – square the each of the results obtained from the step -1.this is to get or remove of any minus signs from the value.
(3) `(sum(x_r - mu)^2)` – this is for adding all the values that is obtained from step-2.
(4) Divide step-3 by N, which is the total number of the given values
(5) For obtaining the standard deviation square root the answer to the step-4.

Example Problems for Standard Deviation Dispersion:

Example problem 1:

Calculate the variance and also standard deviation for the following values: 1, 3, 5, 6, 6, 8, 9, and 10.

Solution:

The mean = `(1+3+5+6+6+8+9+10)/ 8 `
= `48/8`
=6

Therefore (`mu` =6)

Step 1: Subtract the mean value from given values

(1-6)=-5
(3-6)=-3
(5-6)=-1
(6-6)= 0
(6-6)= 0
(8-6)= 2
(9-6)= 3
(10-6)=4

Step 2: In this step we have to calculate the square for the mean values 25, 9, 1, 0,  0, 4, 9, 16.

Step 3: After finding square mean ,sum the square mean 25+9+1+0+0+4+9+16=64.

Step 4: Here , N = 8, therefore variance should be =` 64/ 8` = 8

Step 5: Therefore standard deviation = 2.8284.

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Example problem 2:

Calculate the variance and also standard deviation for the following values: 1, 2, 3, 4, and 5.

Solution:

The mean = `(1+2+3+4+5)/ 5`
= `15/5`
=3

Therefore (`mu` =3)

Step 1: Subtract the mean value from given values

(1-3)=-2
(2-3)=-1
(3-3)=0
(4-3)= 1
(5-3)= 2

Step 2: In this step we have to calculate the square for the mean values 4, 1, 0, 1, 2

Step 3: After finding square mean ,sum the square mean 4+1+0+1+2=8.

Step 4: Here , N = 5, therefore variance should be = `8/5` = 1.6

Step 5: Therefore standard deviation = 1.264.

Eccentricity of Conic Sections

Introduction to eccentricity of conic sections:
A conic is the locus of a point in a plane such that the ratio of its distance from a fixed point in the plane to its distance from a fixed line in the plane is a constant. Here, the fixed point is called the focus, the fixed line is called the directrix and the constant ratio is called the eccentricity of the conic. It is denoted by ‘e’.

(i)     If e = 1, the conic is called the parabola.

(ii)    If e < 1, the conic is called the ellipse

(iii)   If e > 1, the conic is called the hyperbola

(iv)   If e = 0, the conic is called the circle.

Now, let us see few problems on this topic eccentricity of conic sections.

Example Problems on Eccentricity of Conic Sections

Ex 1. Find the eccentricity of conic  x2/ 36 + y 2 / 64 = 1

Soln: Given: `x^ 2 / 64` +` y ^2 / 36 ` = 1 ………… (1)

Here (1) represents an ellipse. To get the eccentricity, are follow this formula: e 2 = 1 – `b ^2 /a^ 2` .

Here a = `sqrt 64` , b = `sqrt 36`     [This we get from the standard form of an ellipse

`x^ 2 / a^ 2` +` y ^2 / b ^2` = 1        Here a >b]

a = 8, b = 6

Therefore, e 2 = 1 – `6 ^2 / 8^ 2`

= 1 – `36 / 64`    = 1 – `9 / 16`    = 16 – `9 / 16`    = `7 / 16`

Therefore e = `sqrt7 / 16`    [Eccentricity is always positive]

Therefore e = `sqrt7 / 4`

Ex 2: Find the eccentricity of the conic:` x ^2 / 49` +` y ^2 / 225` = 1

Sol: Given: ` x^2 / 49` +` y ^2 / 225` = 1

Here a = 7, b = 15,   = a < b

Therefore e 2 = 1 – `a^ 2 / 15^ 2` = 1 – ` 49 / 225`

= `(225 ** 49) / 225`   = `176 / 225`

Therefore  e = `sqrt (176 / 225)` = `sqrt ((16 * 11) / 225)`    =  `(4sqrt 11) / 15`

Ex 3: Find the eccentricity of the conic: ` x^ 2 / 625 **y ^2 / 144`   = 1

Soln: Given:` x^ 2 / 625 ** y ^2 / 144` = 1 ………… (1)

Here (1) represents a hyperbola

The eccentricity is given by e 2 = 1 + `(b^2/ a^2)`

= 1 + `144 / 625`      =` 769 / 625`

Therefore e = `sqrt 769 / 25` .

Ex 4: Find the eccentricity of the conic` x ^2/ 25 ** y ^2 / 36` = 1

Soln: Given:` x ^2 / 25** y ^2 / 36` = 1 ………….. (1)

This is a hyperbola.  a = 5, b = 6

Therefore e 2 = 1 + `b ^2 / a^ 2` = 1 + `36 / 25` = `61 / 25`

Therefore e = `(sqrt 61) / 5`

Note: Here the fraction  b 2 /  a 2 will change as whether a < b or a >b.

Practice Problems on Eccentricity of Conic Sections

Find the eccentricity of the following conics:
A) `x^2/16 + y^2/9` = 1

[Ans: e = `sqrt 7 / 4` ]

B)    ` x ^2 / 25 + y ^2 / 64` = 1

[Ans: e = `sqrt 39 / 8` ]

2. Find the eccentricity of the following conics:

A)    `x ^2 / 64 ** y ^2 / 36` = 1

[Ans: e = `5 / 2` ]

B)    ` x ^2 / 81** y ^2 / 144` = 1

[Ans: `5/3` ]

Sample Algebra Functions

Introduction of sample algebra functions

Algebra is the branch of mathematics deals with finding unknown variable from the given expression with the help of known values. The variable of algebraic are represents alphabetic letters.In algebra the numbers consider as constants, algebraic expression may include  real number, complex number, matrices and polynomials. In algebra several identities need to find the x values by using this we can simply find  the algebraic expression of the particular function. The sample algebra functions may include in the function of f(x), p(x),… to find the x value of the algebra functions.

For example

Sample algebra functions f(x) = 2x2+3x + 4. In this sample algebra function we have  to find the function of f(2).

Sample Algebra Functions Problem

Problems using the square of the sample algebra functions

Problem1:Using  square  problem in sample algebra functions

f(x) = x2 +2x +4 find the f(4).

Solution :

Given the function of f(x) there is x value is given find the sample algebra functions of

f(x) = x2 +2x +4 find the f(4).

The value of x is4 is given

f(4) = 42 + 2*4 + 4

f(4) = 16 + 8 + 4 In this step 4 square is 16 it is calculate and 2*4 is 8 be added

f(4) = 28

Sample algebra functions of f(4) = x2 +2x +4 find the f(4) is 28.

Problem2; Using  square  problem in sample algebra problem

f(x) = x2 + 5x + 5 find the f(5).

Solution :

Given the function of f(x) there is x value is given find the sample algebra functions of

f(x) = x2 + 5x + 5 find the f(5).

The value of x is 5 is given

f(5) = 52 + 5*5 + 5

f(5) = 25 + 25 + 5 In this step 5 square is 25 it is calculate and 5*5 is 25 be added

f(5) = 55

The algebra 2 function of the f(5) = x2 + 5x + 5 find the f(5) is 55.

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Sample Algebra Functions Using the Cubic Equation

Problems 1: using the cubes equation in sample algebra functions:

f(x) = x3 + 2x2 + 2x + 4 find the sample algebra function f(2).

Solution

Given the function of  f(x) there is x value is given find the sample algebra functions

f(x) = x3 +2x2 + 2x + 4 find the f(2).

Here the value of x is given as 2

f(2) = 23 + 2*22 + 2*2 +4

f(2) = 8 + 8 + 4 + 4 In this step 23 is calculated  as 8 and 2 square is 4

f(2) = 24.

The sample algebra functions of the f(x) = x3 +2x2 + 2x + 4 find the f(2) = 24.

Problems 2: using the cubes of the sample algebra functions

f(x) = 2x3 +4x2 + 3x + 4 find the sample algebra functions of f(4).

Solution

Given the function of  f(x) there is x value is given find the sample algebra functions

f(x) = 2x3 + 4x2 + 3x + 4 find the f(4).

Here the value of x is given as 4

f(4) = 2*43 +  4*42 + 3*4 + 4

f(4) = 128 + 64 + 12 + 4 In this step 4 cubes  is calculated  as 64 and multiplied by 2 we get as 128.Then the 4 square is 16 and multiplied by 4 we get 64.

f(4) = 208

Sample algebra functions of the f(x) = 2x3 +4x2 + 3x + 4 find the f(4) = 208.

Multiplication Rule for Independent Events

Multiplication Rule for Independent Events

Probability is the likelihood of the occurrence of an event. An event is a one or more possible outcomes of a certain experiment. An event is called independent event if one event does not affect the other event. For example, choosing a 7 and 8 in the deck of card with replacement is two independent events. An event consisting of more than one simple event is called compound event.

Multiplication rule for two independent events:

If A and B are two independent event then; P(A and B) = P(A) · P(B)

Multiplication rule for three independent events:

If A, B, and B are three independent events then; P(A and B and C) = P(A) · P(B) · P(C)

Multiplication Rule for Independent Events - Example Problems

Example 1: Two coins are drawn at one by one with replacement of previous coin in a bag of 12 nickels and  9 quarters. Find the probability of first coin is nickels and second coin is quarters.

Solution:

Lest S be the sample space, n(S) = 12 + 9 = 21

A be the event of drawing a nickels, n(A) = 12

B be the event of drawing a quarters, n(B) = 9

P(A) = `(n(A))/(n(S))` = `12/21` = `4/7`

P(B) = `(n(B))/(n(S))` = `9/21 ` = `3/7`

P(A and B) = P(A) · P(B) = `4/7` · `3/7` = `12/49`

P(A and B) = `12/49`

Example 2: A jar contains 6 dark, 8 milk, and 10 white chocolates. Three chocolates are drawn one by one with replacement. What is the probability of drawing each kind is one time?

Solution:

Lest S be the sample space, n(S) = 6 + 8 + 10 = 24

A be the event of drawing a dark chocolate, n(A) = 6

B be the event of drawing a milk chocolate, n(B) = 8

C be the event of drawing a white chocolate, n(B) = 10

P(A) = `(n(A))/(n(S))` = `6/24` = `1/4`

P(B) = `(n(B))/(n(S))` = `8/24` = `1/3`

P(C) = `(n(C))/(n(S))` = `10/24` = `5/12`

P(A and B and C) = P(A) · P(B) · P(C) = `1/4` · `1/3` · `1/12` = `1/144`

P(A and B and C) = `1/144`
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Multiplication Rule for Independent Events - Practice Problems

Problem 1:   Two chocolates are drawn one by one with replacement of previous chocolate in a jar of 10 bitter and  5 dark chocolates. Find the probability of first one is bitter and dark chocolate.

Problem 2: A jar contains 11 dark, 12 milk, and 13 white chocolates. Three chocolates are drawn one by one with replacement.What is the probability of drawing each kind exactly one time.

Answer: 1) `2/9` 2) `143/3888`