Equivalent Scale Factor

A scale area is a number which scales, or multiplies, some quantity. In the equation y=Cx, C is the scale factor for x. C is also the coefficient of x, and may be called the constant of proportionality of y to x. For example, doubling distances corresponds to a equivalent scale factor of 2 for distance, while cutting a cake in half results in pieces with a scale factor of ½.
SOURCE: WIKIPEDIA

Example problems of equivalent scale factor:

Equivalent scale factor problem 1:
Find the scale factor to the following figure:

Equivalent scale factor solution:
If we want to find the length of smaller rectangle then we can multiply the length of the one side of larger rectangle and the value of scale factor.
We can find the scale factor of the given rectangles by using the following formula,
Let D_l be the dimensions of larger rectangle and D_s be the dimensions of smaller rectangle and s be the scale factor.
Therefore the formula as,
                D_l*s=D_s
Substitute the values of dimensions into the above formula. Then we get
                  30*s=24
Divide by the value 30 on both sides,
                   `(30s)/30` `=` `24/30`
                       ` s ` `=` `24/30`
Divide by the value of 6 on both numerator and denominator. Then we get the value of scale factor.
                        s = `4/5 ` or 4:5
Therefore scale factor of smaller to larger rectangle= 4:5
Answer: 4:5
Equivalent scale factor problem 2:
Find the larger to smaller scale factor for the following figure:

Equivalent scale factor solution:
If we want to find the length of smaller rectangle then we can multiply the length of the one side of larger rectangle and the value of scale factor.
We can find the scale factor of the given rectangles by using the following formula,
Let D_l be the dimensions of larger rectangle and D_s be the dimensions of smaller rectangle and s be the scale factor.
Therefore the formula as,
                D_l*s=D_s
Substitute the values of dimensions into the above formula. Then we get
                  39*s=26
Divide by the value 48 on both sides,
`(39s)/(30)` `=` `26/39`
` s ``=` `26/39`
Divide by the value of 13 on both numerator and denominator. Then we get the value of scale factor.
                        s = `2/3` or 2:3
Therefore scale factor of larger to smaller scale rectangle= 3:2
Answer: 3:2

Algebra is widely used in day to day activities watch out for my forthcoming posts on Proof of Fundamental Theorem of Calculus and algebra 2 solver step by step. I am sure they will be helpful.

Practice problems of equivalent scale factor:

1. Find the scale factor to the following figure:

2. Find the larger to smaller scale factor for the following figure:

Answer:
1. 3:4
2: 6:5

Mode and Median Calculator

Mode: Mode is the value that takes place most repeatedly in the data set. Measure of central tendency is known as mode. If the data’s are given in the form of a frequency table, the class corresponding to the maximum frequency is called the modal class. The value of the variate of the modal class is the mode.

Median: The median is the middle value when the given values are arranged in an ascending order. Let us see the median and mode calculator.

Median and Mode calculator:

In the calculator enter the set of values in first box, after that clcik the median button it will automatically calculate the median value and it will be displayed in answer box. The same process is done for mode.

Examples on Mode calculator:

Example 1:

            Find the mode of 7, 4, 5, 1, 7, 3, 4, 6, and 7.

Solution:

           The above question is entered in the first box. The calculator doing the follwing process,

           Assemble the data in the ascending order, we get

            1, 3, 4, 4, 5, 6, 7, 7, 7.

            The number 7 occurs many times in the above values.
            Mode = 7 will display the answer box after press the mode button on calculator.
Example 2:

            Find the mode for 12, 15, 11, 12, 19, 15, 24, 27, 20, 12, 19, and 15.

Solution:

           The above question is entered in the first box. The calculator doing the follwing process,

           Assemble the data in the ascending order, we get

            11, 12, 12, 12, 15, 15, 15, 19, 19, 20, 24, 27.

            In the above values 12 occurs 3 times and 15 also occurs 3 times.

            ∴ Both 12 and 15 are the modes for the given data. We observe that there are two modes for the given data.The Mode will be displayed in answer box on calculator
Example 3:

            Find the mode of 19, 20, 21, 24, 27, and 30.

Solution:
            Already the above data are in the ascending order. Each value occurs exactly one time in the series. Hence there is no mode in the above given data.
These are the examples on mode calculator.

Examples on Median calculator:

Example 1:    
            Find the median of the following numbers: 12, 45, 62,10,14,31 and 43.
Solution:
           The above question is entered in the first box. The calculator doing the fololwing process,
            Arranging the given numbers in ascending order we get
            10, 12, 14, 31, 43, 45 and 62.
                            `darr`
                    Middle term
            Median = Middle item = 31.     

         The median 31 will display the answer box

Between, if you have problem on these topics Cubic Equation  please browse expert math related websites for more help on ibsat 2013.

Example 2:         
            Find the median of the following numbers: 3, 7, 4, 10, 22, 16, 21 and 5.
Solution:
            The above question is entered in the first box. The calculator doing the following process,
            Arranging the given numbers in ascending order we get
            3, 4, 5, 7, `darr` 10, 16, 21, 22              
                 Median is here
            Median = Item midway between 7 and 10
                       =` (7 + 10) / 2` = `17 / 2` = 8.5
          The Median 8.5 will display on answer box on calculator
These are the examples on mode and median calculator.

Sine Geometry

Trigonometry is the division of geometry dealing among relationships among the sides also angles of triangles. In geometry sine of an angle is the ratio of the length of the opposite side to the length of the hypotenuse. The  ratio does not depend on size of the particular right triangle chosen, as long as it contains the angle A, since all such triangles are similar

(Source: Wikipedia)

I like to share this sine curves with you all through my article. 

Sine geometry

In the above diagram ,
sin A =opposite/hypotenuse
Examples for sine geometry

A - Right angle of the triangle ABC.
The length of AB, BC and CA are frequently represented through c, a, b.
Obtain point B as middle of a trigonometric circle
Circle with radius = 1.
Now sin (B) are comparative to b, c also a.
sin `(B)/b` =`1/a`
sin (B) = `b/a`

Examples for sine geometry

Example 1
Angle of triangle is 20⁰, opposite side of triangle is 12 apply the sine geometry to find the unidentified side of the triangle?
Solution:

Angle of triangle= 20⁰  

Opposite side of triangle = 12.

sin A =opposite/hypotenuse

sin 20⁰ = `12/x`

sin 20⁰ x = 12

x = `12/sin 20^0`

x =`12/0.3420`    {since the value of sin 20 degree is 0.3420}

x=35.08

Hypotenuse side= 35.08

Example 2
Angle of triangle is 78⁰, hypotenuse side of triangle is 20 apply the sine geometry to find the unidentified side of the triangle?
Solution:

Angle of triangle= 78⁰  

Hypotenuse side of triangle = 20.

sin A =opposite/hypotenuse

sin 78⁰ = `x/20`

sin 78⁰ x 20= x

x = sin 78⁰ x20

x =0.97814x20     {since the value of sin 78 degree is 0.97814}

x=35.08

Opposite side = 19.56

Example 3
If hypotenuse side of triangle 40 and opposite side of triangle 20 find the sine angle?
Solution:

Hypotenuse side of triangle = 40.

Opposite side =20

sin A =opposite/hypotenuse

sin A= `20/40`

sin A =` 1/2`

sin A = 0.5    {sin 30 degree is 0.5}

Therefore the angle is 30 degree

Interval Estimates

Interval Estimate:

• Interval estimation is the process of calculate the interval for possible value of unknown parameter in the population.
• It is calculate in the use of sample data and contrast to the point estimation. It is different from the point estimation. It is the outcome of a statistical analysis.

The most common forms of interval estimations as follows:

• A frequents Method or Confidence interval
• A Bayesian method or credible intervals

The other common methods for interval estimations are

• Tolerance interval
• Prediction interval

And another one is known as the fiducial inference.

Construction of interval estimates parameter:

The normal form of interval estimate of the population parameter is,

• Point estimate of parameter and
• Plus or minus margin of error

Margin of error:

• The amount which is subtracted or added from  the point estimate  of the statistic and produce the parameter interval  estimate is known as the margin of error.
• The margin of error size depends on the following factors:
• Sampling distribution type of sample statistics.
• Area under sampling distribution percentage   that includes the researchers      decision.Usually we consider the confident level as 90%, 95%, 99%.
• The interval of each interval estimates are constructed in the region of the point estimate with its confident level.

My forthcoming post is on Set Interval Notation will give you more understanding about Algebra.

Construction of Interval estimate for Population mean

• Take the point estimate of μ  that is  the sample mean`vecx`
• Define  the mean distribution for the sample.When the  value of n is large we  have to use the central limit  theorem. And   is the normal distribution with the,

                      standard deviation `sigma``vecx` = `sigma/sqrt(n)`  

                      and mean μ.

• Choose the most common confident  level as 95%
• Find the margin of  error  which is related with the confidence level.
• The area  under the curve of  the sample means the normal distribution contains the 95%  of the interval from.

                               z= -1.96 to z= 1.96 

• The interval estimate for 95 % is,   

                            `vecx`- 1.96 (`sigma/sqrt(n)` ) to `vecx``sigma/sqrt(n)`

Common Factors

The common factors of two or more whole digits is the biggest whole digit that equally divides all the whole digits. There are two methods to find common factors in math.

The initial method is to list all the factors of each digit. Then decide the biggest factor.

For Example:

Find the common factors of 12 and 18.

The factors of 12 are 1, 2, 3, 4, 6 and 8.

The factors of 18 are 1, 2, 3, 6, 9 and 18.

The common factors of 12 and 18 are 1, 2 and 3.

Methods for Finding Common Factors

There are two methods to find the common factors of numbers. Listed below are the steps to be followed in finding common factors.
Method I:

• List all the factors of the numbers.
• Collect the common factors among each digit.

Method II:

• Find the prime factors of each digit.
• Merge the common terms of each digit.

Examples

Listed Below are some of the examples in finding the common factors.
Example 1:
What are the Common factors of 34 and 36?
Solution:
The factors of 34 are 1, 2, 17, and 34.
The factors of 36 are 1, 2, 3, 4, 6, 9, 12, 18, and 36.
Now, using method 1,
The common factors to the two numbers are 1 and 2.
The Common factors of 34 and 36 are 1 and 2.
Example 2:
What are the common factors of 40, 45 and 50?
Solution:
Prime factors of 40 are 1, 2, 4, 5, 8, 10, 20, and 40.
Prime factors of 45 are 1, 3, 5, 9, 15 and 45
Prime factors of 50 are 1, 2, 5, 10, 25, and 50.
Now, using method 2,
The common factors of 40, 45 and 50 are 1, 2 and 5.

Practice Problems

Listed below are some of the practice problems in finding the common factors.

Problem 1:

What are the Common factors of 8, 14, 18 and 22?

Answer:    

8   `->` 1, 2, 4, and 8

14 `->` 1, 2, 7 and 14

18 `->` 1, 2, 3, 6, 9 and 18

22 `->` 1, 2, 11 and 22

My forthcoming post is on Series Solutions of Differential Equations will give you more understanding about Algebra.

Problem 2:

What are the Common factors of 15, 30, 45 and 60?

Answer:    

15 `->` 1, 3, 5 and 15

30 `->` 1, 2, 3, 5, 6, 10, 15 and 30

45 `->` 1, 3, 5, 9, 15 and 45

60 `->` 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, and 60

Review of Word Problems

Introduction on review of word problems are first used to identify the variables present in the given problem. Also the word problems are used for the unit numbers and unit variables. Word problems are used for finding the phrases. The consideration includes in this is, we have to analyse the problem present in the statements.

I like to share this Fun Math Problems with you all through my article.

Types of word problems:

There are many types review word problems present. They are,
1.Word problem for numbers
2.Word problem for mixtures
3.Word problem for Age
4.Word problem for time
5.Word problem for linear
Word problem for numbers:
          In word problem for numbers, we  first review the relationships are identified. The variables present in the word problems are reduced.
Word problem for mixtures:
       Different types of concentrations are included in this word problem mixtures.
Word problem for age:
       The relationship of ages are calculated on this word problem.
Word problem for time:
     In this word problem, we review  the various types of train problems are included. Time taken for a train to go and coming back are included in this word problem.
Word problem for linear:
      Linear word problem mostly review the cost problems. For example, the cost for an grapes and the apple is 14.etc..





Algebra is widely used in day to day activities watch out for my forthcoming posts on find the prime factorization of 125 and cbse neet 2013. I am sure they will be helpful.

Example for review word problem:

Question: 1. The difference of twice a number 4 is 8. What is that number?
Solution:
Step 1: First we are changing the above said sentence as difference of twice a number 4 equals 8.
Step 2: Next step is to write the above said equation in the equation form.
     2V-4=8.
This is the solution for a word problem.
Question 2. The difference of twice a number 2 is 4. What is that number?
Solution:
Step 1: First we are changing the above said sentence as difference of twice a number 2 equals 4.
Step 2: Next step is to write the above said equation in the equation form.
     2V-2=4.
This is the solution for a word problem.
Practice to review word problem:
Practice 1.The difference of twice a number 12 is 14. What is that number?
Practice 2.The difference of twice a number 20 is 40. What is that number?

Factoring With Fractions

Factoring is the process of finding the factor of the given function. It is otherwise called as divide the given function as two separate part. We can factor the function more than two times. Fractions are also used in factoring process. Fraction means it has some numerator and denominator values. Both the numerator and denominator values are different. Now in this article, we see about ac factoring with fractions and their example problems.

Example problems for Ac factoring with fractions

Ac factoring with fractions example problem 1:

Factorize the given polynomial fraction equation `(1 / 6)` x2 - `(7 / 6)`x  - 20 = 0

Solution:

Given equation is `(1 / 6)` x2 - `(7 / 6)`x  - 20 = 0

For converting the above fraction equation in normal quadratic function,

Multiply the given equation by 6 on both the sides, we get

x2 - 7x - 120 = 0

Factorize the above equation, we get

x2 - 15x + 8x - 120 = 0

Grouping the first two terms and next two terms, we get

(x2 - 15x) + (8x - 120) = 0

x (x - 15) + 8 (x - 15) = 0

(x - 15) (x + 8) = 0

The factors are (x - 15) and (x + 8)

Answer:

The final answer is (x - 15) and (x + 8)

Ac factoring with fractions example problem 2:

Factorize the given polynomial fraction equation `(1 / 9)` x2 - `(4 / 3)` x + ` (11 / 9)` = 0

Solution:

Given equation is `(1 / 9)` x2 - `(4 / 3)` x + ` (11 / 9)` = 0

For converting the above fraction equation in normal quadratic function,

Multiply the given equation by 9 on both the sides, we get

x2 - 12x + 11 = 0

Factorize the above equation, we get

x2 - 11x - x + 11 = 0

Grouping the first two terms and next two terms, we get

(x2 - 11x) + (- x + 11) = 0

x (x - 11) - 1(x - 11) = 0

(x - 11) (x - 1) = 0

The factors are (x - 11) and (x - 1)

Answer:

The final answer is (x - 11) and (x - 1)

Ac factoring with fractions example problem 3:

Factorize the given polynomial fraction equation `(1 / 2)` x2 - `(11 / 2)` x + 14 = 0

Solution:

Given equation is `(1 / 2)` x2 - `(11 / 2)` x + 14 = 0

Multiply the given equation by 2 on both the sides, we get

x2 - 11x + 28 = 0

Factorize the above equation, we get

x2 - 7x - 4x + 28 = 0

Grouping the first two terms and next two terms, we get

(x2 - 7x) + (- 4x + 28) = 0

x (x - 7) - 4 (x - 7) = 0

(x - 4) (x - 7) = 0

The factors are (x - 4) and (x - 7)

My forthcoming post is on Rules for Dividing Integers and class 11 cbse books will give you more understanding about Algebra.

Answer:

The final answer is (x - 4) and (x - 7)