Fraction Number Sentences

In math, the term sentence defines math sentence. Math sentence replaces the equations and symbols by English terms. We have to convert it into an equation and find the solution. Fraction is a ratio of numbers or variables. In math ,sentence is one of the terms may not be known and needs to be determined. We can convert any math sentence into an equation. In this lesson we will discuss the fraction number sentences .

Fraction Number Sentences – Example Problems

Solved example fraction number sentences problems.

Example 1: `3/5` of a number is equal to 5 less than `4/5` of a number. Find the number.

Solution:

Let unknown number be x.

`3/5` of a number = `(3x)/5`

5 less than 4/5 a number = `(4x)/5 - 5`

`(3x)/5 = (4x)/5 - 5`

`(3x)/5 = (4x-25)/5`

Multiply both sides by 5

3x = 4x – 25

Solve for x to get

x = 25

Example 2: Six more than `5/7` of a number is equal to 3 less than `6/7` of a number. Find the number.

Solution:

Let unknown number be x.

Six more than `5/7` of a number = `(5x)/7 + 6`

3 less than `6/7` of a number = `(6x)/7 - 3`

`(5x)/7 + 6 = (6x)/7 - 3`

`(5x + 42)/7 = (6x - 21)/7`

Multiply both sides by 7

5x + 42 = 6x – 21

Solve for x to get

x = 63

Example 3: Flozia had 75 chocolates in her bag. She sold `4/5` of them at $9 each. How much did she receive?

Solution:

First find the number of chocolates she sold.

`4/5` x 75 = 60

She sold 60 chocolates.

Find how much money she received.

60 × 9 = 540

She received $540.

Example 4: `4/7` of a group of students in a hall is boys. If there is 36 boys in a group, how many students are there in the group?

Solution:

Let x = Total students in a group

So, write an equation as follows

`(4x)/7` = 36

4x = 252

x = 63

Therefore 63 students are there in the group.

Fraction Number Sentences – Practice Problems

Solve these practice fraction number sentences problems.
Problem 1: Flora spent `2/7` of her allowance for food per month. What fraction of his allowance had he left?
Problem 2: In a box there are 25 chocolates and 35 ice creams. What part of the box are ice creams?
Answer: 1)  `5/7` 2)  ` 7/12` 

Sum of Uniform Distributions

Uniform distribution is one of  the part of probability distribution function. The simplest distribution of uniform distribution is also known as uniform random distribution. Two types of uniform distribution:

1.Discrete Uniform distribution,

2.Continues Uniform distribution.

Two types of Continuous variables distributions

Uniform distribution
Exponential distribution.

Uniform distribution:

In probability theory and statistics, the continuous uniform distribution is a family of probability distributions such that for each member of the family, all intervals of the same length on the distribution's support are equally probable.

Exponential distribution:

In probability theory and statistics, the exponential distributions are a class of continuous probability distributions. They describe the times between events in a Poisson process, Study process in which events occur continuously and independently at a constant average rate.

Uniform distribution probability density function (PDF) is:

phi(x) = {(1/ (b - a ) if a
Uniform distribution  cumulative distribution function (PDF) is:

phi(x) =(x-a)/(x-b)

Expectation or mean of uniform distribution:

Let X be a random variable. We represent the probable value (expectation or mean) of X as either or E(X). Mean describes the (probability-weighted) average value for the random variable.

Probability density functions (PDFs) for mean uniform distribution is denoted by two random variables. The mean of each are indicated on the x-axes

If X is discrete random variable means, we define its expectation as

E(x) = sum x phi(x)

Where phi is the probability function.

If X is continuous random variable means, we define its expectation as

E(x) = int_-oo^ooxphi(x)dx

Where phi is the probability density function.

Standard deviation  uniform distribution:

The variance of a random variable X, denoted by sigma^2 or var(X).

Variance of uniform distribution is

Var ( X ) = E [( X- mu )2]

Where mu  = E ( X )

Standard deviation of uniform distribution is denoted by sigma .

mu = ( a + b) / 2

sigma= (b -a) / 2 sqrt(3)

Finally we get

Mean = ( a +b ) / 2

Variance = (b-a)2 / 12

Moment generating function = (etb - eta) / t(b - a)

Types of Uniform Distribution:

Uniform Distributions are classified in to two types,

Continues uniform distributions, and
Discrete  uniform distributions

1. Continues uniform distributions:

The continues uniform distributions is the density function of the random variable. Continues interval between a and b. The density function of  continues uniform distribution is ,

 f(x) ={( 1/(b-a) "when a<= x<=b") , (0 "when xb"):}

2. Discrete Uniform Distributions :

Discrete distributions are also known as  statistical distributions , It is the simplest way of Probability distribution functions.

Probability distributions functions  of p(Xm)  is defined  over m= 1,2,….,N ,

The discrete distribution function as

D(Xn)= sum_(m=1)^n p(Xm)
Uniform Random Distribution General Formula:

The general formula of probability density function of the uniform random distribution function is defined as follows:

f(x) = 1 /b-a               for  a<= x<= b

Where a is the position parameter and (b-a) is the scale parameter. In case where a = 0 and b = 1 is called the standard Uniform random distribution:

Uniform random distribution  is called as simple distribution or rectangular distribution function. In this distribution  their may be chance tooccurring constant probability distribution function. Uniform distribution analysis only numerical data value. That all of the data are  arranged in systematic order is called uniform random distribution.

The equation of the standard uniform random distribution is

f(x) = 1          for   0 <=x <= 1 .

These all are important in the Uniform random distributions.

Examples Problems for Uniform distributions

Example 1: f(x) = 4x for 0 x  1. Find the expected value continuous of given f(x).

Solution:

E (X) = X’ =  x f(x) dx

=   x (4x) dx

=    4x2 dx

= [(4X^3) /3]

=(4/3) - 0

E(X) = 1.333

Example 2: f(x) = x^2 for 0 x  1. Find the expected value continuous of given f(x).

Solution:

E (X) = X’ =  x f(x) dx

=   x (x^2 ) dx

=    x^3 dx

=[(x^3)/3]

=(1/3) - 0

E(X) = 0.333

Example 3: Find the continuous variables of given f(y) = 4e-3y.

Sol:

f(Y) =   ? y (4e -3y) dy

= (-4ye-3y – e -3 / 3)

= ½ (-4ye-3– e -3) – ½ (-4 (0) e0-e0)

f(Y) = ½

Example 4: Find the continuous variable of given f(x) = 4x for 0 x  1.

Sol:

f (X)       =  x f(x) dx

=   x (4x) dx

=    4x2 dx

= [4/ 4  x3]

= 1 - 0

f(X) = 1

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Example 5: Find mean and variance value. From the given value a = 2, b = 4

Solution:

The probability density function F(x) = 1/(4 -2)

F(x) = 1/2 = 0.5

Mean = (a +b ) / 2

Here a = 2 and b = 4

Mean = (2 + 4) / 2

After simplify, we get

= 6 / 2 = 3

Mean = 3

Variance = ( b - a)2 / 12

= (4 - 2)2 / 12

=  (2)2 / 12

After simplify, we get

= 4 / 12 = (1 / 3) = 0.333

Variance = 0.333

Exercise Problems:

Problem -1: f(x) = x^3 for 0 x  1. Find the expected value continuous of given f(x).

Answer:  0.25

Problem -2: f(x) = 2x^3 for 0 x  1. Find the expected value continuous of given f(x).

Answer: 0.5

Problem -3: f(x) = 5x for 0 x  1. Find the expected value continuous of given f(x).

Answer:  2.5

Preparation of Data Analysis

Data analysis is a process of inspecting, cleaning, transforming, and modeling data with the goal of highlighting the useful information, suggesting conclusions, and supporting decision making. Data analysis is a practice in which raw data are ordered and organized so that information can be extracted from it. Data analysis has multiple facts and approaches, encompasses diverse techniques under a variety of names, in different business, science, and social science domains.

Process for preparation of Data Analysis:

Data analysis is a process, in which several phases can be distinguished. The processes for the preparation of data analysis are as follows:

1. Data cleaning:

Data cleaning is an important procedure during in which the data are inspected, and erroneous data are if necessary, preferable, and possible corrected. Data cleaning can be done during the initial stage of data entry. If this is done, it is important that no subjective decisions are to be made. During subsequent manipulations of the data, the given information should always be cumulatively retrievable.

2. Initial Data analysis:

The initial data analysis phase is guided by the following phases,

Quality of Data:

Data quality can be assessed in so many ways, using different kinds of analyses: frequency counts, descriptive statistics (mean, standard deviation, median), normality (skew ness, frequency histograms, normal probability plots), associations (correlations, scatter plots).

Quality of measurements:

Confirmatory of factor analysis.

Analysis of homogeneity, which gives an indication of the reliability of a measurement instrument, i.e., whether all items fit into a one-dimensional scale. During this analysis, we can inspects the variances of the items and the scales, the Cronbach's α of the scales, and the change in the Cronbach's alpha when an item would be deleted from a scale.

3. Main data analysis:

The most important distinction between the initial data analysis and the main analysis is that during initial data analysis it refrains from any analysis.

Basic statistics of important variables

Scatter plots

Correlations

Cross-tabulations

4. Final data analysis:

During the final stage, the finding of the initial data analysis are documented, and necessary, preferable, and possible corrective actions are taken.

Preparation for Types of data analysis:

Preparation for Types of data analysis are as follows:

Qualitative Analysis :

The process of interpreting data which can be collected during the course of qualitative research called as qualitative analysis.

Quantitative Analysis:

The process of presenting and interpreting numerical data are called as quantitative analysis

The quantitative data analysis including often contain descriptive statistics and inferential statistics.

Rectangular Prism

A right rectangular prism is a type of prism in which every lateral face is a rectangle. A right rectangular prism in general is referred to as as cuboids, or in simpler way as rectangular box.

Lateral Surface area of a rectangular prism:  The area of lateral faces of a prism is known as lateral surface area.

Lateral surface area = perimeter * height

= (2l + 2w) * h

where, l = length of base

w = width of base

h = height

Total surface area of a rectangular prism:

Total surface area = Lateral surface area + 2*base

Volume of rectangular prism:   The volume or capacity occupied by the rectangular prism. The volume of the rectangular prism is measured in cube units.

rectangular prism

Volume of the rectangular prism = L*B*H cube units

L is the length of the rectangular prism

B is the breadth of the rectangular prism

H is the height of the rectangular prism.

Model problem to find lateral surface area of the rectangular prism:

Ex 1:A right rectangular prism has a length of  5, width of 2, and the height of 3. Find the lateral area, total area and volume of the right rectangular prism.

Sol: Step 1: Given l = 5.   w = 2,  h = 3.

Step 2:  Lateral area of right rectangular prism  =  perimeter * height

= (2l + 2w) * height

= [2(5) + 2(2)] (3)

= [10 + 4] * 3

Lateral area of right rectangular prism  = 14 * 3 = 42.

Step 3 : Total area of right rectangular prism = Lateral area + 2* base

= Lateral area + 2* (length * width)

= 42 + 2* (5 * 2)

= 42 +20

Total area of right rectangular prism = 62.


Ex 2: A right rectangular prism has a length of 7 , width of 5, and the height of 6. Find the lateral area, total area and volume of the right rectangular prism.

Sol: Step 1: Given l = 7, w = 5, h = 6.

Step 2: Lateral area of right rectangular prism =  perimeter * height

= (2l + 2w) * height

= [2(7) + 2(5)] (6)

= [14 + 10] * 6

Lateral area of right rectangular prism = 24 * 6 = 144.

Step 3: Total area of right rectangular prism = Lateral area + 2* base

= Lateral area + 2* (length * width)

= 144 + 2* (7 * 5)

= 144 + 2*35

Total area of right rectangular prism = 214.

Model problem to find the volume of the rectangular prism:

Ex 1:  Finding the volume of the rectangular prism when the length is 7cm , breadth is 6cm and the height is 10 .

Sol:Step 1:  Length of the rectangular prism = 7cm

Breadth of the rectangular prism = 6cm

Height of the rectangular prism = 10cm

Step 2:  Volume of the rectangular prism = L*B*H cube. Units

= 7*6* 10

= 42*10

= 420 cm3

Volume of the rectangular prism is 420 cm3

Ex 2:  Finding   the volume of the rectangular prism when the length is 9cm, breadth is 8cm and the height is 12cm

Sol:Step 1:Length of the rectangular prism = 9cm

Breadth of the rectangular prism = 8cm

Height of the rectangular prism = 12cm

Step 2: Volume of the rectangular prism = L*B*H cube. Units

= 9*8* 12

= 72*12

= 864 cm3

Volume of the rectangular prism is 864 cm3

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Ex 3: Finding the volume of the rectangular prism when the length is5cm, breadth 4cm and the height is 6cm

Sol: Step 1: Length of the rectangular prism = 5cm

Breadth of the rectangular prism = 4cm

Height of the rectangular prism = 6cm

Step 2:   Volume of the rectangular prism = L*B*H cube. Units

= 5*4* 6

= 20*6

= 120 cm3

Volume of the rectangular prism is 120 cm3

Mean Confidence Interval

It is one of the main topics in statistics. We are going to find the population index in the statistics. Normally confidence level means we have to find the population parameter in between the confidence independence. Confidence level is nothing but the degree of confidence. Here the degree of statistical prediction is mostly accurate. Basically it will reflect the degree of certainty of the true values in between the confidence interval. In this the true statement value lies between the specified ranges.

I like to share this Sample Size and Confidence Interval with you all through my article.

Explanation for confidence level:

We won’t find not only finding the point of mean we have to determine how much it will be accurate. We have to estimate the estimation of accurate. Here we have to use the central limit theorem.  We have to assume the sample standard deviation is nearer to the population standard deviation. Actually the central limit theorem says

Here we are interested to finding the interval which is going to be around x and the large probability where the actual mean falls inside the confidence level interval. We will call this interval as confidence level interval. Here we have to find the confidence level which is lies between the given confidence interval. Let us see the example for finding the confidence level.

Example for finding confidence level:

Suppose a man check for clarity in 50 locations in Mount and he discovers that the average depth of the clarity is 16 feet.  Suppose that he know that the S.D for the entire Mount's depth is 3 feet.  What he will conclude about the average clarity of the mount with a 95% confidence level?

Solution:

Let us see the z –score diagram for this. We are constructed the z - score for the given 95% confidence interval.

We have to move from z = -1.96. now we are going to solve for x

- 1.96 = `(x-16)/(3 / sqrt(50))` = `(x - 14)/(0.42)`

So x - 14 = -0.82

We can say that `+- 0.82` is margin error.

We are having 95% confidence interval so that (15.18, 16.82)

It means the confidence level is lies between 13.18 and 14.82.

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Rounding Numbers

Here in this page we are going to discuss about rounding numbers concept .Sometimes, it is necessary for us to know to round a number. In certain calculations, it is not necessary to keep more number of values after the decimal points.

Rounding a number can be specified as nearest to the integer or to tenths or to hundredths .etc. By doing these rounding of numbers will help us to calculate certain values easier and it will also avoid lots of confusion.

The value of pi is denoted as a rational number 22/7.

But we generally take the value to be 3.14, because it has non terminating decimal values. Hence pi = 3.14 is one of the sample of rounding numbers.

Let us see some more examples on this topic sample of rounding numbers

Examples on rounding numbers

Here are the examples on rounding numbers -

Example 1: Round the following number to a nearest integer:

(i) 25.67

(ii) 32.43

(iii) 52.04

Solution :

(i) To round a number to the desired form, we need to check the following.

Here, we need to round to an integer. So, we have to check the first decimal value. If the first decimal value is 5 or more, then we add 1 to the integral value and give the final rounded value of rounding to the nearest integer. If the first decimal is less than 1, then we take the integral value and give that as the final value of rounding to the nearest integer.

Now, in 25.67, the first integer is 6 which greater than 5. So we will round the value as 25 + 1 = 26.

Therefore, 25.67 is now rounded to the nearest integer as 26. That will be the final answer.

(ii) 32.43. Here the first decimal is 4, which is less than 5. So, we will take only the integral value which is 32. Therefore, the final rounded value will be 32.

(iii) 52.04. Here the first decimal is 0, which is less than 5. So, we will take only the integral value which is 52. Therefore, the final rounded value will be 52

Example 2: Round the following number to a nearest tenth:

25.67

Solution:

With the same concept, here we are going to consider the second digit after the decimal to round the answer to give the first digit after the decimal.

In 25.67, the second digit after the decimal is 7. 7 is greater than 5. So we need add 1 to 6.

Therefore, the final value after rounding will be 25.7.

Example 3: Round the following number to a nearest hundredth:

12.3427

Solution:

Here we need to consider the third value after the decimal. The third value is 2, which is less than 5.

Therefore, the final value after rounding will be 12.34.

Practice problems

Here are the problems on rounding numbers

1: Round the following number to a nearest integer:

345.67

[Answer: 346]

2: Round the following number to a nearest tenth:

125.327

[Answer: 125.3]

3: Round the following number to a nearest hundredth:

572.438

[Answer: 572.44]

I hope you would have understand better about the concept of sample of rounding numbers.

Expected Value and Variance

Let X is a numerically values discrete random variable by the resources of sample space and distribution function m(x). The expected value E(X) is defined by

E(x) =  "sum_(x=0)^(n) (x*m(x))

Provided this calculation covers absolutely. Pass on the expected value as the mean value for the given discrete random variable, and to represent E(X) as  μ. If the above sum does not cover it completely, then we say that X does not have an expected value.

Variance:

Let X be a numerically values discrete random variable with expected value μ = E(X). Then the variance of X, denoted by V(X), is

V (X) = E((X - mu)^2)  .

V (X) can also given by,

V (X) = sum_(x=0)^(n)(x - mu)^2m(x)

here m is the  distribution function of X

Expected value of probability density function:

Let x be the probability density function, integrating x with the given function with using the limits, The expected value for the probability density function is given as,

E(x) = int_a^b x f(x) dx

Variance of probability density function:

Let x be the probability density function, integrating squared difference between x and expected value and with the function using the limits. The variance is given by,

V(x) = int_a^b ((x-E(x))^2) f(x) dx

Expected value and Variance - Example Problems:

Expected value and Variance - Problem 1:

Find the expected value and the variance for the discrete random variable. (1/5). Where x is from 0 to3.

Solution:

Expected value of the discrete random variable,

E(x) =  sum_(x=0)^(n) (x* m_x)

E(x) = 1(1/5) + 2(1/5) + 3(1/5)

E(x) = 1.2

Variance for the discrete random variable,

V (X) = sum_(x=0)^(n)(x - mu)^2m(x)

V (X) = (0-1.2)^2(1/5)+(1-1.2)^2(1/5)+(2-1.2)^2(1/5)+(3-1.2)^2(1/5)

V(x) = 1.07200

Expected value and Variance - Problem 2:

Find the expected value and variance for the probability density function.  2x  0<= x <= 1

Solution:

Expected value,

"E(x) = int_a^b xf(x)dx

E(x) = int_0^1 x (2x) dx

E(x) = int_0^1(2x^2) dx

E(x) = [(2x^3) /3 ]_0^1

E(x) = [2 / 3]

Variance:
"V(x) = int_a^b ((x-E(x))^2)f(x) d(x)

"V(x) = int_0^1 (x-(2/3))^2 (2x)dx

V(x) = int_0^1(x-(2/3))^2 (2x) dx

"V(x) = int_0^1[x^2+(4/9)-(4/3(x))] (2x) dx

V(x) =int_ 0^1[2x^3 +(8/9x)-(8/3)(x^2)]dx

"V(x) = [2(x^4/4)+(8/9)(x^2 /2)-(8/3)(x^3 /3)]_0^1

V(x) = [(2/4) +(4/9)-(8/9)]

V(x) = [0.5+0.44444-0.888888]

V(x) = 0.05555

Correlation Coefficient Online Tutoring

Tutoring is one of the best ways for receiving help in various subjects. Some of the common tools for online tutoring include whiteboard, chat and audio call. There are number of Websites available for online tutoring.(Example: www.tutorvista.com). Correlation is nothing but calculate the strength of the linear relationship between two variables x and y. Correlation will always between -1 and +1. There are two types of correlation. Positive correlation and negative correlation.

Correlation(r) =`[ (NsumXY - (sumX)(sumY)) / sqrt([NsumX^2 - (sumX)^2]xx[NsumY^2 - (sumY)^2])] `

where, N = Number of elements in the set of data. Now, we are going to see some of the problems on correlation coefficient.


Example problem- correlation coefficient online tutoring:

Example problem:

Calculate the correlation coefficient of the following data
x 50 51 52 53 54 55
y 2.1 2.4 2.6 2.8 3 3.3

Solution:

Step 1: First count the number of values.
N = Number of elements in the set of data = 6

Step 2: Next, find XY, X2, Y2
x y X*Y X*X Y*Y
50 2.1 105 2500 4.41
51 2.4 122.4 2601 5.76
52 2.6 135.2 2704 6.76
53 2.8 148.4 2809 7.84
54 3 162 2916 9
55 3.3 181.5 3025 10.89


Step 3: Then, find ΣX, ΣY, ΣXY, ΣX2, ΣY2.
Sum of all values of x = ΣX = 315
Sum of all values of y =  ΣY = 16.2
Sum of all values of xx = ΣXY = 854.5
Sum of square values of x = ΣX2 = 16555
Sum of square values of y =ΣY2 = 44.66

Step 4: Now, Substitute in the above formula given.
Correlation(r) =`[ (NsumXY - (sumX)(sumY)) / sqrt([NsumX^2 - (sumX)^2]xx[NsumY^2 - (sumY)^2])]`


= `((6)xx(854.5)-(315)xx(16.2)) / sqrt([(6)xx(16555)-(315)^2]xx[(6)xx(44.66)-(16.2)^2])`

= `(5127 - 5103) / sqrt([99330 - 99225]xx[267.96 - 262.44])`

= `24 / sqrt(105xx5.52)`

= `24 / sqrt(579.6)`

= `24/24.0749`

= 0.9968

So, the correlation coefficient (r) of the given data= 0.9968.

Practice problem-correlation coefficient online tutoring:

Practice problem:

Calculate Correlation Co-efficient of the following data
x 40 41 42 43 44 45
y 1.1 1.4 1.6 1.8 2 2.3

Answer: Correlation coefficient (r) = 0.9968.

Sum Difference Quotient

Sum is defined as adding two or more numbers that is given. Difference is defined as subtracting two or more numbers. Quotient means the answer getting  after dividing one number by another number. For example: Sum 2 + 2 = 4, here 4 is sum and Difference 6 – 4 = 2, here 2 is the difference and Quotient 8 ÷ 4 = 2, here 2 is quotient.

Example Problems For Sum

Example 1:

Find the sum of two numbers 739 and 248.

Solution:

739    addend ( or even augend )

+ 248    addend

--------------

987     sum

Answer : Sum is 987.

I like to share this Quotient Rule for Exponents with you all through my article.

Example 2 :

Find the sum of six digit numbers 830163 and 294810.

Solution:

830163   addend

+ 294810    addend

-----------------------

1124973

Answer : Therefore, the sum is 1124973.

Example Problems For Difference

Example 3 :

Find the difference of two numbers 523 and 352.

Solution:

523    minuend

- 352    subtrahend

---------------

171

Answer : Difference is 171.

Example 4:

Find the difference of eight digit numbers 72917304 and 52910456

Solution:

72917304    minuend

52910456    subtrahend

----------------------------

20006848

Answer : Therefore, the difference is 20006848.

Example Problems For Quotient

Example 5 :

Find the quotient 82 ÷ 2

Solution:

82 / 2 = 41 => dividend  / divisor = quotient

Answer : Therefore, the quotient is 41.

Example 6 :

Find the quotient 1428 ÷ 14

Solution:

1428 / 14 = 102 => dividend / divisor = quotient

Answer : Therefore, the quotient is 102.

Mote Problems To Practice

1.Find the sum of two numbers 739 and 928

Key : 1667

2.Find the sum of cross digit numbers 720183 and 1294

Key : 721477

3.Find the difference of two numbers 823 and 597

Key : 226

4.Find the diffrence of cross digit numbers 9784 and 7899

Key : 1885

5.Find the quotient of two numbers 48 ÷ 4

Key : 12

6. find the quotient of three digit numbers 1484 ÷ 4

Key : 371

How to Make a Trapezoid

Trapezoid is a type of quadrilateral with two equal opposite pair of sides. Trapezoid has four sides which are in different size. The sum of total interior angle will be 360º and also an exterior angle 360º. The sum of adjacent angle will be 180º. Let us see how to make a trapezoid.



Characteristics of Trapezoid:

The trapezoid is a type of a quadrilateral.
A pair of opposite equal sides is known as trapezoid.
A trapezoid has unequal sides.
Normally the trapezoid has no lines of symmetry.
A trapezoid is an irregular shape.
Two trapezoids can be used to form a Parallelogram.
There will be one pair of parallel lines in a trapezoid
Perimeter of a trapezoid is to sum all the length’s of the trapezoid
Perimeter = a + b + c + B
Area of a trapezoid is ½ h ( B + b )
The sum of the adjacent angles are equal to 180o
The sum of all the interior angles are equal to 360o
Isosceles trapezoid is a type or trapezoid


isosceles trapezoid formula:

Isosceles trapezoid is also a type of trapezoid.

Characteristics of Isosceles trapezoid.

The Isosceles trapezoid is a type of trapezoid.
If non-parallel pair of opposite sides of a trapezium is equal then it is called as isosceles trapezium.
The angles on the both sides of the base are equal
The sum of the adjacent angles are equal to 180o
There will be one pair of parallel lines in a Isosceles trapezoid
There will be one pair of opposite sides which are equal
In a Isosceles trapezoid diagonals are equal
The sum of two adjacent angles are equal to 180o
The sum of all the interior angles are equal to 360o


How to make a trapezoid:

How to make a isosceles trapezoid with a long base B that is 10cm long, a short base d that is 5 cm long and 35 degree angle



Step-1 Take a scale and draw a straight line of it’s length 10 cm and mark it as ‘n’ ‘o’ for the base B.

Step-2 And then take the protractor and keep it in the line n and mark 90o

Step-3 And draw a straight line. Now mark it as ‘m’.

Step-4 Now take ‘m’ ‘n’ as the line and using the protractor mark 90o and draw a straight line.

Step-5 From o mark 35o by using the protractor. So that we can join the line and mark it as p

This is how to make a trapezoid.

Probability Method

Let us see the introduction about probability method.  Probability is used to measures the degree of uncertainty and certainty of an event. The probability theories are used to develop the study of games of chance like as roulette and cards. Probability method is a likelihood that an event will happen.

Probability Methods:

Experiment:

Experiment of an operation will produce some well -defined outcomes.

Random experiment:

The random experiment, do not produce the same outcome in a trail, when repeated under identical condition is one of the several possible outcomes.

Sample space:

In probability sample space of a random experiment is used to the set of all possible outcomes and it is denoted by S.

Event:

In probability an event is the sample space of every subset.

Simple event:

Simple or an elementary event of an event which can contain only a single element of the sample space.

Compound event:

Composite or mixed event of an event which is not a simple.

Let us see the example problems about probability method as given below.

Examples:

1. What is the sample space for toss of a coin?

Solution:

In toss a coin, there are two possible outcomes, namely head (H) and tail (T).

Hence, the sample space in this experiment is given by

S = {H, T}

2. List the sample space in throwing a die.

Solution:

When we throw a die, it can result in any of the six numbers, namely, 1, 2, 3, 4, 5, 6.

Sample space is S = {1, 2, 3, 4, 5, 6}.

3. When two dice are rolled. A is the event that the sum of the numbers shown on the two dice is 4. What is the sample space?

Solution:

When two dice are rolled, we have n(S) = (6 * 6) = 36.

Now, A = {(1, 4)(2, 3)(4, 1)(3, 2)}.

These are the example problems for probability method.

Line Segment

Drawing line segments is a simple activity performed in elementary mathematics. But it is the base of many fields and they are very important. Because lines play a vital role in many real time applications and in our day to day life. Drawing lines in elementary level is very simple which, we can do it with the help of a ruler. But drawing with a ruler is not sure that the line is exactly horizontal or exactly vertical line. Drawing line segments is explained in the following segments.


I like to share this Length of Line Segment with you all through my article.

Line segments:

Line segments are nothing but a part of a line which has end points. Combination of many line segments give many different shapes like, square, triangle, rectangle, etc. The line segment can be vertical or horizontal or inclined as well.

To draw a line segment of particular length, take a ruler with scale on it and place it on the paper as you require whether vertical inclined or horizontal..

Now draw a line for the specified length using the ruler.

The steps involved is explained in the figure,



Examples for drawind line segments:

Here are few example to draw line segments,

Example 1:

Draw a line segment of 10 cm.

Solution:

Take a ruler with scale in centimeter and place it on the paper,

Mark the points '0' and '10' .

Join the points using the ruler.

Example 2:

Draw a line segment of 25 cm.

Solution:

Take a ruler with scale in centimeter and place it on the paper,

Mark the points '0' and '25' .

Join the points using the ruler.

Example 3:

Draw a line segment of 45mm.

Solution:

Take a ruler with scale in millimeter and place it on the paper,

Mark the points '0' and '45' .

Join the points using the ruler.


Example 4:

Draw a line segment of 25 mm.

Solution:

Take a ruler with scale in millimeter and place it on the paper,

Mark the points '0' and '25' .
Join the points using the ruler.

Discrete Math Proof

Discrete mathematics is the study of mathematical structures that are fundamentally discrete rather than continuous. In contrast to real numbers that have the property of varying "smoothly", the objects studied in discrete mathematics – such as integers, graphs, and statements. Discrete objects can often be enumerated by integers. More formally, discrete mathematics has been characterized as the branch of mathematics dealing with countable sets. In this article we shall discuss about discrete math proof.(Source: wikipedia)

Discrete math theorem and proof

Theorem 1:

For any group G, the identity element is the only element of order 1.

Proof:

If a (≠ e) is another element of order 1 then by the definition of order of an element, we have (a)1 = e ⇒ a = e which is a contradiction. Therefore e is the only element of order 1.

Theorem 2:

The identity element of a group is unique.

Proof:

Let G be a group. If possible let e1 and e2 be identity elements in G.

Treating e1 as an identity element we have e1 * e2 = e2 …               (1)

Treating e2 as an identity element, we have e1 * e2 = e1 …              (2)

From (1) and (2), e1 = e2

Therefore Identity element of a group is unique.

Theorem 3:

The inverse of each element of a group is unique.

Proof:

Let G be a group and let a ∈ G.

If possible, let a1 and a2 be two inverses of a.

Treating a1 as an inverse of ‘a’ we have a * a1 = a1 * a = e.

Treating a2 as an inverse of ‘a’, we have a * a2 = a2 * a = e

Now a1 = a1 * e = a1 * (a * a2) = (a1 * a) * a2 = e * a2 = a2

⇒ Inverse of an element is unique.

Discrete math proof example problem

Example: Find the order of each element of the group (G, .) where G = {1, − 1, i, − i}.

Solution: In the given group, the identity element is 1. Therefore 0(1) = 1.

0(− 1) = 2 [ Therefore we have to multiply − 1 two times (minimum) to get 1 i.e.,

(− 1) (− 1) = 1]

0(i) = 4 [ Therefore we have to multiply i four times to get 1, i.e., (i) (i) (i) (i) = 1]

0(− i) = 4 [ Therefore we have to multiply − i four times to get 1].

Example 2:

Find the order of each element of the group (G, .) where G = {1, − 1, i, − i}.

Solution: In the given group, the identity element is 1.Therefore 0(1) = 1.

0(− 1) = 2 [Therefore we have to multiply − 1 two times (minimum) to get 1 i.e.,

(− 1) (− 1) = 1]

0(i) = 4 [Therefore we have to multiply i four times to get 1, i.e., (i) (i) (i) (i) = 1]

0(− i) = 4 [Therefore we have to multiply − i four times to get 1].

Study Calculate Probability

Study calculates probability assumption has its origin in games of opportunity pertaining to gamble. Jerome Cardon an Italian mathematician wrote book on ‘Games of chance’ published in 1663. Study calculates probability is given by ratio of number of favorable outcomes, n to the number of possible outcomes. Power of statistical test is probability that test reject a false null hypothesis.

Frequently used terms in probability

Frequently used terms in probability

Experiment:      
Experiment is defined as a process for which its result is clear.

Sample space:      
Possible outcomes in experiment are called sample space.

Event:          
Each non-empty subset of sample space is an event.

Mutually exclusive events (or disjoint events):        

If two or more events have no simple events is known as mutually exclusive.

Exhaustive events:        

If no event occurs outside set of events it is recognized as exhaustive event of set. And one of events in that set must occur as answer of an experiment.

Equally likely events:        

Equally expected events is the one in which a set of events will not occur in preference to other.

Examples for study calculate probability

Ex 1:   Study calculate probability that a girl get access in Engineering college is 0.80, study calculate probability that she obtain admittance in Medical College is 0.88, and the probability that she will get both is 0.72. Calculate study probability that (i) She will get at-least one of the two seats (ii) She will get only one of two seats

Sol :  Let I be event of receiving admission in Engineering college and M be event of getting

admission in Medical College.

P(I) = 0.80, P(M) = 0.88 and P(I ∩ M) = 0.72

(i) P (at-least one of the two seats)

= P(I or M) = P(I ∪ M)

= P(I) + P(M) − P(I ∩ M)

= 0.80 + 0.88 − 0.72

= 0.96

(ii) P(only one of two seats) = P[only I or only M].

= P[(I ∩ ) ∪ (  ∩ M)]

= P(I ∩ ) + P( ∩ M)

= {P(I)−P(I∩M)}+{P(M)−P(I∩M)}

= {0.80 − 0.72} + {0.88 − 0.72}

= 0.08 + 0.16

= 0.24

Probability is 0.24.

Ex 2:  P(A)=12,P(B)=18 and P(A ∪ B)=10 calculate P(A ∩ B)

Sol :   P(A ∩ B)=  P(A)+P(B) - P(A ∪ B)

=12+18-10

P(A ∩ B)=20

Tree Diagram Sample Space

For representing sample space, we are using tree diagrams, venn diagrams and graphs. A tree diagram is used to find out the elements of sample space of any experiment. The tree diagram is also used to determine the probability of individual outcomes within the sample space. In this article, we are going to see tree diagrams for sample space example problems.

Example 1 - Tree Diagram Sample Space

Illustrate the sample space for tossing of coin and rolling a die.

Solution:

There are 2 faces in a coin. That is, head and tail.

There are 6 faces in a die. That is, 1, 2, 3, 4, 5 and 6.

The tree diagram for this experiment is,

This is the tree diagram for sample space of given data.

The sample space is,

{H1, H2, H3, H4, H5, H6, T1, T2, T3, T4, T5, T6}

The probability for each of outcomes is,

1/2 × 1/6 = 1/12

Example 2 - Tree Diagram Sample Space

There are three children are in a family. How many outcomes are in sample space that represents the sex of children? Assume that probability of male (M) and probability of female (F) are each 1/2 .

Solution:

The tree diagram for this given information is,

This is the tree diagram for sample space of given data.

The sample space for this tree diagram is,

{MMM, MMF, MFM, MFF, FMM, FMF, FFM, FFF}

There are totally 8 outcomes in sample space.

The probability for each outcome in this sample space is,

1/2 × 1/2 × 1/2 = 1/8

Example 3 - Tree Diagram Sample Space

A bag contains red, lavender and yellow color balloons. A ball is picked and coin is tossed at random at same time. Illustrate the sample space tree diagram.

Solution:

The tree diagram for this experiment is,

This is the tree diagram for sample space of given data.

The sample space for this tree diagram is,

{(Red, Head), (Red, Tail), (Lavender, Head), (Lavender, Tail), (Yellow, Head), (Yellow, Tail)}

There are totally 6 outcomes in sample space.

Parallelogram Properties

Parallelogram is one type of quadrilateral. It has four sides, in which the opposite sides are parallel and equal in length. The opposite internal angle of parallelogram is equal in measure. The diagram of parallelogram is shown in below.

parallelogram

Properties of parallelogram:

Length of opposite sides are equal.
Opposite angles are equal in measure.
Area of the parallelogram is the product of base and height.
The diagonals of a parallelogram intersect each other

Having problem with Finding the Area of a Parallelogram keep reading my upcoming posts, i will try to help you.

Area of the parallelogram (A) = base x height square units

= b x h square units

Perimeter of the parallelogram (P) = 2 (a + b)

a and b are adjacent sides of parallelogram

Parallelogram - Example problems:

1. Parallelogram has base 7cm and height 8cm. what is the area of parallelogram?

Solution:

Formula:

Area of the parallelogram (A) = b x h square units

Given:

Base=7cm;       height=8cm

= 7 x 8

= 56 cm2

Area of the parallelogram (A) = 56 cm2

2. Parallelogram has base 12cm and height 8cm. what is the area of parallelogram?

Solution:

Formula:

Area of the parallelogram (A) = b x h square units

Given:

Base= 12cm;    height= 8cm

= 12 x 8

= 96 cm2

Area of the parallelogram (A) = 96 cm2

3. The side length of two adjacent sides of parallelogram are 5cm and 6cm

What is the perimeter of the parallelogram?

Solution:

Given:

a = 5 cm

b = 6 cm

Perimeter of the of parallelogram (P) = 2 (a + b) square units

= 2 (5 + 6)

=2 x 11

= 22

Perimeter of the of parallelogram = 22 cm

4. The side length of two adjacent sides of parallelogram are 10cm and 12cm

What is the perimeter of the parallelogram?

Solution:

Given:

a = 10cm

b = 12 cm

Perimeter of the of parallelogram (P) = 2 (a + b) square units

= 2 (10 + 12)

=2 x 22

= 44

Perimeter of the of parallelogram = 44 cm

Algebra is widely used in day to day activities watch out for my forthcoming posts on prime factorization problems and cbse 9th class syllabus 2012. I am sure they will be helpful.

Parallelogram - Practice problems:

1. Parallelogram has base 8cm and height 10cm. what is the area of parallelogram?

Answer: 80 cm

2. The side length of two adjacent sides of parallelogram is 3cm and 4cm.

What is the perimeter of the parallelogram?

Answer: 14 cm

Angle Degree Conversion

Angle is measured between the two lines which are intersecting each other and the intersecting point is called the vertex of the angle. There are three ways to measure the angle; they are degree, grade and radians. Here we see about the angle measured in degree and how the conversion takes place from degree to radian and grade and vice versa.

Understanding Hexadecimal to Decimal Conversion is always challenging for me but thanks to all math help websites to help me out.

Angle Degree Conversion:

Conversion of degree to radians – Multiply the given degree by `(pi)/(180^0)` to get radians.

Conversion of degree to grade - Multiply the given degree by `(10)/(9)` to get grade.

Conversion of radians to degree – Multiply the given radians by `(180^0)/(pi)` to get degree.

Conversion of grade to degree - Multiply the given grade by`(9)/(10)` to get degree. These are the formula used for the conversion of degree to other measurement. Let see the example problem in angle degree conversion.

Example Problem – Angle Degree Conversion:

Example 1:

Convert the degree 60° to radian.

Solution:

Step 1: Multiply the 60 degree by `(pi)/(180^0)` to get radians.

60° * ( `(pi)/(180^0)`) =  `(pi)/(3)`

Answer: 60 degree = `(pi)/(3)` radian

Example 2:

Convert the radian  `(pi)/(2)` to degree.

Solution:

Step 1: Multiply the  `(pi)/(2)` radians by  `(180^0)/(pi)` to get degree.

`(pi)/(2)` *  `(180^0)/(pi)` = 90°

Answer: radian  `(pi)/(2)`90 in degree

Example 3:

Convert the degree 45° to grade.

Solution:

Step 1: Multiply the 45 degree by  `(10)/(9)` to get grade.

45° * ( `(10)/(9)`) = 10 * (5) = 50 grade

Answer: 45 degree = 100 grades

Example 4:

Convert the grade 100 to degree.

Solution:

Step 1: Multiply the 100 grade by  `(9)/(10)` to get degree.

100 * ( `(9)/(10)`) = 90°

Answer: 110 grades = 90 in degree.

My forthcoming post is on Definition Acute Angle and maths projects for class 10 cbse will give you more understanding about Algebra.

These are example problem in angle degree conversion. Let do the practice problem in angle degree conversion.

Practice Problem – Angle Degree Conversion:

1. Convert the degree 30 to radian.

Answer: π/6

2. Convert the degree 36° to grade.

Answer: 40 grades