Measurement of Angle

• Line joins two positions and which are nonstop in two directions. Lines have one endpoint of two lines and infinite in one direction is called angle.
•  The grouping point of two lines is called the vertex.

               

•  Angles are denoted as degrees. In measurement of angles can diverge in between 0 and 180 angles.

Concept -measurement Angle:

Concept of angles:

      There are three major categories of angles: right angle, obtuse angle, and acute angle.

• A right angle – a measurement of an angle is 90 degrees.
• An acute angle - a measurement of an angle in between 0 to 90 degrees.
• An obtuse angle - a measurement of an angle in between 90 and 180 degrees.

More about angles and measurement of angle:

More about angles:

1)      Vertical angle measurements are two angles with a common vertex and with sides that are two pairs of opposite of opposite rays. (That is, the union of the two pairs of sides is two lines.)  

                              

2)       A right angle measurements has a measure of 90^o.    

     

                       

3)      An acute angle measurement is an angle whose measure is larger than 0^o and smaller than 90^o.  

          

                

4)      An obtuse angle measurement has a measure larger than 90^o and less than 180^o.  

           

               

5)     Straight angle measurement has a measure of 180^o. The rays of a straight angle lie on a line.

6)      Complementary angle measurements are two angles, the addition of whose measures is 90^o.

               

7)      Supplementary angle measurements are two angles, the addition of whose measures is 180^o.

               

Step to measurement of angle:

          

Step 1: To find the angle is obtuse or acute before using the protractor.

          

Step 2: Position the protractor on the center mark on the vertex of the angle

          

Step 3: position the protractor, line up with one ray of the angle and another ray of the angle cross the protractor’s scale

          

Step 4: to check the measure of the angle of the line that crosses the protractor scale.

          

How to calculate volume

Example problems:

Here are some solved examples on How to calculate volume:
Problem 1:

Problem 2:
Calculate the volume of the cone with base radius = 5cm and height at the apex is 4 cm.
Solution:
Volume of a cone: (13)(π)(r2)h
                                 = (13)(π)(52)4
                                = 104.666 cm^3
Problem 3:

Problem 4:
Calculate the length of the rectangular prism, when the volume is 36 cm^3 , width = 3 m, height is = 4m
Solution:
Volume of the rectangular prism = l * w * h.
                                                     36 =  l* 4*3
                                          Therefore,  l= 3m
Students can understand How to calculate volume from the above examples and learn to solve on their own.

Parametric t Test

Conventional statistical procedures are also called parametric tests. In a parametric test a sample statistic is obtained to estimate the population parameter. Because this estimation process involves a sample, a sampling distribution, and a population, certain parametric assumptions are required to ensure all components are compatible with each other. 

For example, in Analysis of Variance (ANOVA) there are three assumptions:

• Observations are independent.
• The sample data have a normal distribution.
• Scores in different groups have homogeneous variances.

In a repeated measure design, it is assumed that the data structure conforms to the compound symmetry. A regression model assumes the absence of collinearity, the absence of auto correlation, random residuals, linearity...etc. In structural equation modeling, the data should be multivariate normal.

Why are they important? Take ANOVA as an example. ANOVA is a procedure of comparing means in terms of variance with reference to a normal distribution. The inventor of ANOVA, Sir R. A. Fisher (1935) clearly explained the relationship among the mean, the variance, and the normal distribution: "The normal distribution has only two characteristics, its mean and its variance. The mean determines the bias of our estimate, and the variance determines its precision." (p.42) It is generally known that the estimation is more precise as the variance becomes smaller and smaller.

Put it in another way: the purpose of ANOVA is to extract precise information out of bias, or to filter signal out of noise. When the data are skewed (non-normal), the means can no longer reflect the central location and thus the signal is biased. When the variances are unequal, not every group has the same level of noise and thus the comparison is invalid. More importantly, the purpose of parametric test is to make inferences from the sample statistic to the population parameter through sampling distributions.

When the assumptions are not met in the sample data, the statistic may not be a good estimation to the parameter. It is incorrect to say that the population is assumed to be normal and equal in variance, therefore the researcher demands the same properties in the sample. Actually, the population is infinite and unknown. It may or may not possess those attributes. The required assumptions are imposed on the data because those attributes are found in sampling distributions. However, very often the acquired data do not meet these assumptions. There are several alternatives to rectify this situation.

Free Sample Trinomials

Trinomials:

In elementary algebra, a trinomial is a polynomial consisting of three terms or monomials.(source : WIKIPEDIA)

The trinomial must be one of the following form .

Examples for free sample trinomial:

1. 9x + 3y + 5z , where x , y, z are variables.

2. xy + x + 2y, Where x, y are variables.

3. x2+x-8, where x is variable.

4. ax + by + c = 0 , Where a,b,c are constants and x,y are variables.

Our  tutor vista website provide opportunity to learn about sample trinomials with free of cost. In this article we are going to see some sample problems on solving and factoring trinomials.

Free sample problems on trinomials:

Problem 1:

Square the following trinomial,

x+3y-2

Solution:

Given, x+3y-2

We need to find the square for the given trinomial,

That is (x+3y-2)2

We can apply the following formukla to find the square for the above trinomial,

( a + b + c)2 = a2+b2+c2+2ab+2bc+2ca

(x+3y-2)2  = x2+(3y)2+22+2(x)(3y)+2(3y)(-2)+2(-2)(x)

= x2+9y2+4+6xy-12y-4x

= x2+ 9y2+ 6xy - 4x -12y + 4

Answer: (x+3y-2)2 = x2+ 9y2+ 6xy - 4x -12y + 4

Problem 2:

Factor the trinomail x2 + x – 156 .

Solution:

Given , x2 + x – 156 .

- 156 (product)

/     \

-12     13

\    /

1 (sum)

= x2 - 12x + 13x -156

= x ( x - 12 ) + 13 ( x -12 )

= (x+13) (x-12)

Answer: (x+13) and  (x-12) are the factors of the given trinomial.

Problem 3:

Solve the trinomial 2x2 - 2x = 12.

Solution:

Given, 2x2 - 2x = 12.

Subtract 12 on both sides,

2x2 - 2x - 12 = 12 - 12

2x2 - 2x - 12 = 0

2(x2 - x - 6 ) =0

Divide by 2 on both sides,

x2 - x - 6 = 0

-6

/  \

-3  2

\  /

-1

x2 -3x + 2x - 6 = 0

x(x - 3) + 2(x-3) = 0

(x+2) (x-3) = 0

(x+2) = 0

x = -2

(x-3) = 0

x =3

Answer: x = 3 , -2

Practice problems on free sample trinomials:

Problems:

1.Find the roots of the trinomial x2 - 5x = -6

2.Factor the trinomial 2x2 + 12 x - 14

Answer Key:

1. x = -1 , x = 6

2. ( x - 1) and (x + 7)

Examples Functions in Math

Functions in math deals with finding unknown variable from the given expression with the help of known values. In algebraic expression the variable are represented in alphabetic letters. Functions in math, the numbers are consider as constants. Algebraic expression deals with real number, complex number, and polynomials. In algebraic expression several identities to find the x values by using this we can easily find the algebraic expression of the particular function. The example function in math  may include the function of p(x), q(x),… to find the x value of the functions.

Examples function in math:

Q(y) = 4y2+12y + 40. In this equation we need to find the variable of Q(2) functions in math her y is 2.

Problems using examples functions in math

Examples functions in math

Problem 1: Examples functions in math using p(y) = y2 +2y +4

p(y) = y2 +2y +4 find the f(6).

Solution :

Given the function of p(y) there is y value is given function in math

p(y) = y2 +2y +4 find the p(6)

The value of x is 2 is given

p(6) = 62 +2*6 +4

p(6) = 36 +12 +4 In this step 6 square is 36 it is calculate and 2*6 is 12 be added

p(6) = 52.

The functions in math p(6) = y2 +2y +4 find the p(6) is 52.

Problem 2 Examples functions in math using q(x) = x2 +2x +40 find the q(3).

q(x) = x2 +2x +40 find the q(3).

Solution :

Given the examples functions in math of q(x) there is x value is given functions in math

q(x) = x2 +2x +40 find the f(3)

The value of x is 2 is given

q(3) = 32 +2*3 +40

q(3) = 9 + 6 +40 In this step 3 square is 9 it is calculate with 2*3 is 6 be added to 40 to find the example function in math q(3).

q(3) = 55.

The examples functions in math q(3) = x2 +2x +40 find the q(3) is 55

Example functions in math using cubic equation

Problems1: examples functions in math using f(x) = x3 +2x2 + 2x + 4 to find the f(3).

F(x) = x3 +2x2 + 2x + 4 find the functions in math f(3).

Solution

Given the function in math of  f(x) there is x value is given as 3. Find example function in math.

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

Here the value of x is given as 3

f(3) = 33 + 2*32 + 2*3 +4

f(3) = 27 +18+ +6 +4 In this step 3cube is calculated  as 27 and  3square is 9.

f(3) = 55.

The math functions of f(x) = x3 +2x2 + 2x + 4 find the f(3) = 55.

Problems 2: Examples functions in math using q(x) = x3 +2x2 + 4x + 25 to find the q(3).

q(x) = x3 +2x2 + 4x + 25 find the functions in math q(3).

q(x) = x3 +2x2 + 4x + 25 find the math function in q(3).

Solution Given the examples functions in math of  q(x) there is x value is given as 3. Find example function in math.

q(x) = x3 +2x2 + 2x + 25 find the f(3).

Here the value of x is given as 3

q(3) = 33 + 2*32 + 2*3 +25

q(3) = 27 +18+ 6 + 25 In this step 3cube is calculated  as 27 and  3 square is 9 multiplied with 2 and add 25 to find the example function in math of q(3).

q(3) = 76 .

The examples functions in math of q(x) = x3 +2x2 + 2x + 25  is the q(3) = 76.

Length of Line Segment

The line segment is the straight line and it has two points. They are starting point and ending point. The starting point is in the starting place of the line and the ending point is in the ending place of the line. The length of line segment is the distance between the starting point and ending point of a line.

Diagram and formula - Length of line segment:

The formula to find the length of the line segment with two points (x1,y1) and (x2, y2) is

Distance d = `sqrt((x2 - x1)^2 + (y2 - y1)^2)`

Example problems - Length of line segment:

Find the length of line segment and the points are (1,1) and (2, 2).

Solution:

Given , (1,1) and (2, 2).

Let us take (1 ,1 ) as (x1, y1) and (2, 2) as (x2, y2).

The formula is Distance d = `sqrt((x2 - x1)^2 + (y2 - y1)^2)`

Now substitute the given values in the formula.

Distance d = `sqrt((2-1)^2 + (2-1)^2)`

=`sqrt(1^2 + 1^2)`

= `sqrt(1 + 1)`

= `sqrt(2)`

Find the length of line segment and the points are (1,1) and (3, 3).

Solution:

Given , (1,1) and (3, 3).

Let us take (1 ,1 ) as (x1, y1) and (3, 3) as (x2, y2).

The formula is Distance d = `sqrt((x2 - x1)^2 + (y2 - y1)^2)`

Now substitute the given values in the formula.

Distance d = `sqrt((3-1)^2 + (3-1)^2)`

= `sqrt(2^2 + 2^2)`

= `sqrt(4+4)`

= `sqrt(8)` .

= `sqrt(4 * 2)`

= `sqrt(4)` `xx` `sqrt(2)`

= 2 `xx` `sqrt(2)` .

Find the length of line segment and the points are (3,3) and (2, 2).

Solution:

Given , (3,3) and (2, 2).

Let us take (3 ,3 ) as (x1, y1) and (2, 2) as (x2, y2).

The formula is Distance d = `sqrt((x2 - x1)^2 + (y2 - y1)^2)`

Now substitute the given values in the formula.

Distance d = `sqrt((2-3)^2 +(2-3)^2)`

= `sqrt((-1)^2 + (-1)^2)`

= `sqrt(1 + 1)`

= `sqrt(2)` .

Find the length of line segment and the points are (2,2) and (5, 5).

Solution:

Given , (2,2) and (5, 5).

Let us take (2 ,2 ) as (x1, y1) and (5, 5) as (x2, y2).

The formula is Distance d = `sqrt((x2 - x1)^2 + (y2 - y1)^2)`

Now substitute the given values in the formula.

Distance d = `sqrt((5-2)^2 + (5-2)^2)`

= `sqrt(3^2 + 3^2)`

= `sqrt(9+9)`

= `sqrt(18)`

= `sqrt(9 * 2)`

= `sqrt(9)` `xx` `sqrt(2)`

= 3 `xx` `sqrt(2)`

Mean Difference Standard Deviation

Definition of mean difference:

Mean difference is defined as the measure of the difference between the given data set and mean value.For finding the mean difference first have to find the mean,

Formula for finding the mean,

`barx = (sumx) / n`

Using the mean value mean difference have to be found Formula for mean difference is,

`x-barx`

Definition of Standard Deviation:

Standard Deviation is the determination of describing the variability and spread of the Data set in the given total values in data set. It is used to take the measurement for the average of numbers in the given Data set. Standard Deviation given by the square root for the summation of the total squared mean difference and it is divided by the total number of values minus one.

Formula for standard deviation,

S =` sqrt(((sum(x - barx))) / (n-1))`

Steps for calculating mean difference and standard deviation:

Get the mean for the given n numbers in the given data set.
Get mean difference of each given numbers in the Data set from the mean.
Take Square for all each deviations. It is called as the squared mean deviation.
Calculate the summation for standard  mean deviations.
Now apply the Standard Deviation formula for finding the Standard deviation form the mean.


Mean difference standard Deviation - Example Problems:

Mean difference standard Deviation - Problem1:

Calculate the mean difference and standard deviation in the following data set.56, 52, 54, 57, 58.

Solution:

Mean:

Formula For finding the mean.

`barx = (sum x) / n`

` barx = (56+ 52+ 54+ 57+ 58) / 5`        

`barx = 277 / 5`

`barx = 55.4`

Mean difference is given below

x                             `(x - barx)`

56                         56 - 55.4 =   0.6

52                         52 - 55.4 =  -3.4

54                        54 - 55.4 =   -1.4

57                        57 - 55.4 =    1.6

58                        58 - 55.4 =    2.6


Standard Deviation: Standard Deviation is given by,

S =` sqrt(((sum(x - barx))) / (n-1))`

S = `sqrt(((0.6)^2+(-3.4)^2+(-1.4)^2+(1.6)^2+(2.6)^2 )/ (5-1))`

S = `sqrt(23.2 / 4)`

S = `sqrt(5.8)`

S = 2.40831892

Mean difference standard Deviation - Problem 2:

Calculate the mean difference and standard Deviation for the given data set. 23, 25, 24, 26.

Mean:

Formula For finding the mean.

`barx = (sum x) / n`

` barx = (23+ 25+ 24+ 26) /4`        

`barx = 98 / 4`

`barx = 24.5`

Mean difference is given below

x                                  `(x-barx)`

23                        23 - 24.5 =   0.5

25                        25 - 24.5 = - 0.5

24                        24 - 24.5 =   1.5

26                        26 - 24.5 = - 1.5

Standard Deviation: Standard Deviation is given by,

S =` sqrt(((sum(x - barx))) / (n-1))`

S = `sqrt(((0.5)^2+(-0.5)^2+(1.5)^2+(-1.5)^2)/ (4-1))`

S = `sqrt(5 /3)`

S = `sqrt(1.66666667)`



S = 1.29099445