Definition of Probability Distribution

Probability distribution function referred as p(x) which is a function that satisfies below properties:

Probability that x can obtain particular value is p(x).

p[X,x]=p(x)=px

p(x) positive for all real x.
Sum of p(x) over all feasible values of x is 1, that is

`sum_(pj)pj=1`

Where, j- all possible values that x can have and pj is probability at xj.

One significance of properties 2,3 is that 0 <= p(x) <= 1.

Steps used for how to determine probability distribution:

Using following steps used to how determine probability distribution:

Step 1 for how to determine probability distribution:

Plot the data for a image illustration of the data type.

Step 2 for how to determine probability distribution:

First steps is to determining what data distribution one has - and thus the equation type to use to model the data - is to rule out what it cannot be.

The data sets cannot be a discrete uniform distribution if there are any crest in the data set.
The data is not Poisson or binomial if the data has more than one crest.
If it contains a single arc, no secondary crests, and contains a slow slope on each side, it may be Poisson or a gamma distribution. But it is not discrete uniform distribution.
If the data is regularly distributed, and it is without a slant in the direction of one side, it is secure to rule out a gamma or Weibull distribution.
If the function has an even distribution or a crest in the center of the graphed outcomes, it is not a geometric distribution or an exponential distribution.
If the incidence of a factor differ with an environmental variable, it probably is not a Poisson distribution.


Step 3 for how to determine probability distribution:

After probability distribution type has been tapering downward, do an R squared examination of each probable type of probability distribution. The one with the maximum R squared value is most possible correct.

Step 4 for how to determine probability distribution:

Remove one outlier data point. Now recalculate R squared. If the same probability distribution form comes up as the neighboring match, then there is high confidence that this is the correct probability distribution to use for group of data.

From above steps we can understand how to use probability distribution.

Polynomial Exponent

Polynomial of a single term is called as Monomial. If the Polynomial has the two terms Sum or Difference then it is called a Bi-nomial, the Sums or Differences of a three-term Polynomial is called a Trinomial. The Sums and/or Differences of polynomial of four or more terms are simply called polynomial.

Two or more terms of a Polynomial exponent that has the equivalent variable and precisely the same whole number polynomial exponent are called Like Terms. For suitable example the terms; 6X3 and -7X3, are Like Terms, since the variable X is the similar variable in both terms, and the Whole number exponent, 3, is absolutely the similar polynomial exponent in both terms. The terms;4X3 and 4X4 are not Like Terms although the variable X is the same in both terms but the Whole number polynomial exponent are completely different from other.

Operation of polynomial exponent:

Addition of polynomial exponent:

Two or more polynomials, adding the terms,

Suitable example adding polynomial exponent,

Example1:

(2x2+3x3)+(x2+7x3)

=2x2+x2+3x3+7x3

The variable and exponent must be same then we add the polynomial exponent,

=3x2+10x3

So the result is  =3x2+10x3

Subtraction of polynomial exponent

Example2:

(3x2+3x3)-(x2+7x3)

=3x2-x2+3x3-7x3

The variable and exponent must be same then we subtract the polynomial exponent,

=2x2-4x3

So the result is =2x2-4x3

Multiplication of polynomial exponent:

Two or more polynomial, so multiply their exponent terms. For reasonable example to multiply the following two Terms, 2X2 and 7X2, Then the following terms then first multiply the coefficients (2) (7) and we multiply (X2) (X2) which can be expressed as the following term 14X4, that is, when we multiply variables that are the same.


Different variables in polynomial exponent:

The variables are different and the exponent also different, so we can write the variables and exponents adjoining to each other.

For Example, (7X2)(3Y2)is equal to 21X2Y2.

So the result is 21X2Y2

Similar to all the different variable, polynomial exponent.

To multiply the two polynomial exponent, for example (X-2Y)(X-2Y), then the result is x2-2xy+4y2 . Here we take the first term X of the first Binomial and multiply each term in the second Binomial, then we take the second term in the first Binomial and multiply each term in the second Binomial, and then we add all the terms. Similar to all the polynomial categories, for example trinomial, binomial exponents.

Inverse Trigonometry Definition

The three trigonometry functions are arcsin(x), arccos(x), arctan(x). The inverse trigonometry function is the inverse functions of the trigonometry, written as cos^-1 x , cot^-1 x , csc^-1 x , sec^-1 x , sin^-1 x , and tan^-1 x .

Sine:

H (x) = sin(x)   where   x is in [-pi/2 , pi/2]

Cosine:

G (x) = cos(x) where   x is in [0 , pi ]

Tangent:

F (x) = tan(x)   where   x is in (-pi/2 , pi/2 )

Understanding Inverse Trig Function is always challenging for me but thanks to all math help websites to help me out.

Cosecant Definition

Csc theta = (hypotenuse)/(opposite)

Secant Definition

Sec theta = (hypotenuse)/(adjacent)

Cotangent Definition

Cot theta = (adjacent)/(opposite)

Examples of Inverse trigonometry functions:

Example 1:

Find the angle of x in the below diagram. Give the answer in four decimal points.

Solution:

sin x =2.3/8.15

x = sin -1(2.3/8.15 )

= 16.3921˚

Example 2:

Solve sin(cos-1 x )

Solution:
Let z = sin ( cos-1 x ) and y = cos-1 x so that z = sin  y. y = cos-1 x may also be marked as


cos y = x with pi / 2 <= y <=- pi / 2

Also

sin2y + cos2y = 1

Substitute cos y by x and solve for cos y to obtain

sin y = + or - sqrt (1 - x^2)

But pi / 2  <= y <= - pi / 2 so that cos y is positive

z = sin y = sin(cos-1 x) = sqrt (1 - x^2)

Example 3:

Evaluate the following sin-1( cos ((7 pi) / 4 ))

Solution:
sin-1( cos ( y ) ) = y only for  pi / 2 <=  y <= -pi / 2 . So we initial transform the given expression noting that cos ((7pi) / 4 ) = cos (-pi / 4 ) as follows

sin-1( cos ((7 pi) / 4 )) = sin-1( cos ( pi / 4 ))  - pi / 4  was selected since it satisfies the condition  pi / 2 <=  y <= - pi / 2 . Hence

sin-1( cos ((7 pi) / 4 )) =pi/4

Example 4:

With the given value find inverse of tan.

Solution:

Since tangent, again, is the trig function associated with opposite and adjacent sides, we apply the inverse of tangent to calculate the measure of this angle.

Tan 16 = (opposite)/(adjacent)

Tan 16 = ((bar(AC))/(bar(CB)))

Tan-1(7/24 ) = 16 o

Associative Property Sum

In mathematics, if x, y, z be any three variables involving the addition operation and then satisfy the following condition,

x + (y + z) = (x + y) + z

This kind of property is called associative property sum or addition. In associative property sum, the sum of algebraic expression provides the same result in changing the order of brackets in this expression. Such as

x + y + z = (x + y) + z = x + (y + z)



Examples for associative property sum:

Example 1 on associative property sum:

Prove the associative property sum of the given expression.

(a) 4 + 6 + 7

Solution:

(a) Let x, y, and z be 4, 6 and 7 respectively.

Therefore, the given expression satisfy the following condition,

x + y + z = x + (y + z) = (x + y) + z

Here,

Prove the following expression

4 + (6 + 7) = (4 + 6) + 7

L.H.S = 4 + (6 + 7)        (first perform the addition operation within the bracket from left to right,

because brackets are higher priority of the order of operations)

= 4 + (13)      (second perform the addition operation outside of the expression from left to right)

= 17                    ------ (1)

R.H.S = (4 + 6) + 7

= (10) + 7

= 17             --------- (2)

From equation (1) and (2) we get,

L.H.S = R.H.S

i.e., 4 + (6 + 7) = (4 + 6) + 7

Hence, the given expressions satisfy the associative property with sum of numbers.

Example 2 on associative property sum:

(a) 23 + 67 + 34, prove the associative property of sum or addition.

Solution:

(a) Let m, n, and l be 23, 67 and 34 respectively.

Therefore, the given expression satisfy the following condition,

m + n + l = m + (n + l) = (m + n) + l

Here,

Prove the following expression

23 + 67 + 34 = 23 + (67 + 34) = (23 + 67) + 34

23 + 67 + 34 = 124       ---------------- (1)

We consider the remaining expression, such as

23 + (67 + 34) = (23 + 67) + 34

L.H.S = 23 + (67 + 34)

= 23 + (101)

= 124                    ------ (2)

R.H.S = (23 + 67) + 34

= (90) + 34

= 124                  --------- (2)

from equation (1) , (2) and (3) we get,

L.H.S = R.H.S = 124

i.e.,  4 + (6 + 7) = (4 + 6) + 7 = 4 + 6 + 7

Hence, the given expression of sum is satisfied the associative property.

Weight of the Cylinder

Cylinder is the three dimensional figure. Weight of the cylinder is same as the volume of the cylinder. The formula to find the weight of the cylinder is pi * radius2 * height. Height is the total height of the cylinder and radius is the radius of the circular face which is at the bottom and top of the cylinder.

Example diagram and formula - Weight of the cylinder:

Formula - Weight of the cylinder:

Weight of the cylinder = pi * r2 * h, where r`=>` radius of the cylinder and h`=>` height of the cylinder.

Example problems – Weight of the cylinder:

Calculate the weight of the cylinder whose radius is 10cm and height is 12 cm.

Solution:

Given, radius of the cylinder = 10cm and height of the cylinder= 12cm.


The formula to find the weight of the cylinder is pi * r2 * h

= 3.14 * 102 * 12

= 3.14 * 100 * 12

= 3768cm3.

The weight of the cylinder is 3768cm3 .

Calculate radius of the cylinder whose weight is 3000cm3 and height is 30cm.

Solution:

Given weight of the cylinder= 3000cm3 and height of the cylinder = 30cm


Weight of the cylinder = pi * r2 * h

3000   = 3.14 * r2 * 30

3000   = 94.2 * r2.

Divide the above expression by 94.2 on both sides,

3000/94.2 = 94.2 * r2/94.2

31.85        = r2

r               = 5.6cm.

Radius  of the cylinder    = 5.6cm.

Calculate the weight of the cylinder whose radius is 5cm and height is 2 cm.

Solution:

Given, radius of the cylinder= 5cm and height of the cylinder= 2cm.


The formula to find the weight of the cylinder is pi * r2 * h

= 3.14 * 52 * 2

= 3.14 * 25 * 2

= 157cm3.

The weight of the cylinder = 157cm3 .

Calculate radius of the cylinder whose weight is 300cm3 and height is 3cm.

Solution:

Given weight of the cylinder= 300cm3 and height of the cylinder= 3cm


Weight of the cylinder = pi * r2 * h

300   = 3.14 * r2 * 3

300   = 9.42 * r2.

Divide the above expression by 9.42 on both sides,

300/9.42 = 9.42 * r2/9.42

31.85      = r2

r             = 5.6cm.

Radius of the cylinder   = 5.6cm.

Percent Loss Formula

Loss is defined as the difference between the cost price and the selling price. Through the loss we can calculate loss percentage. If they give loss percentage we also find the cost price and the selling price. Let us see the formula and examples for loss and loss percent in this article.

Formula for Loss Percentage

Variation between the cost price and the selling price is called as loss, If the cost price is greater than the selling price means we obtain loss.

Loss = cost price- selling price

Selling price is indicated by S.P and cost price is indicated by C.P.

Based on the cost price we determine the loss.

Formula for loss Percent: Loss Percentage formula = (lossxx 100)/(C.P)

Sometimes they give loss percentage and ask cost price or selling price.

Example on Loss Percent

Example 1: Ragu buys pineapple at 5 for 50Rs ech and sold them at 3 for 25 each. Find his loss and loss percent?

Solution: Given

Number of pineapple bought = lest common divisor of 5,3 =15
Investment or cost price = (50 / 5 ) *15 =150
Selling price = (25 /3)  *15 = 125, here cost price is greater then selling price,we can say it is loss.
Loss amount =cost price – selling price
Loss =150 – 125
Loss amount =25
Loss percentage formula = (lossxx 100)/(C.P)
Loss percentage = (25 xx 100)/(150)
Loss percentage = 16.6%
Loss and loss percentage of Ragu were 25 Rs and 16.6%

Example 2: Cost price =50 Rs, Selling price =45 Rs then find the loss and loss percentage?

Solution: Given

Cost price =50
Selling price =45
Loss = cost price - selling price
Loss =50- 45
Loss =5 Rs.
Loss percentage formula = (lossxx 100)/(C.P)
Loss percentage = (5xx 100)/(50)
Loss percentage = 10%

Example 3: Deepen loss 80 cents on $80 .What is the loss percentage of Deepen?

Solution: Given

Loss 80 paisa on 80 rupees
Here Cost price = $80
Loss =80 cents
Loss percentage formula = (lossxx 100)/(C.P)
Loss percentage = (0.8 xx 100)/(80)
Loss percentage = 1 %

Learn Online Linear Equations

A linear equation is an algebraic equation in which each term is either a constant or the product of a constant and (the first power of) a single variable. Linear equations can have one or more variables. Linear equations occur with great regularity in applied mathematics. Source - Wikipedia

Let us Learn linear equation in online. In online we can understand very easily about linear equation.

Learning Concepts of Linear equations in online:

Learning Concepts of Linear equations are as follows:

• In the graph showed, we would like to get the equation of the graphed line. To complete this we start linear equation. We have innocently been functioning with linear equation, so this is NOT a fresh concept for us Most probable you have been working with the standard form of a line, which is Ax + By = C. This is a linear equation with two unfamiliar variables, x and y.

• The ordinary form of the line in the graph is 2x + y = 5, where A =2, B = 1, and C = 5.

• We want to restructure this equation and set it into the y-intercept form which is y = mx + b, in which m is the slope and b is the y-intercept. In this shape, it is very simple to graph. Just explain the standard equation for “y” and you have the y- intercept form.

Example problems for learning linear equations in online:

Example problems for learning linear equations are as follows:

Example 1:

x - 4 = 10

Add 4 to both sides of the equation:

x = 14

The answer is x = 14

Make sure the resolution by substituting 14 in the original equation for x. If the left side of the equation generation the right side of the equation after the substitution, you have found the correct answer.

Example 2:

2x - 4 = 10

Add 4 to both sides of the equation:

2x = 14

Divide both sides by 2:

X = 7

The answer is x = 7.

Algebra is widely used in day to day activities watch out for my forthcoming posts on Unit of Length and Mean Value Theorem Examples. I am sure they will be helpful.

Check the solution by substituting 7 in the original equation for x. If the left side of the equation generation the right side of the equation after the substitution, you have found the correct answer.

2(7) - 4 = 14 - 4 = 10.