Negative Binomial Function

The negative binomial function is having the other name called Pascal distribution. Suppose we are having a sequence of independent trials we have two possible outcomes. One is success and another one is failure. Normally success is denoted as P and failure is denoted as q = 1 – P. we can observe from the sequence we will get r number of failures which is a predefined one. Then the number of success in random is we have seen X, will have the Negative binomial function. It is denoted like the following
                                               X `~~` NB (r, p)

Properties of Negative geometric Distribution:

              Let us take an independent trials we will have the probability for success is r – 1 and we have the probability for failure is x. so the total number of trials is x + r – 1. Then the success is on the (x +r) ^th trials.  Let us see the binomial distribution based on the coin problem. We can define these using two parameters. Where P is the probability of getting head and k is the number of heads where we want to stop the flipping. Here we will see some basic properties of negative binomial function.
Property 1:Probability of mass function:
            The probability of the coin tossing n times before obtaining k heads where P _k (L = n)
                                             P_k {L = n} = `[[n - 1],[k - 1]]`p^kq^{n - k}
               with n = k, k + 1, k + 2, . . .
                          and                  `[[A],[B]]` = `(A!)/(B!(A - B)!)`
              It is the number of combinations of B among A.
Property 2: Mean
              Mean µ = `k / p`
              In negative binomial function the mean is k times of the geometric distribution mean for the value p.
Property 3:Variance:
            Variance σ² = `kq / p^2`
            Variance of the Negative Binomial function is also k times of the variance of geometric distribution.

Relation between Negative Binomial variables and geometric variables:

              Let us take the geometric independent variable G_i , the sum of the geometric variables are
                                      L = `sum` _i G_i     where i = 1, 2, . . . . k
           The negative binomial distribution is having the parameter p with the size k. This is used to provide the simple definition of negative binomial function mean and variance.

Unit Volume Converter

Definition of volume:

Volume is how much three-dimensional space a substance (solid, liquid, gas, or plasma) or shape occupies or contains, often quantified numerically using the SI derived unit, the cubic meters. The volume of a container is generally understood to be the capacity of the container, i. e. the amount of fluid (gas or liquid) that the container could hold, rather than the amount of space the container itself displaces.

(Source: Wikipedia)


Definition of unit converter:

Conversion factors between different units of measurement for the equal quantity is called as Unit converter or conversion.

Volume formulas:

Cube = side3
Rectangular Prism = side1 * side2 * side3
Sphere = (4/3) * `pi` * radius3
Ellipsoid = (4/3) *`pi` * radius1 * radius2 * radius3
Cylinder =`pi` * radius2 * height
Cone = (1/3) * `pi`* radius2 * height
Pyramid = (1/3) * (base area) * height
Torus = (1/4) *`pi` 2 * (r1 + r2) * (r1 - r2)


Example problem for unit volume converter:

Example 1:

Find the volume for the given cylinder with radius = 22 cm and height = 30 cm.



Solution:

Volume formula for cylinder

V = `pi` r2h

Here `pi`= 3.14 or 22/7

Given

Radius r = 22 cm

Height h = 30 cm

V = 3.14 * 222 * 30

= 3.14 * 22 * 22 * 30

= 3.14 * 484 * 30

= 3.14 * 14520

= 45592.8 cm3.

Solution to the volume of the given cylinder is 45592.8 cm3.

Different unit converter for volume:

Here we convert the cubic centimeters into different units such as centiliters, deciliters, Liters, milliliters and barrels.

45592.8 cm3 = 4559.28 centiliters

45592.8 cm3 = 455.928 deciliters

45592.8 cm3 = 45.5928 L

45592.8 cm3 = 45592800 milliliters

45592.8 cm3 = 0.28677 barrels.

Example 2:

Find the volume for the given cone with radius = 12 m and height = 16 m.



Solution:

Volume formula for cone

V = (1/3) *`pi` * radius2 * height

Given

Raidus r = 12 m

Height h = 16 m

Here we use `pi` = 3.14

= (1/3) * 3.14 * 122 * 16

= `(1)/(3)`* 3.14 * 12 * 12 * 16

= `(1)/(3)` * 3.14 * 144 * 16

= `(1)/(3)` * 3.14 * 2304

= `(1)/(3)` * 7234.56

= 2411.52 m3 .

Solution to the volume for the given cone is 2411.52 m3


Different unit converter for the given cone volume:

Here we convert the cubic meters into different units such as centiliters, deciliters, Liters, milliliters and barrels.

2411.52 m3 = 241152000 centiliters.

2411.52 m3 = 2411520000 Liters.

2411.52 m3 = 24115200 deciliters.

2411.52 m3 = 2411520 milliliters.

2411.52 m3 = 15168.004469 barrels.

Circular Functions in Trig

In mathematics, the trigonometric functions (also called circular functions) are functions of an angle. They are used to relate the angles of a triangle to the lengths of the sides of a triangle. The most familiar trigonometric functions are the sine, cosine, and tangent. The sine function takes an angle and tells the length of the y-component of that triangle (Source: Wikipedia)

Circular function in trig

Right triangle for all time contain a 90° angle, now denoted at C. Angles A and B might differ. Trigonometric functions identify the relations between side lengths also interior angles of a right triangle.
Illustrate the trigonometric functions used for the angle A.
Establish through either right triangle to include the angle A.
Three sides of the triangle are,
Adjacent side:
It is the side specifically in contact through both the angle we are concerned within (angle A).
Opposite side:
It is the side opposite toward the angle we are concerned within (angle A)
Hypotenuse side:
It is the side opposed the right angle. The hypotenuse is the highest part of a right-angled triangle.



Trigonometric functions:
 Sin A = opposite / hypotenuse.
Cos A = adjacent/ hypotenuse
Tan A = opposite /adjacent
Csc A =1/Sin A= hypotenuse/ opposite
Sec A =1/Cos A= hypotenuse/ adjacent
Cot A =1/Tan A adjacent / opposite
The circular function has the following equations
Let u = x + y and v = x - y. Then the three equations yields the sums otherwise differences
sin(u) + sin(v) = 2 sin(`(u + v) / 2` ) cos(`(u - v) / 2` )
cos(u) + cos(v) = 2 cos(`(u + v) / 2` ) cos(`(u - v) / 2` )
cos(v) - cos(u) = 2 sin(`(u + v) / 2` ) sin(`(u - v) / 2` )

Examples for circular function in trig

Example 1 for circular function:
Apply the tangent ratio toward solve the unidentified side of the triangle?
Given angle of triangle are 40⁰ and opposite side of triangle are 16.
Solution:
Specified to angle of triangle is 40⁰ and opposite side of triangle is 16. Find out the adjacent side
 tan 40⁰ = opposite/adjacent
tan 40⁰ x = 16
x = 16/tan 40⁰
x =16/0.8391     {since the value of tan 40 degree is 0.8391}
x=19.06
The value of adjacent side =19.06
Example 2 for circular function:
If u=30 and v=30 then find out the sin (u) + sin (v)?
Solution:
We know the formula,
sin (u) + sin(v) = 2 sin(`(u + v) / 2` ) cos (`(u - v) / 2` )
                         =2 sin(`(30+30)/2` ) cos(30-30)
                         =2 sin(60/2) cos 0
                         =2 sin30 cos 0
                         =2x(1/2)x1
 sin (u) + sin(v) =1

Triangular Pyramid Geometry

In geometry, a pyramid is a polyhedron formed by connecting a polygonal base and a point, called the apex. Each base edge and apex forms a triangle. It is a conic solid with polygonal base. A pyramid with an n-sided base will have n + 1 vertices, n + 1 faces, and 2n edge.
Source from wikipedia)

geometry triangular pyramid:

A triangular pyramid is a pyramid having a triangular base.Here we are going to study about triangular pyramid in geometry and example problems.

Formulas:

Surface area of triangular pyramid = A + `(3/2)` s*l
Here A = area of base = `1/2` * a*s
volume of triangular pyramid = `(1/6)` a*b*h

Here,

a represent the apothem

s,b is the sides of the pyramid

l is the slant height

h is the height of the pyramid.

Triangular pyramid geometry - example problems.

Example: 1

Find the surface area of the triangular pyramid with side 7 meter ,slant height is 9 meter and apothem length is 8 meter.

Solution:

We know the formula for surface area of the triangular pyramid = A + `(3/2)` sl

First we have to find the area of base A = `1/2` * a*s

Here a = 8 meter, s = 7 meter substitute the above formula we get

A = `1/2 ` * 8 * 7

A = `1/2` *56

A = 28 meter square

Now we find the surface area of the triangular pyramid

S.A = 28 + `(3/2)` * 7 * 9

= 28 + `(3/2)` *63

Simplify the above we get

= 28 + `189/2`

= 28 + 94.5

= 122.5 meter square

Therefore the surface area of the triangular pyramid = 122.5 meter square


Triangular pyramid geometry - example: 2

Find the volume of the triangular pyramid with height is 11 meter ,base is 9 meter and apothem length is 12 meter .

Solution:

We know the formula for volume = `(1/6)` abh

Here h = 11 meter, b = 9 meter, a = 12 meter substitute this value into the formula we get

Volume =  ` (1/6)` 12 * 9 * 11

Volume = `(1/6) ` 1188

Simplify the above we get

Volume = 198 meter3

Therefore the volume of the triangular pyramid is 198 meter3

Mixed Number Multiplication

Mixed number is a combination of a whole number and a proper fraction. A mixed number can be converted into an improper fraction and vice versa. All of the arithmetic operations can be performed by using the mixed numbers. For example, 2 (1)/(2) is a mixed numbers. In this example, the 2 is called as the whole number and (1)/(2) is called as the improper fraction

It will be easier to perform arithmetic operations on mixed numbers by first converting them to improper fractions

Rules for the multiplication of mixed number

There are many rules are followed for the multiplication of mixed numbers, They are given as follows,

In the first step, we have to convert the given mixed number into the improper fraction.
In the next step, we have to perform the multiplication for the numerator part
In the next step, we have to perform the multiplication in the denominator part
Then, the next step we have put this in an improper fraction.
Then the next step is simplify those improper fractions.
Then the result is converted into a mixed number.


Example problem for the mixed number multiplication

Problem 1: Multiply the given mixed numbers, 2 (4)/(6) with 1  (6)/(9) .

Solution:

Step 1: Write the mixed numbers

2(4)/(6) xx 1 (6)/(9)

Step 2: Convert the given mixed numbers into improper fractions

(16)/(6) xx (15)/(9)

Step 3: Perform the multiplication operation for the numerator terms,

16 xx 15 =240

Step 4: Perform the multiplication operation for the denominator terms,

6 xx 9 =54

Step 5: The result obtained for step 3 and 4 can be combined to an improper fractions

(240)/(54)

= (120)/(17)

This is the required solution for the multiplication of mixed numbers.

Problem 2: Multiply the given mixed numbers, 4 (2)/(7) with 7  (2)/(8) .

Solution:

Step 1: Write the mixed numbers

4 (2)/(7) xx 7 (2)/(8)

Step 2: Convert the given mixed numbers into improper fractions

(30)/(7) xx (56)/(8)

Step 3: Perform the multiplication operation for the numerator terms,

30 xx 56 = 1680

Step 4: Perform the multiplication operation for the denominator terms,

7 xx 8 = 56

Step 5: The result obtained for step 3 and 4 can be combined to an improper fractions

(1680)/(56)

= (840)/(28)

=(420)/(14)

=(210)/(7)

= 30

This is the required solution for the multiplication of mixed numbers.

Differentiate Variable

On the definition of the derivative is a dynamical diagram displaying the derivative as the slope of the tangent to a graph. The necessary preliminaries one should be familiar with are the slope of a straight line and the graph of a function.

When we differentiate any equation we denote them using different variables .these variables can be one or many depending on the equation. These variables are alphabets like x, y, t, u, v,  etc.


Examples of differentiate variables

There are many examples of differentiation. Such as:

Differentiate xx  + (cos x)1/2 with respect to x

Let y =  xx  + (cos x)1/2

We have + or – sign in between the terms, then we should not tale log.

We should write y = u + v

dy/dx = du/dx + dy/dx                                            …..(i)

consider  u = xx

taking log on both sides, we get

log u = x. log x

differentiating with respect to x, we get

1/u . du/dx = x.d/dx (log x) + log x.dx/dx

1/u. du/dx = x.1/x + log x.1

du/dx = u(1 + log x) = x2( 1 + log x)                 …..(ii)

again consider  v = (cos x)1/x

taking log on both sides, we get

log v = 1/x.log(cos x)

differentiating with respect x, we get

1/v.dv/dx = 1/x.d/dx log (cos x) + log(cos x).d/dx(1/x)

1/v.dv/dx = 1/x. 1/ cosx. (-sin x) + log(cos x). (-1/x2)

dv/dx = v[- tanx /x – 1/x2. Log (cos x)]

= (cos x)1/x[- tan x/x – 1/x2.log(cos x)]                    ……(iii)

Substituting from (ii) and (iii) in (i), we get

dy/dx = x2(1+ log x) + (cos x)1/x [-tanx/x – 1/x2. Log(cos x)]

here the variable used are x, y , u and v, and these are called the differentiate variables.


Problems on differentiate variables

Differentiate , y = (x)cos x + (sin x)tan x
y= x.sin x.log x.

Interval Notation Inequality

Interval notation inequality is a process of writting down the set of numbers. Generally interval notation inequalities is used to describe a limit of span or group of spans of numbers along a axis. However, this notation can be followed to describe any group of numbers. For the reasonable example is considered, the set of numbers that are all the numbers greater than 7. The interval notation inequality for this set, letting x be any number in the group of notation, then we it would be given in the notation as shown below,


X >7

Approaches for interval notation inequality:

This similar set possibly will be described in another type of notation is called interval notation inequality. For the collection of numbers that would be written as,

(7,+`oo` )

The interpret notation will be written as:

The span of numbers which as well as in the group is often imagined as being on a number line, normally the x-axis.The 7 on the left then the set of numbers starts at the real number which is suddenly to the right of 7 on the number line. It means we should imagine a number the tinniest bit greater than 7, and that is where the group of numbers begins from the opening stage.
So that the parenthesis to the left of 7 is called a round bracket or an exclusive bracket. Then, 7 is excluded from the group, but the numbers straightly to the right of 5 are included. Shortly put, the numbers greater than 7 are included. The group of interval notation numbers continues to include values greater than 5 all the way to a value which is infinitely greater than 7. That is, the set of all numbers continuously all the way to positive infinity. That is what the positive infinity symbol on the right means, then Infinity symbols are at all times accompanied by round brackets.


Practice problem for interval notation inequality:

Problem 1:

2 < x < 7

The entire numbers between positive two and positive seven, together with the two and the seven.  2 < x < 7,  x is less than or equal to 7 and greater than or equal to 2" { x | 2 < x < 7} "x is between 2 and 7, inclusive" [2,7]

Problem 2:

5 < x < 9

The entire numbers between positive five and positive nine, as well as the five and the nine.  5 < x < 9 x is less than or equal to nine and greater than or equal to five {x | 5 < x < 9} "x is between 5 and 9, inclusive" [5, 9].