Six Grade Probability

Introduction to Six Grade Probability:

Probability is a way of expressing knowledge or belief that an event will occur or has occurred. In mathematics the concept has been given an exact meaning in probability theory, that is used extensively in such areas of study as mathematics, statistics, finance, gambling, science, and philosophy to draw conclusions about the likelihood of potential events and the underlying mechanics of complex systems.

(Source: Wikipedia)

Example Problems for Six Grade Probability:

Six grade probability – Example: 1

Two coins are tossed. Find the probability of getting two heads.

Solution:

Step 1:

n (s) = {HH, HT, TH, TT}= 4

Step 2:

Tossing a coin with two heads:

n (a) = {HH}= 1

Step 3:

Formula:

P (A) = n(a)/n(s)

Answer:

P (A) = 1/4

Six grade probability – Example: 2

Three coins are tossed. Find the probability of getting two tails.

Solution:

Step 1:

n(s) = {TTT, TTH, THT, THH, HTT, HTH,HHT,HHH }= 8

Step 2:

There are 3 tosses with only two tails:

n (a) = { TTT, TTH, THT, HTT}=4

Step 3:

Formula:

P (A) = n(a)/n(s)

Answer:

P (A) = 4/8

P(A) = 1/2.

Six grade probability - Example: 3

When the die is rolled. What is the probability of occurring five?

Solution:

Total Number of possible = n (s) = {1, 2, 3, 4, 5, 6}

n (s) = 6

The number of outcomes n (a) = {5}

n (a) = 1

Formula:

P (A) = n(a)/n(s)

Therefore the probability of getting value = 1/6.

Six grade probability – Example: 4

When the two dice are rolled. What is the probability of occurring four or four?

Solution:

Total Number of possible = n (a) = { 1,1}{1,2}{1,3}{1,4}{1,5}{1,6}

{2,1}{2,2}{2,3}{2,4}{2,5}{2,6}

{3,1}{3,2}{3,3}{3,4}{3,5}{3,6}

{4,1}{4,2}{4,3}{4,4}{4,5}{4,6}

{5,1}{5,2}{5,3}{5,4}{5,5}{5,6}

{6,1}{6,2}{6,3}{6,4}{6,5}{6,6}

n (s) = 36

The number of outcomes n (a) = {1,4}{2,4}{3,4}{4,1}{4,2}{4,3}{4,4}

{4, 5}{4, 6}{5, 4}{6, 4}

n (a) = 11

Formula:

P (A) = n(a)/n(s)

Therefore the probability of getting value = 11/36. I have recently faced lot of problem while learning what is an acute angle, But thank to online resources of math which helped me to learn myself easily on net.

Practice Problems for Six Grade Probability:

1. Three coins are tossed and find the probability of all heads.

[Answer: P(A) = (1)/(8)]

2. Two dice are rolled then What is the probability of occurring six or six?

[Answer: (11)/(36)]

Metric System Distance

Introduction to metric system distance:
The distance is the unit used to represent how far two objects are apart. There are various units for the measurement of the distance like the international standard unit, metric unit, and other conventional units. The metric system of the distance measurement is one among them and the lengths are represented as the 10th multiples of the base unit. In the following article we will discuss in detail about the metric system of distance and the units in the metric system of distance. I like to share this Distance Time Formula with you all through my article.

More about Metric System Distance:
As described before, in the metric system of the distance the units of the distance are represented as the 10th multiples of the base unit in the metric system of distance. The base unit in the metric system of distance is the meter. The various units of the distance measurement in the metric system are,

1 Kilometer = `1000` meters

1 Hectometer = `100` meters

1 Decameter = `10` meters

1 Decimeter = `1/10` meters

1 Centimeter = `1/100` meters

1 Millimeter = `1/1000` meters

The above metric system distance relations can also be used for the conversion from the meters to the other units by reversing the above relations. Please express your views of this topic Frequency Polygon by commenting on blog.

Example Problems on Metric System Distance:

1. Convert the distance of 1.26 meters into centimeters and decimeters.

Solution:

1 Centimeter = `1/100` meters

1 meter = 100 centimeters

1.26 meters = 1.26*100 centimeters

1.26 meters = 126 centimeters

1 Decimeter = `1/10` meters

1 meter = 10 decimeters

1.26 meters = 1.26*10 decimeters

1.26 meters = 12.6 decimeters

2. Convert the distance of 0.25 kilometers and 0.75 decameters into meters.

Solution:

1 Kilometer = 1000 meters

0.25 Kilometer = 0.25*1000 meters

0.25 Kilometer = 250 meters

1 Decameter = 10 meters

0.75 Decameter = 0.75*10 meters

0.75 Decameter = 7.5 meters

Practice problems on metric system distance:

1. Convert the given distance of 14.56 decimeters into meters.

Answer: 145.6 meters

2. Convert the given distance of 0.98 kilometers into meters.

Answer: 980 meters.

3. Convert the given distance of 15600 millimeters into meters.

Answer: 1.56 meters

Standard Exponential Form

Introduction to Standard Exponential Form

Notation : `a^b` , where a is known as base and b is known as exponent which must be a positive and integral value. I like to share this Exponential Function Definition with you all through my article.

Explanation of `a^b` : The standard exponential form, `a^b` can mathematically be expressed as `a`   multiplied by itself `b` times. That is to say

`axxaxxaxxaxx.... b "times"`. This yeilds a single number if we know the value of a and b.

eg. Assuming a = 2 and b = 3 , we can express `a^b` = `2^3` = `2xx2xx2`  = 8(as b = 3 so we multiply a = 2 with itself 3 times). Note that the answer 8 results in single number.

Sample Examples of the Exponential Form:

1. `3^2` = `3xx3` = `9` (3 multiplied by itself 2 times)

2. `4^3` = `4xx4xx4` = `64` (4 multiplied by itself 3 times)

[Note: Here in standard exponential form, if we assume a and/or b are variables, they do not yeild any number. Instead they remain in the variable format. like

3. `x^3` = `x xx x xx x` (`x` multiplied by itself 3 times)

4. `x^a` = `x xx x xx x xx ... ` (`x` multiplied by itself a times) ]

[ Notes 2. If we take `b` as a decimal value such as 0.3, it comes under the section of nth root of a number, while here we are discussing how to compute standard exponential form ]

Abstracts of Standard Exponential Form:

Following are the rules and abstracts to calculate expressions involvint standard exponential forms.

1. `(a^b)^c` = `a^(bc)`

eg. a. `(2^3)^2` = `2^(3xx2)` = `2^6` = `2xx2xx2xx2xx2xx2` = 64

b. `(3^2)^4` = `3^(2xx4)` = `3^8` = `3times3times3times3times3times3times3times3`

Abstracts of Standard Exponential Form (continued)

2. `a^b xx a^c` = `a^(b+c)` (constraints : Both exponential form must have base that are equal)

`a^b//a^d` = `a^(b-d)` (For multiplication, both exponents are added, while for division, exponent for numerator is subtracted by exponent of denominator.Please express your views of this topic 7th grade math problems online by commenting on blog.

Examples :

1. `5^2xx5^3` = `5^(2+3)`= `5^5` = 3125

2. `3^5//3^2` = `3^(5-2)` = `3^3` = 27

3. `a^(-c) = 1/a^c` (when exponent is a negative number, one can make it positive by reciprocating the expression.)

such as `2^-3` = `1/2^3` = `1/8`

Example Problems on Standard Exponential Form

Can you compute these expressions

`3^5 = `
`x^3=`
`x^b=`
`(2^2)^4 = `
`2^5 xx 2^3 = `
`2^8//2^5 =`
`2^-3 = "(convert to positive exponential form)"`
Compute `(2^5xx2^7)/2^4`

Distributive property

The Distributive Property is one of the number properties. It says that when number is multiplied to an addition of two or more numbers, the result is the same as the sum of the products of the same number and each addend. To define distributive property algebraically, it is expressed in formula form as, a*(b + c) = a*b + a*c. That is, the number is ‘distributed’ to each of the addend and then the addition can be done. Thus, it can also be referred as distributive property of addition.But one must clearly understand that while a*(b + c) = a*b + a*c is true, a/(b + c) is ?  (a/b) + (a/c). Hence to stress this point some emphatically refer this property as ‘Distributive Property of Multiplication over Addition’.

Thus in general, the definition of distributive property is when a term is multiplied to the sum of group of terms, then the result is same as the sum of the products of the first term with each of the terms of the sum. It may be noted this property is applicable to subtraction of terms also, because in the general formula any of the three ‘a’. ‘b’ and ‘c’ can also be negative.

The distributive property greatly helps calculations and provides easier methods of solution. For example I need to multiply 51*101. If one tries the actual multiplication he/she has to take a paper and pencil and do the work. But the easiest way is by applying the property we discussed and it may be amazing to note that you can find the answer by mental calculation!. That is,51*101 = 51*(100 + 1) = 51*100 + 51*1 = 5100 + 51 = 5151 which is as fast as a calculator. Please express your views of this topic cbse 10th question papers by commenting on blog.

This property is also a great tool in factorization. In such cases, we use the property the other way round. That is we do, a*b + a*c = a*(b + c). For example let us take a quadratic expressionx2 + 6x + 8. Let us study how the distributive property helps in factoring.
Splitting the middle term, x2 + 6x + 8 = x2 + 4x + 2x + 8
Identifying the common factor x in the first two terms and the common factor 2 in the last two terms and using this property, x2 + 4x + 2x + 8 = x(x + 4) + 2(x + 4)
Again finding a common factor (x + 4) and applying the property once more,
x(x + 4) + 2(x + 4) = (x + 4)(x + 2)

Poisson Distribution Variance Help

Introduction to Poisson Distribution Variance Help:

In Poisson distribution, x is called as discrete random variable. Lambda is called as the Poisson distribution parameter in a Poisson experiment. In affixed period of time interval, the number of events is occurred when the events happened in random with separate time and constant rate. While lambda is huge, the Poisson distribution is related to the standard normal distribution with lambda be the occurrence rate of proceedings per unit time. Let us see about Poisson distribution variance help in this article.

General Formula for Poisson Distribution Variance Help:
Formula for Poisson distribution problem is:

Formula:

`Poisson distribution =` `(e^(-lambda) lambda^(x))/(x!)`

Where,

x means Poisson value
`lambda` means rate of change
e means log funct ion


Variance of Poisson Distribution:

In poisson distribution, `lambda` represents the variance of Poisson distribution.

Variance Formula

`V(X) = sigma^(2) = lambda`

Please express your views of this topic Multivariate Regression Analysis by commenting on blog.

Worked Examples to Poisson Distribution Variance Help:
Example 1 for Poisson Distribution Variance Help

Solving the variance of Poisson distribution if `P(X = 2) = P(X = 3)` and also find `P(X = 6)`

Solution

Given `P(X = 2) = P(X = 3)`

Therefore, we have `(e^-lambda lambda^2)/(2!)` = `(e^-lambda lambda^3)/(3!)`

`=>`  `3 lambda^(2) = lambda^(3)`

`=>`  `lambda^(2) (3 - lambda) = 0 `  as `lambda!= 0` .

Variance of the Poisson distribution is `lambda = 3`

`P(X = 6) =` `(e^-lambda lambda^6)/(6!)`

` = (e^-3 (3)^6)/(6!)`

` = ((0.049787)(729))/(720)`

`= (36.2947)/(720)`

`P(X = 6) = 0.05040`

Therefore, the variance of poisson distribution for above given values are 0.05040.

Example 2 for Poisson Distribution Variance Help

Solving the variance of poisson distribution if `P(X = 3) = P(X = 4)` and also find `P(X = 9)`.

Solution

Given `P(X = 3) = P(X = 4)`

Therefore, we have `(e^-lambda lambda^3)/(3!)` = `(e^-lambda lambda^4)/(4!)`

`=>`  `4 lambda^(3) = lambda^(4)`

`=>`  `lambda^(3) (4 - lambda) = 0 `  as `lambda!= 0` .

Variance of the poisson distribution is `lambda = 4`
`P(X = 9) =` `(e^-lambda lambda^9)/(9!)`

` = (e^-4 (4)^9)/(9!)`

`= ((0.01831)(262144))/(362880)`

`= (4799.8566)/(362880)`

`= 0.01323`

Therefore, the variance of poisson distribution for above given data is 0.01323.

Riemann Problem

Riemann problem - Introduction:

In math, a Riemann sum is a technique for similar to the sum area below a curve on a graph, or else identified as an integral. It may also be used to describe the integration operation. A function is definite to be Riemann integrable if the minor and higher Riemann sums get ever nearer as the partition obtain better and finer. This information can also be used for numerical integration.

Riemann Problem - Definition:

For a function f: S -> R, where S is a subset of the real numbers R,

I = [a, b] is a closed interval contained in S. A finite set of points `{x_0, x_1, x_2, ... x_n}` such that `a = x_0 < x_1 < x_2 ... < x_n = b` creates a partition `S = (x_0, x_1), (x_1, x_2), ... (x_(n-1), x_n] of I.`

Because T is a partition with n elements of I, the Riemann sum of f over I with the partition T is defined as

`M = \sum_{i=1}^{n} f(y_i)(x_{i}-x_{i-1})`

Where `x_(i-1) <= y_i <= x_i.`

The choice of `y_i` in this interval is arbitrary. If `y_i = x_(i-1)` for all i, then M is called a left Riemann sum. If `y_i = x_i` , then M is called a right Riemann sum. If `y_i = (x_i+x_(i-1))/2,` then M is called a middle Riemann sum. The average of the left Riemann sum and right Riemann sum is the trapezoidal sum.

Riemann Problem – Examples:

Riemann problem – Example 1:

`\int_{1}^{e^{\pi}} \frac{\sin (\ln x)}{x}\,dx\,`

Solution:

`u=\ln x\,`

`du=\frac{1}{x}dx\,`

So we have

`\int_{1}^{e^{\pi}} \frac{\sin (\ln x)}{x}\,dx=\int_0^{\pi}\sin u\,du\,`

Notice, the limits of integration changed because when `x=1\, and x=e^{\pi}\,, "we have "u=0\, and u=\pi`, respectively.

` \int_0^{\pi}\sin u\,du=-\cos u |_0^{\pi}=-\cos \pi+\cos 0=2\,`

Riemann problem – Example 2:

`\int \frac{x}{\sqrt{4+x^2}}\,dx\,`

Solution:

` u=4+x^2\,`

`du=2x\,dx\,`

` \int \frac{x}{\sqrt{4+x^2}}\,dx=\frac{1}{2}\int \frac{du}{\sqrt{u}}=u^{\frac{1}{2}}+C=\sqrt{x^2+4}+C\,`

I have recently faced lot of problem while learning Differentiation Rules, But thank to online resources of math which helped me to learn myself easily on net.

Riemann problem – Example 3:

`\int \frac{x}{1-x^2}\,dx\,`

Solution:

We could do this integral with partial fractions, but for instructive purposes let's use the method of trigonometric substitution.

`x=\sin \theta\,`

` dx=\cos \theta\,d\theta\,`

` \int \frac{x}{1-x^2}\,dx`

`=\int \frac{\sin \theta\cos \theta}{1-\sin^2\theta}d\theta`

`=\int \frac{\sin \theta\cos \theta}{\cos^2\theta}d\theta`

`=\int \frac{\sin \theta}{\cos \theta}d\theta\,`

At this point, we have to do a substitution again.

` u=\cos \theta\,`

`du=-\sin \theta\,d\theta\,`

` \int \frac{\sin \theta}{\cos \theta}d\theta=-\int \frac{du}{u}=-\ln |u|+C=-\ln |\cos \theta|+C\,`

Now, we know x = sinθ and we need to know the value of cosθ. Using the well known trig identity, sin2x + cos2x = 1, we get that` \cos \theta=\sqrt{1-x^2}` . So

` \int \frac{x}{1-x^2}\,dx=-\frac{1}{2}\ln |1-x^2|+C\,`

Unbounded Solution

Introduction to unbounded solution:

Unbounded solution of an objective function is a feasible set of points, which are unbounded in a particular direction. An unbounded solution may or may not have a maximum or a minimum value and if it has a maximum or a minimum then it occurs at the extreme points. Unbounded region have infinite set of solutions.

Unbounded solution set is feasible and may extend beyond positive or negative infinity. There can be unbounded solution for both maximizing as well as minimizing problems. Only the solution or set of points can be unbounded but the constraints, which are defined, are never unbounded. Simply put unbounded solutions are not in an enclosed area.

Modes of Unbounded Solution

FOR MAXIMIZING PROBLEM: When maximizing is to be done the solution set may have an infinitely large value, which makes the solution set unbounded at the positive infinity.
FOR MINIMIZING PROBLEM: When minimizing is to be done the solution set may have infinitely small values, which make the solution, set unbounded at the negative end.

Unbounded Solution-causes and Illustrations

Causes: One of the major causes of unbounded solution is the improper formulation of the problem. Unbounded solution set of a problem may occur if one of the constraints of the problem is inadvertently removed.

An unbounded solution can be converted to a bounded one by changing the objective function. Real world problems usually do not have unbounded solutions. Sometimes an unbounded region may not have an optimal solution. Addition of a constraint can also make an unbounded region into a bounded one. Inorder to find an optimal solution of an unbounded solution usually z line is drawn and no solution is considered to be optimal beyond the z line.

Illustration: Minimize C = 3x + 4y subject to the constraints
3x - 4y ≤ 12,
x + 2y ≥ 4
x ≥ 1, y ≥ 0.
The feasible region of this problem is unbounded with points at (1,1.5) and (4,0)
Although the feasible region is unbounded, we can minimize C = 3x + 4y at x=1,y=1.5 so that C=9