Primary Number Line

Introduction to primary number line:

The line is marked with the number is called primary number line. The number line has both positive and negative number. The center point of the primary number line is zero. The zero will not be neither positive nor negative. The positive may or may not have the positive sign on primary number line. But the negative numbers should have negative sign(“-“)

Example for Primary Number Line:

Example 1:

Point the expression x < 3 and x > -2 on the primary number line

Solution:

The expressions x < 3 and x > -2 can be written as -2
Example 2:

Point the expression n > -1 and n < 4on the primary number line

Solution:

The expressions n > -1 and n < 4on can be written as -1 < n < 4. This means, n is greater than -1 but less than 4. Please express your views of this topic second order differential equation by commenting on blog.

Practice Problems for Primary Number Line:

1. Point the expression x < 3 and x > -2 on the primary number line

2. Point the expression x < 5 and x > -5 on the primary number line

3. Point the expression x < 4 and x > -7 on the primary number line

5. Point the expression x < 2 and x > -8 on the primary number line

6. Point the expression x < 9 and x > -11 on the primary number line

7. Point the expression x < 12 and x > -1 on the primary number line

8. Point the expression x < 13 and x > -15 on the primary number line

9. Point the expression x < 4 and x > -8 on the primary number line

10. Point the expression x < 9 and x > -1 on the primary number line

The Bun in Math

Introduction to the bun in math:

Mathematics is the one of the greatest subject in the world, which acts as a fundamental for all subjects. It is generally used to study about the rate of change, its space, and their quantities. In this article, we shall discuss the word problems in math. The following example problems will make the students to get more experts in basic arithmetic problems in math.

Examples for the Bun in Math:

Example problem 1:

In a baker’s shop, there are three buns in the box and two buns in the glass. Find the total amount of buns in baker’s shop.

Solution:

Given that, three buns in the box.

In addition, two buns in the glass.

Therefore the total amount of buns = 3 buns+2 buns

= 5 buns

Finally the total amount of buns in baker’s shop = 5 buns as shown in the following figure.

Example problem 2:

There are 5 buns in each box. How many buns are there in 9 boxes?

Solution:

Given that, one box contains 5 buns.

Therefore 9 boxes contain = 9 x 5 buns

= 45 buns

Totally, there are 45 buns in all boxes.

Example problem 3:

Mani bought 120 grams of buns in a pack. He equally divided the buns in 5 boxes. Calculate the amount of bun in each box.

Solution:

Given that, one pack contains 120 grams of buns.

120 grams is equally divided into 5.

Therefore the amount of bun in each box = `120/5`


= 24 grams.

Finally, each box contains 24grams of bun.

Additional Examples for the Bun in Math:

Example problem 4:

Mike ate `3/6 ` of bun and George ate `2/6` of bun. Calculate the amount of bun that they eat jointly.

Solution:

Given that, Mike ate `3/6` of bun and George ate `2/6` of bun.

Jointly they eat = `3/6+2/6` bun

= `(3+2)/6` bun

=` 5/6` bun.

Finally they jointly eat 5/6 bun.

Example problem 5:

Last month Rajesh sold 100kgs of buns. This month he sold 10 more kgs of bun than last month. What is the percentage of increase in weight of buns?

Solution:

The formula for the percent of increase = (Number of increase) x 100

Original number.

Here increased weight of bun = 10 kgs.

Actual weight of bun in previous month = 100 kgs.

Therefore the percentage of increase of weight of bun `=10xx100/100`

= 10%

That is 10% of buns increased in selling.

Regression Analysis Formula

Introduction to regression analysis formula:

In mathematics, one of the most important topics in statistics is regression. Regression analysis formula is determining the relationship between two variables. Regression analysis formula are used to analysis the several variables. Regression is one of the statistical analysis methods that can be used to assessing the association between the two different variables. Here we are going to discuss about regression analysis formula with suitable example problems.

Formula for Regression Analysis:

Here we study about calculate the regression analysis formulas are,

Formula for regression analysis:

Regression Equation (y) = a + bx

Slope `(b) = (NsumXY-(sumX)(sumY))/(NsumX^2-(NsumX)^2)`

Intercept `(a) = (sumY-b(sumX))/N`


Where,
x and y are the variables.
b = the slope of the regression line is also defined as regression coefficient
a = intercept point of the regression line where is in the y-axis.
N = Number of values or elements
X = First Score
Y = Second Score
`(sumXY)` = Sum of the product of the first scores and Second Scores
`(sumX)` = Sum of First Scores
`(sumY)` = Sum of Second Scores
`(sumX^2)` = Sum of square First Scores.

Example Problem for Regression Analysis Formula:

Problem for regression analysis formula:

Example 1:

Find the regression slope coefficient, intercept value and create a regression equation by using the given table.

X Values	Y Values
5	8
10	16
15	24
20	32
30	40

For the given data set of data, solve the regression slope and intercept values.

Solution:

Let us count the number of values.
N = 5
Determine the values for XY, X2

X Value	Y Value	X*Y	X*X
5	8	40	25
10	16	160	100
15	24	360	225
20	32	640	400
30	40	1200	900
`(sumX = 80)`	`(sumY = 120)`	`(sumXY)= 2400`	`(sumX^2) = 1650`

Determine the following values `(sumX), (sumY), (sumXY), (sumX^2).`
`(sumX) = 80`
`(sumY)= 120`
`(sumXY)= 2400`
`(sumX^2) = 1650`

Plug values in the slope formula,

Is this topic how to solve linear equations in one variable hard for you? Watch out for my coming posts.

Slope `(b) = (NsumXY-(sumX)(sumY))/(NsumX^2-(NsumX)^2)`


`= (5 xx(2400)-(80)xx(120))/((5)xx(1650)-(80)^2)`


`= (12000 - 9600)/(8250-6400)`


`= 2400/1850`

`b= 1.28`

Plug the values in the intercept formula,


Intercept `(a) = (sumY- b(sumX))/N`


`= (120-(1.28xx80))/5`


`= (120 - 103.78)/5`


`= 16.22/5`


`a = 3.24`

Plug the Regression coefficient values and intercept values in the regression equation,
Regression Equation(y) = a + bx
= 3.244 + 1.28x

Answer:

Slope (or) Regression coefficient (b) = 1.28

Intercept (y) = 3.24

Regression equation y = 3.24 + 1.28x

Poisson Distribution Sample Problems

Introduction to Poisson distribution sample problems:

In mathematics, Poisson distribution is a term used to find the number of events happened within a certain time interval. We can also say the Poisson distribution as a discrete probability distribution. A formula is used for finding the Poisson distribution. In this article we are giving some sample problems for Poisson distribution.

Explanation to Poisson Distribution Sample Problems:

The general for calculating the Poisson distribution is as follows.

f(x) = `(e^(-lambda)lambda^x)/(x!)`

Here,

`lambda` – average rate

x – Poisson random variable

e – base value for logarithm having a value = 2.718

Let us see some sample problems using this formula of Poisson distribution.

Example Problems to Poisson Distribution Sample Problems:

Example: 1

In a class room, 2 students are absent today. Find the possibility for exactly 3 students to be absent on tomorrow.Please express your views of this topic irrational numbers examples by commenting on blog.

Solution:

Given:

`lambda` = 2

x = 3

Step 1:

The general formula used for Poisson distribution is,

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

Step 2:

e-2 = (2.718)-2

= 0.135

Step 3:

`lambda`  = 2

x = 3

`lambda^x` = (2)3 = 8

Step 4:

Substitute the obtained values on the formula.

`((e^-lambda)(lambda^x))/(x!)`  = `((0.135)(8))/(3!)`

= `1.08/(6)`

= 0.18

Answer: The possibility of getting 3 absentees on tomorrow is 0.18

Example: 2

In a book shop, 2 customers arrived today. Find the possibility for exactly 5 customers to be arrived on tomorrow.

Solution:

Given:

`lambda` = 2

x = 5

Step 1:

The general formula used for Poisson distribution is,

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

Step 2:

e-2 = (2.718)-2

= 0.135

Step 3:

`lambda`  = 2

x = 5

`lambda^x` = (2)5 = 32

Step 4:

Substitute the obtained values on the formula.

`((e^-lambda)(lambda^x))/(x!)`  = `((0.135)(32))/(5!)`

= `4.32/(120)`

= 036

Answer: The possibility of getting 5 customers on tomorrow is 0.136

Practice Problems to Poisson Distribution Sample Problems:

Problem: 1

In a call center, 2 students are absent today. Find the possibility for exactly 4 students to be absent on tomorrow.

Answer: 0.09

Problem: 2

In a shop, 2 customers arrived today. Find the possibility for exactly 6 customers to be arrived on tomorrow.

Answer: 0.012

Ratio Statistics

Introduction for ratio statistics:

A relation is getting form the comparison of two quantities like in some wisdom is called a ratio. Ratio will be show by the terms of fraction, that is, a:b is equal to `a/b` For example 25 is `1/4` of the hundred, Therefore, the ratio of 25 to 100 is `1/4` . We can write it in ratio as 1:4.

Example Problems for Ratio Statistics:

Example 1: Find the ratio of 90 centimeters to 7 meters in its simplest form.

Solution:

7 m = 7 `xx` 100 cm

Therefore, 7 m = 700 cm

Therefore, the ratio of 90 cm to 7  m = 90cm: 700 cm

Now we have to cancel out the units. So we get

90: 700

Now we have to divide both terms by 10. So we get

9:70

Example 2: Find the ratio of 700 m to 1.7 km in its simplest form.

Solution:

1 km = 1000 m

1.7 km = 1.7 `xx` 1000

Therefore, 1.4 km = 1700 m

Therefore, the ratio of  700 meter to 1.7 km = 700 m : 1700 m

Now we have to cancel out the units. So we get,

700:1700

Now we have to divide both terms by100. So we get

7:17

Example 3: Find the ratio of 20 minutes to 7 hours in its simplest form.

Solution:

1 hour = 60 minutes

7 hours = 7 `xx` 60 minutes

Therefore, 5 hours = 420 minutes

Therefore, the ratio of 20 minutes to 7 hours = 50 min : 420 min

Now we have to cancel out the units. So we get

20: 420

Now we have to divide both terms by 10. So we get

2:42

Now we have to divide both terms by 10. So we get

1:21

Having problem with how to divide mixed numbers keep reading my upcoming posts, i will try to help you.

Practice Problems for Ratio Statistics:

Problem 1 for ratio statistics:

Find the ratio of 900 milliliters to 1.9 liters in its simplest form.

Solution: 9:19

Problem 2 for ratio statistics::

Find the ratio of 900 milligrams to 9 kilograms in its simplest form.

Solution: 1:10

Line Segments in Geometric

Introduction to line segment in geometric:

A line segment is a part of a line that is bounded by two distinct end points and contains every point on the line between its end points. Depending on how the line segment is defined, either of the two end points may or may not be part of the line segment. Two or more line segments may have some of the same relationships as lines, such as being parallel, intersecting, or skew        (source: Wikipedia)

Line Segment in Geometric

Euclidean geometric:

Within the Euclidean geometric, a line is a directly curve.

Line is established to contain assured properties which transmit it to other lines also points.

For instance for either two different points, here be a unique line having them, in addition to either two different lines meet at most one point within two dimensions, that is the Euclidean plane, two lines which perform not meet are called similar. Within higher dimensions, two lines do not intersect might be similar if they are limited within a plane.Please express your views of this topic Rectangular Coordinate System by commenting on blog.

Any groups of finitely various lines separations the plane into convex polygons.

These separations are recognized like a collection of lines.

Coordinate geometric:

Lines in a Cartesian plane are able to be illustrated algebraically through linear equations in addition to linear functions.

Two dimensions, the distinguishing equation is frequently known through the slope-intercept form

y = mx + c

Examples for Line Segment in Geometric

Example 1:

Given two line equations are y = 6x-2, y =4 x+8.  Find the intersecting point?

Solution:

Point of intersection they will equally contain the similar y-coordinate value.

Set the equations equivalent to all other

6x - 2 = 4x + 8

We can solve:

Rearranging x terms

6x - 4x = 8 + 2

2x = 10

x = 10/2

x = 5

To find y to substitute x value for any given equation

y = 6x - 2

y =  6(5) - 2

= 30 - 2

y = 28

The intersecting point is (5, 28)

Example 2:

Given point of the line is (20, 17) and slope of the line is 7. Find the equation of the line?

Solution:

Given point of the line is (20, 17)

The slope of the line is 7

The equation of the line is y-y0 =m(x-x0)

y - 17 = 7(x - 20)

y - 17 = 7x - 140

7x - y = -140 + 17

7x - y + 123 = 0

Therefore the equation of the line is 7x - y + 123

Weighted Mean Calculator

Introduction to weighted mean calculator:

Weighted mean is nothing but the same as that of arithmetic mean, where as each of the data given is contributed equally to the weighed value. If all the weighted values are equal, then the weighed mean is equal to the arithmetic mean.  Weighted mean basically behaves like a similar fashion to the arithmetic mean. The term weighted mean can also be called as weighted arithmetic mean or weighted harmonic mean . Here we are going to see about the weighted mean calculator and the example problems related to it.

Weighted Mean Calculator:

Here given a calculator for calculating the weighted mean statistics for the given set of data.




Procedure for calculating the weighted mean calculator.

Give the input for the observed field and the weighted value field.

Now click on the calculate button to find the result of the given data set.

Definition of Weighted Mean:

Weighted mean statistics is defined as the summation of the observed values multiplied with the allocated weights which is divided by the summation of the observed values. This is called as the statistics weighted mean.

Formula for the weighted mean statistics is given by,

`""""barx_w = (sum_(i=1)^ n (w_i x_i)) / (sum_(i=1)^n w_i) `

where as

`barx_w`  This is the weighted mean variable

`w_i`  This is the allocated weighted value for the given data.

`x_i`  This is the observed values for the given data.

Steps for solving Statistics weighted mean using calculator:

Calculate the summation for the allocated weighted value and observed value of the given data set.
Calculate the summation for all observed values of the given data set.
Now make the division for the above two calculated values from the given data set.

Weighted Mean Calculator - Example Problems:

weighted mean calculator - Problem:

Calculate the weighted mean for the given data as  the observed values  22, 24, 26, 28 and the allocated weights are 10, 20, 30, 40.

Solution:

Weighted Mean:

Formula for Statistical weighted mean is given by,

`"barx_w = (sum_(i=1)^ n (w_i x_i))/(sum_(i=1)^n w_i) `

= `((10)(22)+(20)(34)+(30)(26)+(40)(28))/ (10+20+30+40) `

= `2800 / 100`

`barx_w`  = 28

I am planning to write more post on 9th class cbse sample papers. Keep checking my blog.

Weighted Mean Calculator - Practice Problems:

Problem 1:

Calculate the weighted mean from the observed values  65, 70, 85, 90 and the allocated weights are 60, 70, 80, 90.

Answer: weighted average = 81.3333

Problem 2:

Calculate the weighted mean for the given data as  the observed values  10, 20, 30, 40 and the allocated weights are 15, 20, 25, 30.

Answer: weighted average = 25

Generating Function Method

Introduction for generating function method:

In generating function method, a moment generating function is a real valued function from which one can generate all the moments of a given random variable. In many cases, it is easier to compute various moments of X using the moment generating function.

In other words, the moment generation function may be considered an expression containing all the moment of a probability distribution function, p(|x).

Definition for Generating Function Method:

In generating function method, Let X be a random variable whose probability density function is f(x). A real valued function M : `RR` `->` `RR` defined by

M(t) = E(`e^(tX)`)

is called the moment generating function of X if this expected value exists for all t in the interval −h < t < h for some h > 0.

In general, not every random variable has a moment generating function. But if the moment generating function of a random variable exists, then it is unique. At the end of this section, we will give an example of a random variable, which does not have a moment generating function or generating function method,

Using the definition of expected value of a random variable, we obtain the explicit representation for M(t) as

`M_x`(t) `=`  E(`e^(tX)`) `=` `sum_x` `e^(tx)` f (x)

if X is a discrete random variable and

`M_x(t)` `=`  E(`e^(tX)`) `=`int_(-oo)^(oo)` `e^(tx)` f(x) dx

if X is a continuous random variable.Having problem with free homework help math keep reading my upcoming posts, i will try to help you.

Example for Generating Function Method:

In generating function method, let X is a random variable whose moment generating function is M(t) and n be any natural number. What is the nth derivative of M(t) at t = 0?

Solution:

`d/(dt)`M(t) `=` `d/(dt)` E(`e^(tX)`)

`=` E(`d/(dt)` `xx` `e^(tX)`)

`=` E(`X` `e^(tX)`).

Similarly,

`((d)^(2))/(d(t)^(2))`M(t)         `=` `((d)^(2))/(d(t)^(2))` E(`e^(tX)`)

`=` E(`((d)^(2))/(d(t)^(2))` `xx` `e^(tX)`)

`=`  E(`X^2` `e^(tX)`).

Hence, in general we get

`((d)^(n))/(d(t)^(n))`M(t)         `=` `((d)^(2))/(d(t)^(2))` E(`e^(tX)`)

`=` E(`((d)^(n))/(d(t)^(n))` `xx` `e^(tX)`)

`=`  E(`X^n` `e^(tX)`).

If we set t`=` 0 in the `n^("th")` derivative, we get

`((d)^(n))/(d(t)^(n))`M(t) `|_(t=0)` `=` E(`X^n` `e^(tX)` `|_(t=0)` `=` E(`X^n`).

Hence the `n^("th")` derivative of the moment generating function of X evaluated at t= 0 is the `n^("th")` moment of X about the origin.

In generating function method, this example tells us if we know the moment generating function of a random variable; then we can generate all the moments of X by taking derivatives of the moment generating function and then evaluating them at zero.

Three Types of Probability

Introduction to three types of probability:

Three types of probability article deals with the definition of three types of probability and the model problems related to it.

Types of probability:

The probability is classified in to

Probability as counting.
probability as geometry
Probability as algebra, are the three types and the explanation is given below

Three Types of Probability:

Probability as counting:

This probability is very simple and easy to understand. This is used to find the probability of number of times an event occurs. This probability is calculated by dividing the number of successive out come by the total number of outcomes.

Probability as counting =` (("number of sucessive outcome")/ ("total number of outcomes"))`

Probability as geometry:

This probability is suitable for one or more continuous variable. We can find the probability dividing the length of successive part by the length of total parts in the event.

Probability as geometry = `(("length of successive part")/ ("length of total part"))` .

Probability as algebra:

This is applicable to problems where we can find the enough relationship between two probabilities, at that time we can solve the problem algebraically.Having problem with 10 cbse sample papers keep reading my upcoming posts, i will try to help you.

Model Problems to three Types of Probability:

Problem 1: what is the probability of getting 5, when a dice is thrown?

Solution:

The sample space S = {1, 2, 3, 4, 5, 6}

n(S) = 6

Getting 5 = {5}

n (5)= 1

Probability of getting 5 = `(n (5))/(n(S))`

= `(1/6)`

Probability of getting 5 is `(1/6)` .

Problem 2:

The probability of getting mark in next test is x, which is likely, equal to three the marks he will not get.what is the probability of marks in next test.

Solution:

Getting mark in next test is x

Probability he will not get is 1-x

By given data:

x =2(1-x)

x= 2-2x

x+2x=2

3x =2

x=`2/3`

The probability of getting mark in next test is `2/3`

Problem 3:

John is working from 8am to 5pm, every 15minutes he will go out. What is the probability of going out from 4pm to 5pm?

Solution:

John goes out every 15 minutes.

Probability of going out from 4pm to 5pm = `(15minutes)/(60 minutes)`

= `(15/60)`

= `(1/4)`

Probability of going out from 4pm to 5pm is `(1/4)`

Sum of Poisson Random Variables

Introduction to sum of Poisson random variables:
The sum of the Poisson random variables is the mean value for the Poisson distribution. Poisson distribution is the limited form of the binomial distribution. The sum (mean) is calculated by using the number of the possibilities and the probability for the occurrence of the event. The example for the Poisson distribution is the number of cars passing through the certain street in the time period of t. This article has the information about the sum of the Poisson random variables.

Formula for the Sum of the Poisson Random Variables:

The formula used to find the sum (mean) of the Poisson distribution is

`lambda` = n p

Where n is the number of trial for the event and the p is the probability of the success.

Examples for the Sum of the Poisson Random Variables:
Examples 1 for the sum of the Poisson random variables:

Two coins are tossed for 33 times. Using the Poisson distributions calculate the sum (mean) of head for the coin.

Solution:

The number of times the coin tossed is 33. So the value of n is 33.

The probability for getting one head with one coin is `1/2` .

The probability for getting two head with two coins are p = `(1/2) ^2` = `1/4` .

The sum (mean) `lambda ` = n p

`lambda ` = 33 (1/4)

` lambda` = `33 /4`

` lambda ` = 8.25

The sum (mean) for getting the head for the two coins is 8.25.Having problem with practice pre algebra keep reading my upcoming posts, i will try to help you.

Examples 2 for the sum of the Poisson random variables:

Four coins are tossed for 122 times. Using the Poisson distributions calculate the sum (mean) of head for the coin.

Solution:

The number of times the coin tossed is 122. So the value of n is 122.

The probability for getting one head with one coin is `1/2` .

The probability for getting four head with four coins are p = `(1/2) ^4` = `1/16` .

The sum (mean) `lambda` = n p

`lambda` = `122 (1/16)`

`lambda` = `122 /16`

`lambda` = 7.625

The sum (mean) for getting the head for the four coins is 7.625.

Examples 3 for the sum of the Poisson random variables:

Five coins are tossed for 150 times. Using the Poisson distributions calculate the sum (mean) of head for the coin.

Solution:

The number of times the coin tossed is 150. So the value of n is 150.

The probability for getting one head with one coin is `1/2` .

The probability for getting five head with five coins are p = `(1/2) ^5` = `1/32` .

The sum (mean) `lambda` = n p

`lambda` = `150 (1/32)`

`lambda ` = `150 /32`

`lambda` = 4.6875

The sum (mean) for getting the head for the five coins is 4.6875.

Vertices of a Hyperbola

Introduction to vertices of a hyperbola:

A hyperbola is defined as a smooth curve which tends to lie down in the plane and it can be described by the geometric properties or by the type of equation for which it may be the solution set. A hyperbola can be defined in the form of a curve that is represented by the function f(x) = 1 / x in the Cartesian plane.

In this article we are going to see about the vertices of a hyperbola.

How to Find the Vertices of a Hyperbola:

In the case of ellipse, there exists a relationship between x, y, and z axes with ellipses, the computations are long and painful. So that, for hyperbolas (where x < z), the relationship is z2 – x2 = y2 or, which means the same thing, z2 = y2 + x2.  This has been proved using the Pythagorean Theorem.
When the transverse axis is in the horizontal plane (when the center, and vertices line up side by side which is parallel to the horizontal axis), then the x2 goes with the part of the hyperbola's equation, and the y part is subtracted.
((a – y) 2 / x2) – ((b – c) 2 / y2) = 1.

Example to Find the Vertices of a Hyperbola:

Example 1: Given a hyperbola (x2 / 16) – (y2 / 4) = 1. Find the vertices of the given hyperbola.

Solution:

The given hyperbola opens right or left because it is of the form x – y

Here a2 = 16, b2 = 4, c2 = 16+4 = 20.

Hence a = 4, b = 2 and c = 4.5

The vertices of a hyperbola are given by (a, 0), (-a, 0)

Thus the vertices are (4,0), (-4,0).

Example2: Given a hyperbola (y2 / 25) – (x2 / 9) = 1. Find the vertices and foci of the given hyperbola.

Solution:

The hyperbola opens up and down because it is of the form y – x.

Here a2 = 25, b2 = 9, c2 = 25 + 9 = 34.

a = 5, b = 3, c = 5.8

The vertices of a hyperbola are given by (0, a), (0,-a)

Vertices of a hyperbola: (0, 5) and (0,-5)

The foci of a hyperbola are given by (0, c), (0,-c)

Foci of the hyperbola: (0, 5.8), (0,-5.8)

Acute Triangle Calculator

Introduction about triangle:
A triangle is three side polygons. Triangle is one of the important shapes in geometry.

Types of triangle

Equilateral triangle
Scalene triangle
Right triangle
Isosceles triangle
Acute triangle
Obtuse triangle
In this article we shall discuss about the acute triangle. A triangle has three sides and three angles. A triangle angles are less than 90 degree, such an angle is called acute triangle.

Acute means the angle should be less than 90 degree.

To find area of the acute triangle = `1/2` * base * height

Acute Triangle Calculator:




In calculator you have to enter the value of height and base of the triangle after pressing the calculate button area of the triangle is calculated and display on answer box.

Problem 1:

Calculate the area of the acute triangle whose base is 10cm and height is 12 cm

Solution:

Area of the acute triangle = `1/2` * base * height

Base =10 cm Height = 12 cm

Area = `1/2` * 12 *10

= 60 cm2

The answer is 60 cm2

Problem 2:

Calculate the area of the acute triangle whose base is 8cm and height is 11 cm

Solution:

Area of the acute triangle = `1/2` * base * height

Base = 8 cm Height = 11 cm

Area = `1/2` * 8 *11

= 44 cm2

The answer is 44 cm2

Problem 3:

Calculate the base of the acute triangle whose area is 121cm2 and height is 6 cm

Solution:

Area of the acute triangle = `1/2` * base * height

Area =144 cm2 Height = 6 cm

Area = `1/2` * b *16

121 = `1/2` * b *16

B=121*`2/16`

B= 15.13 cm

The answer is 15.13 cm

Problem 4:

Calculate the height of the acute triangle whose base is 12cm and area is 196 cm2

Solution:

Area of the acute triangle = `1/2` * base * height

Base =12 cm Area = 196 cm2

Area = `1/2` * 12 *h

196 = `1/2` * 12 *h

h=196*`2/10`

B=39.2 cm

The answer is 39.2cm

Normal Sum Distribution

Introduction to normal sum distribution:

Let us see about normal sum distribution,

The normal distribution is also known as the normal probability distribution, happens to be the mainly helpful theoretical distribution for continuous variables. A lot of statistical data sum regarding business and economic problems are display in the type of normal sum distribution.In fact normal distribution is known as the ‘corner stone’ of  Modern statistics.Just like the Poisson distribution, the normal sum  distribution  may also be regarded as a limiting case of binomial distribution.

Definition of Normal Sum Distribution:

Let we see about definition of normal sum distribution,

A variable X is told that a normal distribution with parameter µ and s (or µ and s2) if the probability function is,

Here, Z =(X – µ) s

Constants of Normal distribution:

Mean = µ

Variance = s2

Standard deviation = s

The graph sum of the normal curve is shown below,

The distribution with µ = 0 and s2 = 1 is known as the standard normal distribution.

Understanding problem solving with proportions is always challenging for me but thanks to all math help websites to help me out.

Normal Sum Distribution-example Problems:

1) If X is normally distributed having mean 7 and standard deviation 6 find. P(0 = X = 10).

Solution :

Given, µ = 7, s = 6

(i) P(0 = X = 10)

We know that Z =(X – µ) / s

When X = 0, Z =(0 – 7) / 6 = -1.16

When X = 10, Z =(10 – 6) / 6 = 4/6 = 0.67

? P(0 = X = 10) = P(-1.2 < Z < 0.67)

= P( 0 < Z < 1.2) + P( 0 < Z < 0 .67) (due to symmetry)

= 0.3849 + 0.258

= 0.516

2) If X is normally distributed having mean 10 and standard deviation 8 find. P(0 = X = 20).

Solution :

Given, µ = 10, s = 8

Find, P(0 = X = 20)

We know that Z =(X – µ) / s

When X = 0, Z =(0 – 10) / 8 = -1.3

When X = 10, Z =(20 – 10) / 8= 10/6 = 1.3

? P(0 = X = 20) = P(-1.3 < Z < 0.3)

= P( 0 < Z < 1.3) + P( 0 < Z < 0 .3) (due to symmetry)

= 0.4032 + 0.4032

= 0.8064

Constant Rate of Change Math

Introduction to constant rate of change math:

Rate on which reliant output modify by value to vary within independent input is describing the rate of change. Functional association be usually indicated through y = f(x). Alteration within y is indicated through ?y also alteration within x is indicated through ?x. Consequently, as of the function y = f(x), we contain y + ?y = f(x + ?x).  In addition to constant rates of alteration are known through method, ?y / ?x.

Constant Rate of Change in Math:

In math consider the function is the linear function that is graph of y also x is a line after that the function is known through y = mx + x. In linear function rates of change is a constant which is known through its slope, m. Slope of the line can be find out through the following

m = ?y / ?x =  (y2 – y1/(x2 – x1)       

In math consider the function is non-linear then its rates of alteration are dissimilar on dissimilar points. Therefore, for such functions derived are estimated through distinguish the function toward get the rates of modify. Derivative of y through value to x can be written as dy/dx.

Examples for Constant Rate of Change Math:

Example 1

Solve the rate of change for the line to exceeds through points (5, 4) and (7, 10)

Solution:

Here, x1= 5, x2=4, y1=7 and y2=10

so, slope m = (y2 -  y1)/(x2- x1)

m = (10 - 4)/(7 - 5) = 6/2 = 3

Consequently, rate of alteration otherwise slope of line is 3.

Example 2

Solve the rate of change of function, f(x) =4x2 + 10x at x = 6

Solution:

f(x) =4 x2 + 10x

Derivative f '(x) = 8x + 10

Substitute x = 6 in f '(x) which gives,

f '(6) = 8*6 + 10 = 48 + 10 = 58

Therefore, rate of change of function f(x) at x = 6 is 58.

Example 3

Solve the rate of change of function, f(x) = 3x3 + 3x2 - 7x at x = 3

Solution:

f(x) =3x3 + 3x2 - 7x

Derivative f '(x) = 9x2+6x -7

Substitute x = 3 in f '(x) which gives,

f '(6) = 9(32)+6(3) - 7 = 81 + 18-7 = 92

Therefore, rate of change of function f(x) at x = 3 is 92.

Precalculus Math Final Help

Introduction to Precalculus Math Final Help:

The Precalculus is one of the foundation class of mathematics. The topics involved in Precalculus test are real  complex numbers, binomial theorem, composite & polynomial functions, vectors, parametric equations, polar coordinates, matrices, rational functions, solving inequalities & equations and trigonometry. In this article we shall discuss about the examples involved in Precalculus math final help. The following are the examples involved in Precalculus math final help.

Precalculus Math Final Help Problems:

Example 1:  Express the real and imaginary parts for 28 - i `sqrt(14) `

Solution:        Let z= 28 - i `"sqrt(14)`      

Re(z) = 28     Im(z) = `sqrt(14)`

Example 2: For what value of m the vectors `veca` and `vecb` are perpendicular to each other

`veca` =m`veci` + `vecj`` + ``veck`   `vecb`= 7`veci` - 9 `vecj` + 2 `vecj`

Solution: Given `veca` `_|_``vecb `

So, `veca` `vecb` = 0

=> (m`veci` +  `vecj`` + ``veck`) . (7`veci` - 9 `vecj``veck` )

=>  7m - 9 + 2

=>  7m  = 7

m =`7/7`  =1

Example 3: Solve for x,  ln(x-3) + ln(x-2) = ln(2x -6 )

Solution:   Simplify left side, So

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

Therefore the given equation can be rewritten as

ln(x - 3)(x - 2) =  ln (2x - 6)

Taking exponent on above equation

eln(x - 3)(x- 2) = eln(2x - 6)

Simplifying the above equation

(x - 3) (x - 2)  = (2x - 6)

x2  - 5x + 6 = 2x - 6

x2  - 7x +12 = 0

Factoring the above equation, we get

(x - 4) (x - 3) = 0

When products of two factors are equal to 0, then one of the factors must be equal to 0.

So, x-4 = 0 => x=4

x-3 = 0 => x=3

Here x = 4 & 3 is the solution.

Example 4:   If f(x) = 14x-1, find f-1(y)?

Solution:Let y = 14x - 1

The given equation can be rewritten as

y + 1 =14x

=> x = y + 1
14

Therefore the f-1(y) = y + 1
14

Example 5: Write the given number in complex form `sqrt(-69)`

Solution: `sqrt(-69)` = `sqrt((-1)(69))`

= `sqrt(-1)` * `sqrt(69)`

= i `sqrt(69)`

Precalculus Math Final Help Practice Problems:

Problem 1:Write the given number in complex form `sqrt(-1415)`

Answer: = i `sqrt(1415)`

Problem 2:Write the real and imaginary parts for 89 - i `sqrt(75) `

Answer:   Re(z) = 89     Im(z) =`sqrt(75)`

Problem 3:   If f(x) = x-9, find f-1(y)?

Answer: y + 9

Radical Perfect Prime Numbers

Introduction to radical perfect prime numbers:
A radical perfect prime number is a symbol and it’s denoted as root of a number, the thought of a radical the root of a number can best be understood by first tackling the design of exponentiation, or raising a number to a given power. We point to a number raised to the nth power by writing xn. This expression denotes that we are multiplying x by itself n number of times. Example for radical perfect prime numbers, v5 =2.236, and v3 = 1.732. a radical perfect prime number is a number that has exactly 2 factors...1 and itself , and one thing is 1 is not a prime number ,so  that 2,3,5,7,11,13,17,19,23 etc  are all primes.

Examples for Radical Perfect Prime Numbers:

Addition for radical perfect prime number:

Example 1: Find the value of 2v3 +3 v3

Solution:

=v3 (2+ 3)
=5v3

Here the root is square root, only take the base are same so that we can add the roots easily.

So the result is =5v3

Example 2: Find the value of v11 +6 v23+5v11 +7 v23

Solution:

= v11 (1+5)+ v23 (6+7)

= v11 (6)+ v23 (13)

= 6 v11 + 13 v23

Here the root is square root, the base are same in different terms so that we can add the roots easily.

So the result is 6 v11 + 13 v23

Examples for different radical perfect prime numbers:

Example for square root:

v3 = 1.732

v5 = 2.23

v7 =  2.64

v11 = 3.31                              

Some other Examples for Radical Perfect Prime Numbers:

Multiplication for radical perfect prime numbers:

Example 1:

Solution:

To find multiplication of v3x . v1y    

= v(3 x y)            

=  v3x y   

The result is =  v3x y    

Both the radicals have an same index of two, so we are multiply mutually under one radical keeping the 2 as its index number.

Example 2:

Solution:

To find multiplication of v(23xy2). vy    

= v(23 x y3)            

= v(23 x y3)

The result is v(23 x y3)

Both the radicals have an same index of two, so we are multiply mutually under one radical keeping the 2 as its index number.


Factorization for Radical Perfect Prime Numbers Examples:

Example 1:

To find the value of v75

v75 = v5*3*3

= 3v5

Radical perfect prime factors for v75 = 3v5

So the result is in the form of prime numbers and radical.

Integral Trapezoidal Rule

Introduction to integral trapezoidal rule :

The integral trapezoidal rule is a method which has been used to calculate the definite integral that has been given by

a
∫ f(x) dx
b

A definite integral is defined as the integral which has both the upper and the lower limits. In this topic we are going to see about the integral trapezoidal rule. The integral trapezoidal rule is also called as the trapezium rule or the trapezoid rule.

Integral Trapezoidal Rule:

The trapezoidal rule has been used by considering the region that has been under the graph of the given function f(x) as a trapezoid and then finding its area.

a
∫ f(x) dx ≈ (b-a) f(a)+f(b)
b                        2

The next step is to split the interval of the integration (a, b) into a number of smaller intervals and then apply the trapezoidal rule on every subintervals. This is given by

a                                            n-1
∫ f(x) dx ≈ (b-a)   (f(a)+f(b))/2   + ∑ f(a+k((b-a)/n)
b                 n                         k=1

This integration can also be written as

a
∫ f(x) dx ≈ ((b-a)/2n)(f(x0)+2f(x1)+2f(x2)+…+2f(xn-1)+f(xn))
b

Where the value of xk = a + k(b-a)/n, for the values of k = 0, 1, …. N

Between, if you have problem on these topics Derivative of Parametric Equations, please browse expert math related websites for more help on Integration by Parts Table.

Error Analysis of the Integral Trapezoidal Rule:
The error of the integral trapezoidal rule is given by the difference of the integral and the result.

b                                                n-1     
Error = ∫ f(x) dx -  b-a   (f(a)+f(b)/2) + 2   + ∑ f(a+k((b-a)/n)
a                 n                             k=1

This can be simply written as

Error = - (b-a)3    f′′(ξ)
12n2

Integral trapezoidal rule for non uniform intervals:

Let us consider the given values are x1 ….xn and y1,…..yn, then according to the integral trapezoidal rule

n
∫ydx ≈ ½ ∑ (xi – xi-1)(yi+yi-1)
i=2

The integral trapezoidal rules are said to be exact when the periodic functions are said to be integrated with their periods and the trapezoidal rule converges very rapidly for the periodic functions. When the function is said to be periodic and it integrates over one complete period then there exists a number of graphs which makes the error of the trapezoidal rule to be cancelled.

Integers Rational Irrational Numbers

Introduction to integers rational irrational numbers:

Let us study about integers rational irrational numbers. Integer numbers are commonly seemed as the set of numbers that starts from infinite negative side to the infinite positive side.
Rational numbers are defined as numbers that can express themselves in the form of ratios as `m/n` where ‘n’ should not be equal to equal to ‘0’.
The numbers which cannot express themselves in the simple ratio are named as irrational numbers.

Integers Rational Irrational Numbers:

Example 1: Add the following numbers: 0.27, `3/2` , 1.52, -3, 2.5, 7.


Solution:

The given numbers are 0.27, `3/2` , 1.52, 3, 2.5, 7.
To add the given numbers the following steps are to be considered:
= 0.27 + `3/2` + 1.52 - 3 + 2.5 + 7

= 0.27 + 1.3 + 1.52 - 3 + 2.5 + 7 (do conversions of the fractions to decimal numbers)

= 9.59

Example 2: Subtract the following numbers by using same given order: 3.9, `4/5` , 0.13, 4, 1


Solution:

The given numbers are 3.9, `4/5` , 0.13, 4, 1.
To subtract the given numbers the following steps are to be considered:
= 3.9 – `4/5` – 0.13 – 4 – 1.

= 3.9 – 0.8 – 0.13 – 4 – 1. (do conversions of the fractions to decimal numbers)

= -2.03

Example 3: Do the multiplication process for the following numbers:  1.1, 0.4, 3, 9, `6/5` .


Solution:

The given numbers are 1.1, 0.4, 3, 9, `6/5` .
To add the given numbers the following steps are to be considered:
= 1.1 * 0.4 * 3 * 9 * `6/5` .

= 1.1 * 0.4 * 3 * 9 * 1.2 (do conversions of the fractions to decimal numbers)

= 14.256

Example 4:Do the division process for the numbers: `9/4.5` .


Solution:

= `9/4.5`
= 2


Exercises:

Add the following numbers: 0.2, 6, `11/5` , 4 (Answer: 12.4)
Subtract the following numbers by using same given order: 7, `5/8` , 1.3, 2. (Answer: 3.075)
Do the multiplication process for the following numbers:  0.09 0.3, `15/2` , 1, 7 (Answer: 1.4175)
Do the division process for the numbers: `2/0.5` (Answer: 4)

Adding Radicals Calculator

Introduction to adding radicals calculator:

Radicals:

The symbol used to indicate root of a number is known as radicals.  The square root, cube root can be expressed by using the radical symbol.  The radicals also mean that the root.  To understand the radicals, first we need to know about exponentiation. That is exponential xn = x * x * x * x …….n times x. The root of xn can be expressed as root (n)(x^n) = x .

The radical symbol is,

Fig(i) Radicals

Calculator:

A calculator is a small (often pocket-sized), usually inexpensive electronic device used to perform the basic operations of arithmetic.(source : Wikipedia).

Fig(ii) Adding radicals calculator

In this article we are going to learn about how to adding radicals using calculator.

Rules to Adding Radicals :

Before we get into problems , we need to know the rules for adding radicals.

Rules :

We can add the radicals with the same index and radicand . For example the radicals of index 2 can only add. We can add the radical of index 2 with the radical of index 3.

For example , `sqrt(4)` + `sqrt (4)` = ( 1 + 1) `sqrt(4)`

= 2 `sqrt(4)`

Let us see some problems on adding radicals by using calculator.

I am planning to write more post on Factorization Calculator, Equivalent Fractions Calculator. Keep checking my blog.

Problems on Adding Radicals Calculator:

Problem 1:

Add the radicals `5sqrt(36)` + `2sqrt(81) `

Solution:

Given, `5sqrt(36)` + `2sqrt(81) `

We need to add the given radicals.

Before that we can simplify the given radicals.

We know that, `root(n)(x^n) ` = x

`sqrt(36)` = `sqrt(6^2)` = 6

`sqrt(81)` = `sqrt(9^2) ` = 9

`5sqrt(36)` + `2sqrt(81)` = 5*6 + 2*9

= 30 + 18

= 48

Answer: 5`sqrt(36)` + 2`sqrt(81)` = 48

Problem 2:

Add the radicals `8sqrt(8)` + `sqrt(8) `

Solution:

Given, `8sqrt(8)` + `sqrt(8) `

We need to add the given radicals.

Before that we can simplify the given radicals.

We know that,` root(n)(x^n)` = x

`sqrt(8)` = `sqrt(4 * 2 )`

= `sqrt(4)` * `sqrt(2)`

= `sqrt(2^2)` * `sqrt(2)`

= `2sqrt(2)`

`sqrt(8)` = `2sqrt(2)`

8`sqrt(8)` + `sqrt(8)` = 8 * `(2sqrt(2)) ` + 2`sqrt(2)`

= 16 `sqrt(2)` + 2`sqrt(2)`

= (16 + 2)`sqrt(2)`

=  18`sqrt(2)`

Answer: 8`sqrt(8)` + ` sqrt(8) ` = 18`sqrt(2)`

Problem 3:

Add the radicals 3`root(3)(9)` + 2`root(2)(8) `

Solution:

Given,3`root(3)(9)` + 2`root(2)(8)`

We need to add the given radicals.

Here , given two radicals are has two different index as well as different radicand.

So we can add it further.

My Previous Blog :- http://online-math-problems-solved.blogspot.in/2012/10/solving-literal-equation.html

Polynomials in One Variable

Introduction to polynomials in one variable
Polynomials in one variable, here we need to understand each terms very clearly. Polynomial here means, a mathematical expression consisting of  a sum/subtraction/multiplication but not division of variables , constant sand exponents.One variable means  expression consisting of one letter. Polynomial with one term is called as a "Monomial", two terms as "Binomial" and three terms as "Trinomials". Terms can be separated by addition sign or subtraction sign only but not by multiplication sign.So we can say in otherwords that polynomial is a sum of monomials . And each monomial in polynomial is called as a Term.


Examples of Polynomials with One Variable
Let us see some examples to understand better the polynomials with one variable:

1. 4x

here we have one term only with single variable x

2. 2x^3 + 4x^2 - x

Here we have three term with one variable x.

3. 2x^2

Here we have one term with variable x and the exponent as 2.

Polynomial the word poly means "many" and the word nomial means "terms". so polynomial means many terms. But we are learning here polynomials with one variable.

In polynomial we have constants , variables and exponents. we need to understand these with examples.

Constants can be any number as 3 , 2 , -20 , 3/2 etc.

Variables can be any letter as x , y , z , a , b etc

Exponents will be always positive integers such as 2x^2 , 2a3, 4y5 etc

Algebra is widely used in day to day activities watch out for my forthcoming posts on what s an algebraic expression and what is an example of an algebraic expression. I am sure they will be helpful.

Degree of a Polynomial with One Variable

The largest exponent of the variable is called the degree of a polynomial with one variable.

Let us take an example to understand better:

1. 2x4 + 4x^3 - x^2 + 8

This is a polynomial with one variable with the highest degree as 4 which is also called as the highest power or exponent of the given variable.

2. 3a^2 - 5

This  is a polynomial with one variable with degree 2.

3. x

Here the degree is 1. A variable without a power is considered to be as 1 always.

Standard Form of a Polynomial in One Variable
Standard form means writing the polynomial in one variable with the highest degree first.

Let us take an example:

3x^3 - 2x^2 +5

here the polynomial in one variable is written on that order of degree( highest degree first).

It is not necessary to write it in standard form , but it helps us in doing simplification.

Let us take few more examples:

1.Add: 3x + 4x^2+5 and 8x^2 +2x

writing it in standard form

=(4x^2+3x+5) + (8x^2+2x)

add the like terms

=4x^2+8x^2+3x+2x+5

=12x^2+5x+5

2. subtract: (3a3+ 8a^2+3) -(2a^2- 3a3+8)

write it in standard form

=(3a3+8a^2+3)-(-3a3+2a^2+8)

Reverse the sign

=3a3+8a^2+3 +3a3-2a^2-8

now add/ subtract the like terms

=3a3+3a3+8a^2-2a^2+3-8

=6a3+6a^2-5

3. Multiply: 3x^2 by 2x^3

Multiply the constant and add the exponent of the variable

=6x5

Conclusion

Polynomials in one variable is easy to simplify.

If we add / multiply polynomials in one variable we get the results as a polynomial itself.

Always while adding polynomials simply add the like terms( variable with same degree) together.

In case of multiplication , multiply the constants first then the variable and then add the degree of the variable.

While doing the subtraction change the sign to reverse sign of each term you are subtracting.