Probability of a Intersection B

Probability of A Intersection B

Probability is the possibility of the outcome of an event of a particular experiment. Probabilities are occurs always numbers between 0 (impossible) and 1(possible). The set of all possible outcomes of a particular experiment is called as sample space. For example probability of getting a 6 when rolling a dice is 1/6. In this lesson we will discuss about probability problems using intersection rule.

Probability of a Intersection B – Example Problems

Example 1: A jar contains 5 red candies, 4 orange candies. If three candies are drawn at random, find the probability, that 1 is red candy and 2 are orange candies

Solution:

We have to select 3 candies, from 9 (5 + 4) candies.

n(S) = 9C3 = (9!)/(3!xx6!) = (9xx8xx7)/(3xx2xx1) = 84

Let A = Event of getting 1 red candy

B = Event of getting 2 orange candies

n(A) = 5C1 = `(5!)/(1!xx4!) ` = `5/1` = 5

n(B) = 4C2 =` (4!)/(2!xx2!) ` = `(4xx3)/(2xx1)` = 6

P(A) =` (n(A))/(n(S))` = `5/84`

P(B) = `(n(B))/(n(S))` = `6/84`

P(A intersection B) = P(A) ∙ (B) = `5/84` ∙ `6/84` = `30/7056`

P(A intersection B) = `5/1176` .



Example 2: A box contains 6 yellow marbles, 6 orange marbles. If four marbles are drawn at random, find the probability, that 2 are yellow marbles and 2 are orange marbles.

Solution:

We have to select 4 marbles, from 12 (6 + 6) marbles.

n(S) = 12C4 = `(12!)/(4!xx8!)` = `(12xx11xx10xx9)/(4xx3xx2xx1)` = 495

Let A = Event of getting 2 yellow marbles

B = Event of getting 2 orange marbles

n(A) = 6C2 = `(6!)/(2!xx4!)` = `(6xx5)/(2xx1)` = `30/2` = 15

n(B) = 6C2 = `(6!)/(2!xx4!)` = `(6xx5)/(2xx1)` = `30/2` = 15

P(A) = `(n(A))/(n(S)) ` = `15/495` = `1/33`

P(B) = `(n(B))/(n(S)) ` = `15/495` = `1/33`

P(A intersection B) = P(A) ∙ (B) = `1/33` ∙ `1/33` = `1/1089`

P(A intersection B) = `1/1089` .

Probability of a Intersection B – Practice Problems

Problem 1: A jar contains 4 lemon candies, 4 orange candies. If two candies are drawn at random, find the probability, that 1 is lemon and 1 is orange candy.

Problem 2: If P(A) = `1/5` , P(B) = `1/7` , P(A or B) = `1/9` , find (A and B)?

Answer: 1) `1/49 ` 2) `73/315`

Area of a Square Inscribed in a Circle

A square is a four sided figure; all the four sides are equal. If we situate a square inside the circle, all the four edges or vertices may touches the circle. A square is a quadrilateral, if the four sides of a square touch the circle, the four sides act as a four chords of a circle. The area of a square is measured in square units such as feet, inch, meter etc. In this article we shall discuss about the area of a square inscribed in a circle.

Area of a Square Inscribed in a Circle

If a square is inscribed in a circle and four sides of a square touches a circle.



The diagonals of the circle act as a diameter of the circle. We can find the area of a square is by:

Area of square = side times side

Area of the square = side side

= s2

Example:

Find the area of the square, sides of the square is 9ft.

Solution:

Given: Side = 8ft therefore diameter = 9ft

Area of a circle = side × side

= 9 × 9

= 81square ft.

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If the square does not touch the circle but inscribed in a circle, In such case we can also find the area of the square.

Example:

Find the area of a square inscribed in a circle, side of the square is 6ft.

Solution:

Given: side = 6t

Area of a square = side × side

= 6 × 6

= 36square ft.

The one or two sides of a square may touch the circle.

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Example for Area of a Square Inscribed in a Circle:

Example1:

Find the area of the square inscribed in a circle, sides of the square is 12m.

Solution:

Given: Side = 12

Area of a square = side × side

= 12 × 12

= 144m2.

Example2:

Find the area of the square inscribed in a circle, sides of the square is 23ft.

Solution:

Given: Side = 23ft

Area of a square = side × side

= 23 × 23

= 529square ft.

Poisson Probability Table

Introduction:

The allocation was first introduced by Simeon-Denis poisson (1781–1840) and in print, together with his probability theory, in 1838 in his work (Research on the Probability of Judgments in Criminal and Civil Matters). The job focused on sure random variables N that tally, among other things, the number of separate occurrences that take place during a time-interval of given length. In probability hypothesis and statistics, Poisson division is a discrete probability division expresses probability of a number of actions.

Poisson Probability Table Conditions:

Poisson probability distribution states two conditions

1.The amount of successes in two disjoint time intervals is autonomous.

2.The probability of a success during a small time interval is comparative to the entire length of the time interval.

Poisson probability formula:

The Probability distribution of a Poisson random variable, X  instead of the number of successes happening in a given time interval or a specified area of space is given by the formula:

P(X)=e-`nu` `nu`X / X!

where

X=0,1,2,3....

e = expected value

µ = mean figure of successes in the given time interval or area of space.

Mean and Variance for poisson probability:

If µ is the standard number of successes happening in a given time interval or area in the Poisson distribution, then the mean and the variance of the Poisson distribution are both equal to µ.

E(X) = µ

and

V(X) = s2 = µ

Note: In a Poisson distribution, only one factor, µ is needed to determine the probability of an occurrence.
Poission Probability Table:

The below shows the value of poisson probability for the mean and X values of

`lambda` =0.1 to 5.5

X  =0 to 17

poisson probability table

Examples of Poisson Probability Distribution:

Example 1:

A life assurance salesman sells on the usual 3 life insurance policies per week. Use Poisson's law to work out the possibility that in a given week he will sell

(a) a few policies

(b) 2 or supplementary policies but less than 5 policies.

(c) assume that there are 5 working days per week, what is the probability that in a given day he will sell single policy?

Solution:

Here, µ = 3

(a) "Some policies" means "1 or more policies". We can work this out by finding 1 minus the "zero policies" probability:

P(X > 0) = 1 - P(x0)

Now P(X)=e-`nu`    `nu`X / X!    So P(X`@` )= e-3 30 / 3! = 4.9787 `xx` 10-2

So

Probability= P(X > 0)

=1 - P(x0)

=1-4.9787`xx`10-2

=0.95021

(b)     P(2`<=` X `>=` 5) = P(X2)+P(X3)+P(X4)

= (e-3 32 / 2!) + (e-3 33 / 3!) + (e-3 34 / 4!)

=0.61611

(c)    Average number of policies sold per day:  `3/5 ` = 0.6

So on a given day   P(X)=e-0.6 (0.6)1 / 1!

= 0.32929

Example 2:

A business makes electric motors. The probability an electric motor is imperfect is 0.01. What is the probability that a model of 300 electric motors will contain precisely 5 defective motors?

Solution:

The average number of defectives in 300 motors is µ = 0.01 × 300 = 3

The probability of getting 5 defectives is:

P(X ) = e-3 35 / 5!

= 0.10082

Intersection of Two Sets

Set is a fundamental part of the mathematics. This set concept is applied in every branch of mathematics. Sets are used in relations and functions. The application of sets are geometry,  sequences, Probability etc. Sets are used in everyday life such as a volleyball team, vowels in alphabets, various kinds of geometry shapes etc.

There are two main important operations in sets. They are  union and intersection of two sets. Let us learn the concepts and properties of intersection of two sets. We will learn some example problems and give practical problems about intersection of two sets.

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Intersection of Sets:

Let A and B be any two sets.  A intersection B is the set of all elements which are similar to both A and B. The symbol `nn` is used to denote the intersection. A intersection B is the set of all those elements which belong to both A and B. Symbolically, we write A `nn` B = { x : x `in` A and x `in` B }.

Ex:

Consider the two sets X = { 1, 5, 8, 9 } and Y = { 5, 6, 9, 15 } . Find X intersection Y.

Sol:

We see that 5, 9 are the only elements which are similar to both X and Y. Hence  X ∩ Y = { 5, 9 }

Some Properties of Operation of Intersection:

(i) Commutative law:  A ∩ B = B ∩ A

(ii) Associative law: ( A ∩ B ) ∩ C = A ∩ ( B ∩ C )

(iii) Law of identity: φ ∩ A = φ, U ∩ A = A (Law of φ and U).

(iv) Idempotent law: A ∩ A = A

(v) Distributive law: A ∩ ( B ∪ C ) = ( A ∩ B ) ∪ ( A ∩ C ) i. e., ∩ distributes over ∪

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Practice Problems on Intersection of Two Sets:

Problem 1:

Find the intersection of each of the following two sets:

1. X = { 1, 3, 5 }   Y = { 1, 2, 3 }

2. A = [ a, e, i, o, u ]   B = { a, b, c }

3. A = { 1 , 2 , 3 }    B = `Phi`

Sol:

1. X `nn` Y = { 1, 3 }

2. A `nn` B = [ a ]

3. A `nn` B = `Phi`

Problem 2:

If A = { 3, 5, 7, 9, 11 }, B = {7, 9, 11, 13}, C = {11, 13, 15}and D = {15, 17}; find

(i) A ∩ B  (ii) B ∩ C (iii) A ∩ C ∩ D

(iv) A ∩ C (v) B ∩ D (vi) A ∩ (B ∪ C)

(vii) A ∩ D (viii) A ∩ (B ∪ D)

Sol:

i) A ∩ B = { 7, 9, 11 }

ii) B ∩ C = { 11, 13 }

iii) A ∩ C ∩ D = Nill

iv) A ∩ C = { 11 }

v) B ∩ D = nill

vi) A ∩ (B ∪ C) = { 7, 9, 11, 13 }

vii) A ∩ D = nill

(viii) A ∩ (B ∪ D) = { 7, 9, 11}

Diameter of Intersecting Lines

Diameter of Intersecting Lines means we have to find the diameter of the lines which are intersecting the circle. Here we are going to learn about the diameter of the intersecting lines. Basically diameter refers the length of the line from one side of the circle to the other side of the circle which passes through the center. If a line intersecting the circle through its center means we ca say that line is the diameter.

Examples for Diameters of Intersecting Lines:

From the above diagram the line AB intersect the circle. This is passes through the center of the circle. The radius of the circle is r then the diameter of intersecting lines are d = 2r. We will see some example problems for it.

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Example 1 for diameter of intersecting lines:

A line AB is intersecting the circle through its center from the one side to another side. The circumference of the circle is 14 cm. Find the diameter of the intersecting lines.

Solution:

A line AB is intersecting the circle through its center from its one side to another side. We have to find the diameter of the intersecting lines. The diameter is d = 2r

Circumference of the circle C = 2πr

Here 2πr = 14 cm

So 2r = `14 / 3.14` = 4.45 cm

So the diameter of the intersecting lines is 4.45 cm

More Examples for Diameters of Intersecting Lines:

Example 2 for diameter of intersecting lines:

A line XY is intersecting the circle through its center from the one side to another side. The area of the circle is 22 cm2. Find the diameter of the intersecting lines.

Solution:

A line XY is intersecting the circle through its center from its one side to another side. We have to find the diameter of the intersecting lines. The diameter is d = 2r

Area of the circle A = πr2

Here πr2 = 22 cm

So r2 = `22 / 3.14` = 7 cm2

Radius r = 2.645 cm

So the diameter of the intersecting lines is 2r = 5.29 cm

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Average Deviation Calculator

The average deviation in which it is defined to calculate the average for the deviations for the given data. It is calculated by taking sum for all values of the deviation divided by the total number  of values in the given data set.

For finding the average deviation value, find the mean for the given data is using the formula,

`barx = (sum_(k = 1)^n (x_n))/ (n)`

Calculate the deviation value by using the calculated mean value, by using the formula given below,

`(x - barx)^2`

Average Deviation is calculated by taking average for the founded deviation value of the data set.

Average Deviation = `"(sum_(K=1)^n (x-barx^2)) / N`

` T It is nothing but the above shown formulas the population of the variance formula in which it is also called as the average deviation.`

Steps to Calculate the Average Deviation:

1. Give the average for all the given dimensions of the data set .
2. Give the difference of the initial value of the data and the average value we have found which is called as mean difference.
3. Take the all absolute value from this mean difference of the given data.
4. Repeat the steps 2 and 3 for all the other given values and find the mean difference to the data set.

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Screen Shot for the Calculator to Find the Average Deviation:

For finding the average for the squared mean difference which is known as the average deviation of the given data set.



Select the formula to find the result and put the values in input field  then press calculate button.



Average Deviation Calculator - Example Problems:

Average deviation calculator - Problem 1:

Calculate the average deviation for the given data set. 35, 36, 37, 38.

Solution:

Mean: Formula for finding mean,

`barx = (sum_(k = 1)^n (x_n))/ (n)`

`barx = (35+36+37+38) / 4`

` barx = 146 / 4`

` barx =` 36.5

Calculate the deviation for the given data set from the mean,

Deviation  =  `(x - barx)^2`

= `((35-36.5)^2+(36-36.5)^2+(37-36.5)^2+(38-36.5)^2)`

Deviation = 5

Calculate the average deviation for the given data,

Average Deviation = `(sum (x-barx)^2) / (n)`

= `5 / 4`

Average Deviation   = 1.25

Average deviation calculator - Problem 2:

Calculate the average deviation for the given data set. 33, 35, 36, 38.

Solution:

Mean: Formula for finding mean,

`barx = (sum_(k = 1)^n (x_n))/ (n)`

`barx = (33+35+36+38) / 4`

` barx = 142 / 4`

` barx =` 35.5

Calculate the deviation for the given data set from the mean,

Deviation  =  `(x - barx)^2`

=`((33-35.5)^2+(35-35.5)^2+(36-35.5)^2+(38-35.5)^2)`

Deviation = 13

Calculate the average deviation for the given data,

Average Deviation = `(sum (x-barx)^2) / (n)`

= `13 / 4`

Average Deviation   =  3.25

Average deviation calculator - Problem 3:

Calculate the average deviation for the given data set. 2, 5, 6, 5, 4, 7, 8, 5.

Solution:

Mean: Formula for finding mean,

`barx = (sum_(k = 1)^n (x_n))/ (n)`

`barx = (2+5+6+5+4+7+8+5) / 8`

` barx = 42/ 8`

` barx =` 5.25

Calculate the deviation for the given data set from the mean,

Deviation  =  `(x - barx)^2`

= `((2-5.25)^2+(5-5.25)^2+(6-5.25)^2+(5-5.25)^2+(4-5.25)^2+(7-5.25)^2+(8-7.25)^2+(5-5.25)^2)`

Deviation = 16.5

Calculate the average deviation for the given data,

Average Deviation = `(sum (x-barx)^2) / (n)`

= `16.5 / 8`

Average Deviation   = 2.0625

Average Deviation Calculator - Practice Problems:

1 Calculate the average deviation for the following 55.3, 56.6, 50.9 and 54.0.

Answer: Average Deviation = 4.47500

2. Calculate the average deviation for the following data

Answer: Average Deviation = 2

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Statistical Power Table

Statistical power is important in math. In math statistical power means probability of reject a false null hypothesis. In statistics to test hypotheses and also test the null hypothesis. The power is equal to 1-beta. In odd position we want to reject our null hypothesis in favor of the alternative. It is more helpful for exam preparation.

Basic Statistical Power of Negative Z-score Table:

In the following statistical power of negative z score table to explain the how to calculate the probability value




For example,

P(0 > Z > 2.31) = P( 0> Z > `oo` ) − P(-1.13 > Z >0)

= 0.5 − 0.1292 (Use statistical power of negative z score table to calculate the probability value)

= 0.3708

Select the first column value -1.13 then choose the right side eight column value 0.03 in the same direction we got an answer as 0.1292

Example Problems for Statistical Power Table:-

Problem 1:-

Calculate the standard deviation range of P(0 > Z > 1.9) by using statistical power table

Solution:

P(0 > Z > 1.9) = P(-1.9< Z<0)

= 0.0287 (Use statistical power of negative z score table to calculate the probability value)

Select the first column value -1.9 then choose the right side eleventh column value 0.0 in the same direction we got an answer as 0.0287

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Problem 2:-

Calculate the standard deviation range P(-2.2 > Z >3.1) by using statistical power table

Solution:

P(-1.2 > Z >2.1) = P(− 2.2 > Z > 0) + P(-3.1 > Z > 0)

= 0.0139+ 0.0010 (Use statistical power of negative z score table to calculate the probability value)

= 0.133

Select the first column value -2.2 then choose the right side eleventh column value 0.0 in the same direction we got an answer as 0.0139

Select the first column value -3.1 then choose the right side eleventh column value 0.0 in the same direction we got an answer as 0.0010

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Adding and Subtracting Vectors

The addition may be represented graphically by placing the start of the arrow b at the tip of the arrow a, and then drawing an arrow from the start of a to the tip of b. The new arrow drawn represents the vector a + b.



To subtract b from a, place the end points of a and b at the same point, and then draw an arrow from the tip of b to the tip of a. That arrow represents the vector a − b.
Source: Wikipedia



Adding and subtracting of vectors operation and example problems are given below.

Operations for on Vectors:

1. Addition of vectors:

Let `vec(OA)` = `veca` , `vec(AB)` =` vecb` . Join OB.

Then `vec(OB)` represents the addition (sum) of the vectors veca and vecb.



This is written as `vec(OA) ` +` vec(AB)` = `vec(OB)` Thus `vec(OB)` = `vec(OA) ` + `vec(AB) ` = `veca ` +` vecb`

This is known as the triangle law of addition of vectors which states that, if two vectors are represented in magnitude and direction by the two sides of a triangle taken in the same order, then their sum is represented by the third side taken in the reverse order.

Applying the triangle law of addition of vectors in




ΔABC, we have BC + CA = BA ⇒ BC+ CA = − AB

⇒ AB + BC + CA = 0

Thus the sum of the vectors representing the sides of a triangle taken in order is the null vector.

2. Subtraction of vectors:

If `veca` and `vecb` are given two vectors, then the subtraction of `vecb` from `veca` is defined as the sum of `veca` and − `vecb` and  denoted by `veca` − `vecb` .



`veca ` − `vecb` = `veca` + ( − `vecb` )

Let `vec(OA)` = `veca` and `vec(AB)` =` vecb `

Then `vec(OB)` = `vec(OA)` + `vec(AB)` = `veca` + `vecb`

To subtract `vecb` from `veca` , produce BA toAB' such that AB = AB'.

∴ `vec(AB')` = − `vec(AB)` = −` vecb`

Now by the triangle law of addition

` vec(OB')` = `vec(OA)` +` vec(AB')` = `veca` + ( `-vecb` ) =` veca ` − `vecb`

Example Problems for Adding and Subtracting of Vectors:

Example problem 1:

The position vectors of the points A, B, C, D are `veca` , `vecb` , `2veca` + `3vecb` ,`veca` − `2vecb` respectively. Find `vec(DB) ` and `vec(AC)`

Solution:

Given that

` vec(OA)` = `veca` ; `vec(OB)` =` vecb` ; `vecOC` =` vec2 ` + `3vecb` ; `vec(OD)` = `veca ` − `2vecb `

`vec(DB)` = `vec(OB)` − `vec(OD)` = `vecb ` − (`veca` − `2vecb` ) = `vecb ` − `veca` + `2vecb` = `3vecb` − `veca`

`vec(AC)` = `vec(OC)` − `vec(OA)`

= (`2veca` + `3vecb` ) − `veca` = `veca` + `3vecb `

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Example problem 2:

` vec(OA)` = `2veca` + `3vecb` - `vecc` , `vec(OB)` = `4veca` + `2vecb` + `2vecc` , Find adding and subtracting of vectors

Solution:

Subtracting the two vector,

`vec(AB) ` = `vec(OB)` - `vec(OA)`

`vec(AB)` = `4veca` + `2vecb ` +` 2vecc` - `2veca` - `3vecb` +` vecc`

`vec(AB)` = `2veca` - `vecb ` + `3vecc`

Adding the two vector,

` vec(AB)` = `2veca` + `3vecb` - `vecc` + `4veca` + `2vecb` + `2vecc`

` vec(AB)` = `6veca` +` 5vecb` + `vecc`

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Accurate to Two Decimal Places

Decimal numbers contains decimal places. Here it is very important to know about the place value in decimal numbers.

So what is Decimal Place Value?

It can  be explained by taking example:



let us discuss it using whole number which does not have decimal point in it.

As we go from  right to left the value, the position or the place value gets 10 times bigger.

Similarly as we move from left to right the place value gets 10 times smaller.

What do we get as we move beyond unit place?

Here is where the decimal point comes into picture. If we move beyond unit place the value becomes 1/10, 1/100, 1/1000 and so on. These number which contains decimal point is called DECIMAL NUMBER.

Below is the example for decimal number:



here we can see that the place value of number goes on decreasing as we move beyond decimal points. The value of one in the above example is 1/10 , the place of 2 become 1/100

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Conversion of Decimal Number to Whole Number (accurate to Two Decimal Places)

HOW TO CONVERT  DECIMAL NUMBER TO WHOLE NUMBER?

This can be explained using eg given below:

consider 125.43 and also 125.56

First check the number immediately after the decimal  place. If it is greater than or equal to 5 then add 1 to the unit place and write it as whole number otherwise discard the numbers after decimal point and write the whole number as it is.

In the above example 1)  the number 4 appears immediately after decimal point which is less than 5 .So there is no change and we have to discard all the number proceeding decimal point i,e 4 and 3 and write the number as it is.

whereas in case of example 2) the number which appears immediately is 5 so add one to unit place which becomes 126


Accurate Writing of the Decimal Numbers to 2 Decimal Places

How to write numbers accurate to 2 decimal places?

Even here we have to follow the same rules mentioned above.

examples:

123.4612 ----------------------->123.46

123.4689 ------------------------> 123.47


In the above example 1) check the third number after the decimal point. In this case the third number is 1 which is smaller than 5 so don't add 1 to the number which in 1/100's  place and discard the 3rd digit after the point and write as it is. Here  the final number accurate to 2 decimal point becomes  1 2 3 4 . 4 6

example 2) here the 3rd digit after decimal point is 8 which is greater than 5 so add 1 to the second digit after decimal. Here  the final number accurate to 2 decimal point becomes  1 2 3 4 . 4 7

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Intersection of Two Lines Calculator

Straight line :-

A straight line is generally termed as line. The  curvature of a straight line is 0 . The orientation of a straight line is given by its slope.General representation of straight line is given by  AB.

Explanation to Intersection of Two Lines Calculator

Intersecting lines:

Two lines are said to be intersecting if and only if the have a common root  or  solution.The general form of equation of a line is given by Y=mX +c Where   m= slope, c= y intercept of line .

Features of straight line  of form Y= mX +c :-

i) The straight line is parallel to X axis of m = 0 ie slope of line is 0.

ii) The straight line passes through orign when constant "c" =0.

iii)The straight line makes an angle 45owith "X" axis when m=1 and c=0.

iv) The straight line makes an angle 135o with "X" axis when m=-1 and c=0.

v) The straight line parallel to "y" axis has slope infinity .

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Condition for two line to  intersect :-

For two lines to intersect, their slopes must not be equal .Lines having same slopes are called as parallel lines and intersection point of two parallel lines is 0 or we can say that they meet at infinity. There can be  only one point of intersection for two distinct lines  if they  are not parallel.Multiple intersection of two lines is not possible.

Finding point of intersection of two lines :-

The point of intersection of two lines-can be found by two methods :-

i) By solving the equations

ii) By graphing the equation

Both of these approaches lead to same answer

lets us take two lines  y = 2x+3 ; y= -0.5x + 7 and find the point of intersection

here  primary examination of slopes is to be done.

Slopes are 2& -0.5 so these lines are not parallel lines

So we go by first method solving equations

y = 2x+3 ; ------ equation 1

y= -0.5x + 7 --------equation 2

=> 2x+3  = -0.5x + 7

=> 2.5x = 7-3

=> x= 4/2.5

=> x=1.6

substituting in equation 1 or 2 we get

 y= 2*(1.6) +3

=> y= 6.2

so (1.6 , 6.2) is the point of intersection of  y = 2x+3 ; y= -0.5x + 7

on graphing the equations and plotting them



we can observe that the point of intersection is (1.6, 6.2).

Hence both the approaches give the same result .

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Examples to Intersection of Two Lines Calculator

 EX 1:-

find point of intersection of  lines 3y= 6x +3 and y= 2x+3

Solution:-

Given equations 2y= 4x +2 and y= 2x+3

comparing with standard form of equation y=mx +c

slope of 3y= 6x +3 => y= 2x+1 is 2

slope of line y= 2x+3 is 2

here slopes of lines are equal so they are parallel

so there will be no point of intersection for lines  3y= 6x +6 and y= 2x+3.

 EX 2 :-

FInd the point of intersection of lines y= 2x+1  & x=2

solution :-

Here both lines are not parallel as on comparing with standard equation y=mx+c

slope of  y= 2x+1  is 2 and   x=1 is infinite as its parallel to y axis

substituting x=2 in y= 2x+1

=> y= 2*2+1

=> y= 5

so point of intersection of  y= 2x+1  & x=1 is (1,5)

EX 3:

Find the point of intersection of "X" axis and "Y" axis

solution:

equation of  x axis is  y=0

equation of y axis is x=0

So point of intersection of x, y axis is (0,0) ie origin.

Solve Real Life Algebra Problems

Introduction:

In this branch of mathematics, we use the alphabetical letters in problems like a, b, x and y to denote numbers. In real life algebra, the various operations are addition, subtraction, multiplication, division or extraction of roots on these symbols and real numbers are called algebraic expressions. In the real life a Greek mathematician Diaphanous has developed and solve this subject to a great extent and hence we call him as the father of algebra.

Solve Real Life Algebra Problems-solved Problems :

Example1:

A woman on tour travels first 160 km at 60 km/hr and the next 160 km at 80 km/hr. The average speed for the first 310 km of the hour. Find the total time?

Solution:

The given data’s which are taken and can be solve as follows,

Total time taken = (160 / 60+160 / 80) hr.

=12/3 hr.

Example 2:

The ratio between the speeds of two trains is 5 : 6. If the second train runs 300 km in 3 hours, then what is the speed of the first train?

Solution:

This can be solve as below,

Speed of two trains 5 and 6 km/hr

6x = (300 / 3)

=100

X = (100 / 6) =16

Speed of first train = 16 × 5km/hr

= 80km/hr

Average speed= (340 ×12 / 3)km/hr

= 87.62km/hr.

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Solved Problems for Real Life Algebra:

Example1:

A man traveled a distance of 30 km in 6 hours. He traveled partly on foot at 2 km/hr and partly on bicycle at 6 km/hr. What is the distance traveled on foot?

Solution:

The problem which can be solve as,

Distance traveled on bicycle = (30 - x) km

x / 2 + (30-x) / 6 = 6

6x + 2 (30-x) = 6 × 12

x = 12 km

Example 2:

A bus crosses a 600 m long street in 2 minutes. What is his speed in km per hour?

Solution:

This is a real life problem in algebra which can be solve as,

Speed = (600 / 2 × 60)m/sec

= 5 m/sec

We have to convert  m/sec to km/hr

= (5 × 18/2)

= 45 km/hr

Example 3:

The ratio between the speeds of two trains is 3 : 5. If the second train runs 300 kms in 3 hours and average speed is 50 km/hr.what is the speed of the first train?

Solution:

The problem in the real life algebra which can be solve as follows,

Speed of two trains 3 and 5 km/hr

5x = (300 / 3)

=100

X = (100 / 5) = 20

Speed of first train = 20 × 5 km/hr

= 100 km/hr

Average speed = (50 ×9 /3 ) km/hrs

= 150 km/hr.

These are all the solve problems which are in the real life algebra.

Envision Math California

Envision MATH California is Based on problem-based, interactive knowledge and theoretical understanding. What does that mean? Students with theoretical thoughtful know more than inaccessible details and techniques, they are capable to study fresh thoughts by linking to the thoughts they previously be acquainted with. They study ideas by interact with problems; this communication takes place on a daily basis and is exact to the idea being trained.lets see some problems on envision MATH California.

Envision Math California Problems:-

Problem 1:-

Find the perimeter of the following figure:-  

                                                                                                         
Solution:-

We need to find the perimeter of the following figure

Perimeter can be found by adding lengths of the given figure.

Here the length of each sides are the following

4.8 yards, 4.3 yards, 1.1 yards, 4.5 yards, 4.6 yards, 1.6 yards.

Perimeter = sum of all given sides

= 4.8 yards + 4.3 yards + 1.1 yards + 4.5 yards  + 4.6 yards + 1.6 yards.

= 20.09  yards.

So the perimeter of the given figure is 20.09 yards.

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Problem 2:-

Find the Perimeter of the given figure.

Solution:-

We need to find the perimeter of the following figure

Perimeter can be found by adding lengths of the given figure.

Here the length of each sides are the following

4.7 yards, 2.3 yards, 1.1 yards, 1.1 yards,  4.3 yards, 2.4 yards, 3 yards.

Perimeter = sum of all given sides

= 4.7 yards+ 2.3 yards+1.1 yards+ 1.1 yards+  4.3 yards+ 2.4 yards+ 3 yards.

= 18.9 yards.

So the perimeter of the given figure is 18.90 yards.

Problem 3:-

Find the Perimeter of the given figure.

Solution:-

We need to find the perimeter of the following figure

Perimeter can be found by adding lengths of the given figure.

Here the length of each sides are the following.

2 feet, 2feet, 5 feet, 5 feet, 2 feet.

Perimeter = sum of all given sides

= 2 feet+ 2feet+ 5 feet+ 5 feet+ 2 feet.

= 16 feet

So the perimeter of the given figure is 16 feet.

Envision Math California Practice Problems:-

Find the area of the given parallogram:-

Answer:- 84 cm2


Between, if you have problem on these topics Business Math, please browse expert math related websites for more help on area of hexagon formula.

Function Derivative Calculator

Solving 2nd order derivatives of a function

1) Solve the derivative for the function  f(x) =  x^2 + 8x + 9

Solution :  The given function is  f(x) = x^2 + 8x + 9

Differentiate the above equation with respect to 'x' . It is represented as f'(x) .

f'(x)  =   `(d(x^2))/dx`   +  `(d(8x))/dx`   +  `(d(9))/dx` .

f'(x)   =     2x   +   8   +   0

f'(x)    =     2x    +   8   .

The answer is   f'(x)  =  2x  +  8 .

2) Solve the derivative for the function   f(y)  =  y^2  +  10y  + 3

Solution :   The given function is  f(y)  =  y^2  +  10y  +  3

Differentiate the above equation with respect to 'y' . It is represented as f'(y)  .

f'(y)  = `(d(y^2))/dy`   +  `(d(10y))/dy`   +   `(d(3))/dy`

f'(y)  =   2y   +  10   +   0

f'(y)  =   2y   +   10

The answer is  f'(y) = 2y  +  10

Solving third Order Derivative Functions

1) Find the derivative of the function  f(x)  =  x3 + 3x^2 + 18x  +  20

Solution : The given   function  f(x)  =  x3 + 3x^2 + 18x  +  20

Differentiate the above function with respect to 'x' .

f'(x)  =  3 x^2   +  3  ( 2 ) x   +  18   +  0

f'(x)   =  3x^2  +  6x  +  18  .

The answer is   f'(x)   =  3x^2  +  6x  +  18  .

2) Find the derivative for   f(x)  =  6x3  + 5x^2  +  3x  +  1

Solution : The given function is f(x)  =  6x3  + 5x^2  +  3x  +  1

Differentiate the above f(x) with respect to 'x'  .

f'(x)  =  6 (3)x^2  +  5 (2)x   +  3  + 0

f'(x)   =  18x^2   +  10x   +  3

The answer is  f'(x)   =  18x^2   +  10x   +  3

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Solving 4th Order Derivative Function

1) Solve the derivative for the function  f(y) = y^4  + 3y^3  +  5y^2  + 4y  +  9

Solution : The given function  f(y) = y^4  + 3y^3  +  5y^2  + 4y  +  9

DIfferentiate the above equation with respect to 'y' .

f'(y)  =  4y^3  +  3(3)y^2  +  5(2)y  +  4  +  0

f'(y)  =  4y^3   +  9y^2  + 10y   +  4

The answer is    f(y) = y^4  + 3y^3  +  5y^2  + 4y  +  9

2) Solve the derivative for the function f(y)  =  6y^4  +  y^3   +  y^2  + 10 y  + 3

Solution :  The given function  is    f(y)  =  6y^4  +  y^3   +  y^2  + 10 y  + 3

DIfferentiate the above equation with respect to 'y' .

f'(y)   =  6(4)y^3   +  3y^2    +  2y   +  10  +   0

f'(y)   =   24y^3   +  3y^2   +  2y   +  10

The answer is    f'(y)   =   24y^3   +  3y^2   +  2y   +  10

Algebra is widely used in day to day activities watch out for my forthcoming posts on Derivative Calculator and Derivative of Natural log. I am sure they will be helpful.

Free Algebra Classes

Algebra is a cluster of mathematics, which is used to create mathematical problems of valid-globe actions and control problems that we cannot explain using arithmetic.

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Free algebra mainly different kind of lesson handled here.

Algebra uses the cipher as calculation for addition, subtraction, multiplication and division and it includes constants, operating signs and variables.

Algebraic equations represent a collection, what is finished on one side of the range with a number to the other side of the range.

Below lesson topics explain the algebra classes.

Algebra Free Algebra Classes Topics:-

Exponents

Radicals

Polynomials

Factoring

Division of Polynomials

Solving Equations

Solving Inequalities

Lines

Solving Quadratic Equations

Complex Numbers

Graphing Quadratics (Parabolas)

Systems of Equations (2x^2's)

Systems of Equations (3x^3's)

Determinants and Cramer's Rule

Functions

Inverse Functions

Exponentials and Logarithms

Absolute Value Equations and Inequalities

Sequences and Series Combinatory

Advanced Graphing

Graphing Polynomials

Graphing Rational Functions

Matrices

Above the topics free algebra classes based process used here. And then free algebra mainly different kind of classes handle through now.

Free Algebra Classes - Problems:

Problem 1:

Solve for x: 3x+6 = 4x+8.

Solution:-

3x+6 = 4x+8

Subtract by 4x on both sides,

3x-4x+6 = 4x-4x+8

-x+6 = 8

Subtract by 6 on both sides,

-x+6-6 = 8-6

-x = 2

So, x = -2.

Order of Operations problem:-

Problem 2:

5 + 4 * 3y (x + 2)

Solution:

- 5 + 12xy + 6

Negative Exponents of Numbers:-

3 -3

Solution:- 1/27

Negative Exponents of Variables:-

K -3

Solution:- 1/ k^3

Negative Exponents in Fractions

X -2 y/x

Solution:- y/x 3

Substitution method:-

Problem:

Evaluate: w + a^2, where w = -12 and a = z + 3

-12 + (z + 3 )2 = z2 + 2(3(z)) + 3

Solution: Z2 + 6z + 9

Finding a Greatest Common Factor (GCF) of Two Terms:-

9 x^2y, 3 xy^3

3xy

Factoring a GCF from an Expression:-

4x^2 + 16x^3

4x^2(1 + 3x)

Factoring a Difference between Two Squares

x^2 – 9

(x + 3) (x – 3)

Factoring a Trinomial Completely:-

X^4 + 7x^3 + 12 x^2

x^2(x + 3) (x + 4)

Solve By Factoring

x^2 + 3x + 2

Solution:- -1, -2.


Practice Problems for Free Algebra Classes:

1. Find factors and root of the equation x^2 + 10x + 24

Answer: -6, -4

2.Find factors and root of the equation x^2 +11x + 28

Answer: -4, -7

3.Find factors and root of the equation x^2 – 2x + 8

Answer: 4, -2

4.Find factors and root of the equation 2x^2 – 15x + 8.

Answer: 1, 15

I am planning to write more post on long division polynomials, Types of Polynomials. Keep checking my blog.

Rate Histograms Tutoring

In rate histograms tutoring the word rate is a Greek term. In rate histograms the interval with each frequency value are equal.  In mathematical the rate histogram is used to counting the number of observations in the histogram. Tutor is a person who shares his knowledge with student which is called tutoring. Now the tutor will discuss about the study histogram in this tutorial.

Example for Rate Histograms Tutoring:

Example: 1

While doing a project for P.E. class, Alonzo researched the number of athletes competing in an international sporting event. By using this data complete the histogram below.

Solution:
In the above histogram the missing bar is for the range 41 to 50 athletes and count the number of values in this range. Two countries have between 41 and 50 athletes so set the height of the bar to 2.

Example: 2

Prepare a histogram  by using the following data in the frequency table

Solution:
From the above table take the class intervals on the X-axis and frequency on the Y-axis  and mark the Y-axis scales as 0,5,10,15,20 To avoid errors in our histogram we marking the class intervals so that  the histogram be more perfect  and draw the rectangle bars .

 Example: 3

Prepare a histogram  by using the following data in the frequency table

Solution:
From the above table take the class intervals on the X-axis and frequency on the Y-axis  and mark the Y-axis scales as 0,5,10,15,20,25,30,35. To avoid errors in our histogram we marking the class intervals so that  the histogram be more perfect  and draw the rectangle bars .

These are the examples of rate histograms tutoring.

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Mathematical Information of Rate Histogram:

In the mathematical terms, the rate histograms are commonly used for displaying and mapping of mi items. These items are generally used  for counting the number of observations that are present in the histogram. The mathematical formula for calculating the rate histogram is given below,

`sum_(P=i=1)^k`

where,

p is used to represent the sum of  number of observations in the given table.

mi is used to represent the certain conditions in the table,

k is used to represent the total number of bins present in the table.

This is the general information of rate histograms tutoring in mathematics. Tutoring is one of the process. Student can easily develop their knowledge through online tutoring . Tutoring is the correlation between student and tutor.

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Solve Equations with Exponent

Introduction to solving equations with exponents

Exponent equations are the equations in which variable appear as an exponent.
To solve these equations rules and laws of exponents are used. Exponent equations are of two types
(1) Exponent equations in which bases are same
(2) Exponent equations in which bases are different.

Steps to Solve Equations with Exponent

Solving Exponential Equations of the same base

1) Ignore the bases, and simply set the exponents equal to each other

2) Solve for the variable

 When the bases of the terms are different

1) Ignore the exponents; rewrite both of the bases as powers of same number. For    example if there are 2 and 4 in the bases, then convert base 4, in to base 2 by writing it again as (2)^2

2) once the bases are same , ignore them

3) Equalize the exponents

4) Solve for variable

Simple Problems of Equations with Exponents

Solve for variable Answer 1. 3m  =  35 Since the bases are the same, set the exponents equal to one another: m = 5 2. 5t   = 125 125can be expressed as a power of 5: 5t  = 53 t = 3  3.  493y=343 49 and 343 can be expressed as a power of 7: [(7)2]3y = 73 76y = 73 6y = 3 y = 1/2

More Problems of Equations with Exponents

Solve for x. Answer 1.  52x+1  =  53x-2 Since the bases are the same, set the exponents equal to one another: 2x + 1 = 3x - 2 3 = x 2.  32x-1  = 27x  27 can be expressed as a power of 3: 32x-1  = 33x 2x - 1 = 3x -1 = x 3.   43x-8  = 162x 16 can be expressed as a power of 4: 43x-8= [(4)2]2x 3x - 8 = 4x   -8 = x I like to share this multiplying rational expressions solver with you all through my article.

Ratios and Unit Rates

Introduction to Ratios and Unit rates:
                  In mathematics, there are more topics which perform arithmetic and binary operations. Ratios are the mathematical terms which are used to make a relationship between any two numbers or variables. They are used to compare two values of any quantities. A rate is a ratio which is used to compare any two terms. Unit rate is also a rate in which one quantity is corresponding to the unit of another quantity. Let us discuss about the ratios and unit rates with definitions.

Definitions for Ratios and Unit Rates:

           Here we see the definitions for rate and ratios.

Rate:

          The rate is defined as a ratio which is used for comparing any two different numbers. It is a ratio between two or more numbers.

Formula for rate:

              The term rate can be calculated or found as the distance travelled (d) to a time period (t) per hours or minutes in the form of proportion.

Unit rate:

          A unit rate is used to check the given equation or to compare the given quantity. It defines how many total units are there in one quantity to the other quantity.

Example for common unit rates:

         Salary per month, cost per item, kilometers per hour, dollars per pound.

Ratios:

        Ratios are defined as comparing of two numbers specified in a form by the use of a colon. It is a relationship between two or more things.

Example for ratio:

        The numbers 4 and 8 can be represented in ratio as 4: 8. It is also given as fraction or proportion form as `4/8` .

These are the definitions and explanation for ratios and unit rates.

Problems for Ratios and Unit Rates:

       We see some problems for ratios and unit rates.

Problem 1:

          Reena has bought 6 apples, 5 chocolates and 2 bananas.

 i) What is the ratio of apples to bananas?

ii) Mention the ratio of chocolates to bananas.

Solution:

      i) The ratio of apples to bananas can be written as one quantity of number to other.

So, the ratio is 6 : 2.

      ii) In this, the ratio of chocolates to bananas can be mentioned as number of chocolates to number of bananas.

That is, 5 : 2.

Problem 2:

     Find the unit rate for 52 kilometers per 3 hours.

Solution:

        The unit rate can be found as one quantity to other.

Therefore unit rate = `52/3`

= 14

Hence the unit rate = 14 kilometers per hour.

These are the problems for ratios and unit rates.

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Monomial Factor with Exponents

Introduction to monomial factor with exponents:

"Monomial factor with exponent" means the monomial term with exponents need to factorise. Here we need to understand each and every terms.

From the term mono we can understand it is talking about a single term, so "monomial" means it is an algebraic expression which is having only one single term. We can also say that monomial is a product of numbers and variables, where variables can be any letter or a power of letter( power or index is called as exponents).Monomial can be a single letter or a number also.So, Monomial factor with exponents can be a single term consist of a number with variable and also with exponents and which can be factorise also. Here the factor means when we multiply two elements together we can get the final product.
Now we come to the term "Exponent" here it means the number can be multiply how many times which depends upon the power. Now we have a small idea about the term "Monomial factor with exponent".

Examples on Monomial Factor with Exponents:

1. 12x²

This a monomial, here we have one single term which is 12x². Here 12 is a number with a variable x having the power as 2. x² is  called  x to the second power. Here x is the base and 2 is the exponent.

2. -24abc

This is also a monomial having a single term. The exponent here is 1 for each variables.

Above are the examples of monomial with exponents.

Let us take an example and learn to facrorise:

1)12 x⁴

first we will find the factors of 12 which is

1 x 12

2 x 6

3 x 4

Now we will factor x² .This can be written as

x.x³

x^2.x²

x⁰.x⁴

As per the exponent law (a)^m.(a)ⁿ=a^m+n

so if we add the power we get the final results as x⁴.

2) 6a⁴b¹⁰= (2a³b²) .(  ? )

Here we need to find the missing terms. As we have already one part, to find the other part let us find the factor.

6.............can be written as

1 x 6

2 x 3

As we have already 2 in the question so we have to consider the factor for 6 as 2 x 3.

Now factor a⁴b¹⁰

a⁴=a¹.a³                           and  b¹⁰=b¹.b⁹

a⁴=a².a²                                    b¹⁰=b².b⁸

a⁴=a⁰.a⁴                                    b¹⁰=b³.b⁷

                                                     b¹⁰=b⁴.b⁶

                                                     b⁵=b⁵.b⁵

In the question we have a³b². so from the above factorisation we can find as the final results as 3ab⁸

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Conclusion on Monomial Factor with Exponents:

In monomial factor with exponents always we have to remember that whenever we multiply the monomials we add the exponents also. Degree of monomials is the sum of exponents of all the letters.