Least to Greatest Calculator

The order of the least number to the greatest number is the called the ascending order. The ascending order is doing on arrange the give number for the given order and also the least to greater calculator is used to give a input to mixed order. Then click the calculate button to arrange the given order. In this article id discuss about the least to greatest calculator.

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Least to Greatest Calculator - Examples:

Least to greatest calculator - Example 1:

          75, 710, 15, 425, 235, 145, 505 Arrange the order of least to greatest order

Solution:

Alter input as fractions if necessary:

          75/1, 710/1, 15/1, 425/1, 235/1, 145/1, 505/1

The least common denominator (LCD) is: 1.

Alter as equivalent fractions with the LCD:

75/1, 710/1, 15/1, 425/1, 235/1, 145/1, 505/1

Ordering these fractions by the numerator:

          15/1  <   75/1  <   145/1  <   235/1  <   425/1  <   505/1  <   710/1

Therefore, the order of your input is:

          15  <   75  <   145  <   235  <   425  <   505  <   710

Least to greatest calculator - Example 2:

705, 7110, 115, 4025, 2135, 1435, 5705

Solution:

Alter input as fractions if necessary:

          705/1, 7110/1, 115/1, 4025/1, 2135/1, 1435/1, 5705/1

The least common denominator (LCD) is: 1.

Alter as equivalent fractions with the LCD:

          705/1, 7110/1, 115/1, 4025/1, 2135/1, 1435/1, 5705/1

Ordering these fractions by the numerator:

          115/1  <  705/1  <  1435/1  <  2135/1  <  4025/1  <  5705/1  <  7110/1

Therefore, the order of your input is:

          115  <  705  <  1435  <  2135  <  4025  <  5705  <  7110

Least to Greatest Calculator - more Examples:

Least to greatest calculator - Example 1:

13, 19, 16, 112, 118, 211, 214, 218, 310, 411

Solution:

Alter input as fractions if necessary:

          13/1, 19/1, 16/1, 112/1, 118/1, 211/1, 214/1, 218/1, 310/1, 411/1\

The least common denominator (LCD) is: 1.

Alter as equivalents fractions with the LCD:

          13/1, 19/1, 16/1, 112/1, 118/1, 211/1, 214/1, 218/1, 310/1, 411/1

Ordering these fractions by the numerator:

          13/1  <  16/1  <  19/1  <  112/1  <  118/1  <  211/1  <  214/1  <  218/1  <  310/1  <  411/1

Therefore, the order of your input is:

          13  <  16  <  19  <  112  <  118  <  211  <  214  <  218  <  310  <  411

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Least to greatest calculator - Example 2:

`2/3, 5/2, 6/7, 7/4, 2/8, 6/5, 8/6, 2/3`

Solution:

 Alter input as fractions if necessary:

          `2/3, 5/2, 6/7, 7/4, 2/8, 6/5, 8/6, 2/3`

The least common denominator (LCD) is: 840.

Alter as equivalents fractions with the LCD:

          `560/840, 2100/840, 720/840, 1470/840, 210/840, 1008/840, 1120/840, 560/840`

Ordering these fractions by the numerator:

         ` 210/840<560/840=560/840<720/840<1008/840<1120/840<1470/840<2100/840`       

Therefore, the order of your input is:

         ` 2/8<2/3=2/3<6/7<6/5<8/6<7/4<5/2`

Value of an Integral Type Expected

Expected value is one of an important concept in probability. In probability, expected value of a given real-values are chance the variables as present a compute of the center of the distribution of the variable. In online, few websites are providing math tutoring. Tutor, will give step by step explanation for the expected value problems. Expected problems are deals with, probability, geometry distribution, etc. in this article we shall discuss for value of an integral type expected.


Sample Problem for Value of an Integral Type Expected:

Value of an integral type expected problem 1:

Evaluate the expected value from the given continuous random variable using uniform distribution. Value is expected from the interval value is 3 < x < 7.

Solution:

Given:

A given interval value for the uniform distribution is a = 3 and b = 7

Formula for finding the expected value for uniform distribution is

E(X) = `int_a^bx f(x)dx` --------------- (1)

Here, the value of f(x) is `1/(b - a)` , for 3 < x < 7

= `1/(7 - 3)`

= `1/4`

f(x) = `1/4`

In the next step we put the f(x) value in the above equation, we get

= `int_3^7x (1/4) dx`

= `1/4int_3^7x dx`

= `1/4[ x^2/2]_3^7`

= `1/4[(7)^2/2-(3)^2/2 ]`

= `1/4 ` [1/2 (7)2 - (3)2]

= `1/8` [49 - 9]

= `1/8` (40)

= `40/8`

= 5

We get expected value E(x) is 5.

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Value of an Integral Type Expected Problem 2:

Evaluate the expected value from the given probability density function using exponential distribution `e^-2x` with the interval of [0, `oo` ].

Solution:

Given:

A given probability density function f(x) is `e^-2x` .

Formula for finding the expected value for uniform distribution is

E(X) =` int_0^ooxf(x)dx` --------------- (1)

In the first we find out the expected value of the exponential distribution function


=` int_0^oox(e^(-2x))dx`

=` int_0^ooxe^(-2x)dx`

Here we use `int udv = uv - int vdu`

u = x       dv = `e^(-2x)`

u' = 1      v = `(e^(-2x))/-2 `

u'' = 0      v' = `(e^-2x)/4`

in the next step we substitute the above values, we get

= ` [(xe^(-2x))/-2- (e^(-2x))/4]_0^oo`

= ` [((ooe^(-2oo))/-2-(e^(-2oo))/4 )- ((0 - e^(0))/4)]`

Here we use `e^-oo` = 0, e0 = 1/4

= 0.25

We get the expected value E(X) is 0.25.

Solve Graphing Trigonometric Functions

Graphs of Trigonometric Functions


The relation (variation) between the angles and the values of the trigonometric ratios at them are plotted by graphs .

Pro 1:    Graph of  y  =  sinx

x    -`pi`    `-pi/2`    0    `pi/2`    `pi`    `(3pi)/2`    `2pi`    `(5pi)/2`    `3pi`
y    0    -1    0    1    0    -1    0    1    0
Plot the graph in the coordinate plane by taking angles x in radian measure on X-axis and the values of sinx  =  y on Y-axis .

Sol :



By choosing a suitable scale , plot and join the points of  y = sin x with a smooth curve to get the graph .

This curve passes through the origin . The values of sin x vary between -1 and +1 which are respectively the minimum and the maximum . It is in the shape of a wave whose wave length is 2`pi` . This wavelength is nothing but the period .

SInce  `-1<=sinx<=1`   `AA`  x  `in`  R , the sine function is bounded . It can be proved that it is a continuous function on  R .

Pro 2:  Graph of  y  =  cosx

x    `-pi`    -`pi/2`    0    `pi/2`    `pi`    `(3pi)/2`    `2pi`    `(5pi)/2`
y    -1    0    1    0    -1    0    1    0

Sol :



By choosing a suitable scale , plot and join the points of y = cos x with a smooth curve to get the graph .

This curve does not pass through the origin . It is evident that the maximum and the minimum values are  +1 and -1 respectively .

Since  `-1<=cosx<=1`   `AA`  x  `in`  R  ,  the cosine function is bounded . It can be shown that is a contnous function and periodic with `2pi` as the period .

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Solve Graphing Trigonometric Functions : Tanx and Cotx

Pro 3:  Graph of  y  =  tanx

x    `-pi/2`    0    `pi/2`    `pi`    `(3pi)/2`
y    not defined    0    not defined    0    not defined

Sol :



The curve nearly  touches the vertical lines  at  x  =  ...........`-pi/2`  , `pi/2` , `(3pi)/2` , . .. . . . .

The curve has nreakes at  x  =  (2n + 1)`pi/2`   ,  n  `in`  Z  and passes through the origin . it is not bounded .

The tan function is periodic and `pi`  is the period of it .

Pro 4:  Graph of  y  =  cotx



The curve nearly touches the vertical lines at  x  =  . . . . . . . `-pi` , 0 , `pi`  . . . . . . .  .

The curve has breakes at  x  =  n`pi`   ,    n  `in`   Z   and does not pass through the origin . It is not bounded .

The cot function is periodic and  `pi`  is the period of it .

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Solve Graphing Trigonometric Functions : Secx and Cosecx

Pro 5:   Graph of  y = secx



The curve nearly touches the vertical lines at x  =  . . . . . . . . `-pi/2` , 0 , `pi/2` , `pi` , `(3pi)/2` . . . . . .

The curve has breakes at  x  `in`   (2n+1) `pi/2` ,  n `in`  Z . It is not bounded . The values  of  secx  lie  in  `(oo,-1]uu[1,oo)`

The secant function is periodic and 2`pi`  is the period of it .

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Pro 6:   Graph of  y  =  cosecx



The curve nearly  touches the vertical lines at  x  =  . . . . . . 0 , `pi/2` , `pi` , `(3pi)/2` , 2`pi` , . . . . . .

The curve has taken

Multiplication and Division of Whole Numbers

Division:

For numbers, the operation of assert how many times one number, the divisor, is contained in a second, the dividend. The result is known as quotient and if the divisor is not contained an integral number of times in the dividend; any number left over is called the remainder. The symbol of division is / or ÷.

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Multiplication:

The product of two numbers is known as multiplication. Here we are going to learn about some example problems of multiplication and division whole numbers.

Example Problems of Multiplication Whole Numbers:

Example 1:

12*14

Step 1: First we need to take 4 and multiply with 12 = 48.

Step 2: Now we need to take 1 and multiply with 12 = 12.

Step 3:

Step 4: Therefore the answer is 168.

Example 2:

23*42

Step 1: First we need to take 2 and multiply with 23 =46.

Step 2: Now we need to take 4 and multiply with 23 =92.

Step 3:



Step 4: Therefore the answer is 966.

Example 3:

231* 14

Step 1: First we need to take 4 and multiply with 231 = 924.

Step 2: Now we need to take 1 and multiply with 231 = 231

Step 3:



Step 4: Therefore, the answer is 3234.

These are the examples problems of multiplication of whole numbers.

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Examples of Dividing Whole Numbers:

Example 1:

Solve: `(143)/(3)`

Solution:

Step 1: Here we need to perform division operation `(143)/(3)` .

Step 2: First we need to take first two digits that is 14.  And we know that 4 times 3 = 12.

Step 3: Now we need to minus 12 from 14 = 2.

Step 4: Here we need to get the next term.

Step 5: The next term is 3. So presently we have 23.

Step 6: We know that 7 times 3 = 21.

Step 7: So the quotient is 47 and remainder is 2.



Example 2:

Solve:  `(276)/(12)`

Solution:

Step 1: Here we need to perform division operation `(276)/(12)` .

Step 2: First we need to take first two digits that is 27.  And we know that 2 times 12 = 24.

Step 3: Now we need to minus 24 from 27 = 6.

Step 4: Here we need to get the next term.

Step 5: The next term is 6. So presently we have 36.

Step 6: We know that 3 times 12 = 36.

Step 7: So the quotient is 23 and remainder is 0.



These are the examples of division of whole numbers.

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Triangles Venn Diagram

Venn diagrams are used to describe the relationship of the sets. The description of some sets is given and you are asked to draw a Venn diagram to illustrate the sets. You are well known about the triangles that they are having three sides. Based on their sides and the angles, it is classified into different types. The relationship of the triangles can also be described using Venn diagrams.

Triangles and Venn Diagrams

Venn diagrams:

Venn diagrams to determine the relationships between the sets such as subsets and intersections.

Venn diagram for A within B :



Here, all members of A belongs to B or A ⊂ B or A ∪ B = B or A ∩ B = A or n(A ∩ B) = n(A).

Venn diagram for  A overlap B:



Here, some members of A belongs to B or A ∩ B ≠ Ø or n(A ∩ B ) ≠ 0

Venn diagram for disjoint sets of A and B :



Here, no members of A belongs to B or A ∩ B = Ø or n(A ∩ B ) = 0

Triangles:

scalene triangle:

A triangle is said to be scalene triangle, if all the sides are of different lengths.Some scalene triangles are also right triangles.



Right triangle:

A triangle is said to be right triangle, if one interior angle is exactly 90°. Some right triangles are also scalene triangles.



Equilateral triangle:

A triangle is said to be equilateral triangle, if all the sides are of same length.



Isosceles triangle:

A triangle is said to be isosceles triangle, if two sides are of same length and two angles are equal.



Examples

Let U is the set of triangles, A is the set of isosceles triangles, B is the set of equilateral triangles and C is the set of right-angled triangles.

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Draw a Venn diagram.

Solution:

We have to determine the relationships between the sets.

All the equilateral triangles are said to be isosceles triangles, so B ⊂ A. (within)

Some of the right-angled triangles are said to be isosceles triangle. C ∩ A ≠ Ø (overlap)

No right-angled triangles are equilateral triangles. C ∩ B = Ø (disjoint)

Venn diagram for the triangles:



Let U is the set of triangles, S is the set of scalene triangles and R is the set of right triangles.

Draw the Venn diagram

Solution:

The relationship between scalene and right triangle to be defined.

Some of the right triangles are scalene triangles.

Integer Divide by Zero

Integer is the number which is greater than the zero or less than the zero and a number greater than zero is a positive and less than the zero is called negative. Zero has no sign like positive and negative sign. In number line two integers are same distance from the zero in opposite directions are the opposites.In this section we are going to see about any integer which is divided by zero and the integer divided the integers.


Division Property of Integers:

For integers we have some properties, let assume a, b be the two integers for that positive and negative integers. But with the zero it has some special property any integer which divided by the zero we get infinity and the zero divided by the any integer we get zero.


Let a & b be the two integers, where  `a/b` is not always an integer.

Examples: `(-4)/5` ; ` 5/ (-3)`  are not integers ;  `1/2 ` is an integer.

For any integer 'a' is not equal to zero.  `a/a ` = 1; and  `a/1` = a

Examples:
`5/5` = 1

`5/1` = 5

` 1/5`  is not  an integer

For any integer 'a' is not equal to zero.
`a/(-1)` = -a; ` a/(-a)` = -1.

For every non zero integer a; `0/a` = 0.

Examples:

` 3/ (-1)` = -3

`3/ (-3)` = -1

`1/ (-3)` is not an integer

Algebra is widely used in day to day activities watch out for my forthcoming posts on Integers and Absolute Value and greatest integer function. I am sure they will be helpful.

Problems with Integer Division:

Problem 1:
Divide +95 by 5.

Solution:
=    ` 95/5`

=19 is an integer


Problem 2:
Divide  -65 by 0.

Solution:
We know that the any integer divided by 0 is infinity by division property. Here, one twenty five by zero
= ∞

Problem 3:

Divide 106 by 0.

Solution:
We know that the any integer divided by 0  is infinity by division property. Here,thousand divided by zero.

= ∞

Problem 4:

Divide 0 by 55

Solution:
We know that the zero divided by any number is zero by division property

=0

Fraction Chart for Math

Fraction can be defined as part of something bigger than that. Suppose your class has 40 boys and 60 girls thus totaling 100 students. Boys are the part of the class and so are girls. But the class consists both boys and girls and is bigger set. In the class, we say 40 out 100 are boys. We can write this as a Fraction as below



We observe that number 40 is written at the top and then a bar below it and then number 100 below the bar. This is how the fractions are represented.

The boys in the class form a part of the class and they are 40 in number. The class is bigger and has 100 students. So, the part here is the boys and the whole is the class. So, this fraction represents that 40 boys are a part of a class of 100 students. Thus we see, fraction represents the part of a whole.

It must also be noted that there could be fractions where the part can be bigger than the whole.

Fractions Illustration

Let us illustrate how fractions can be formed as a part of the whole. Let us a draw a full circle first as below
 

Now let us divide the circle into two different parts and shade one part green as shown below



So we have total two parts and one part is green. This is represented as ½.

Now let us divide the original circle into three and shade two parts of it green as below



So we have total three parts and two parts are green. This is represented as `(2)/(3)`

 Thus fractions represent a part of the whole.

As we saw a fraction has one number at the top of a bar and one number below it. The top number is called the numerator and the bottom number is called the denominator as shown below.



The chart below shows fractions in pictorial form



Types of Fractions

• If the numerator is less than the denominator, it is called a proper fraction.

All the fractions we saw above like `(1)/(2)` , `(2)/(3)`  etc are proper fractions because the top number or numerator was smaller then the bottom number or denominator

• If the numerator is greater than the denominator, it is called an improper fraction.

Example. `(4)/(3)` , `(7)/(2)`  etc. These fractions represent where the part is greater than the whole. We will illustrate with an example. Suppose your teacher is conducting an examination for you. The examination had 25 questions and the teacher asked you to solve any 20 questions to get the maximum mark of 20. Suppose you managed to solve all the 25 correctly then your score in the test will be `(25)/(20)` . That means that you solved more than the maximum required.

• There are another kind of fractions, which has both a whole number and a proper fraction. These fractions are called mixed fractions.

Example 3 `(1)/(2)` ,4`(2)/(3)`  etc.  Thus these fractions are a mix of whole number and a proper fraction and hence they are called mixed fractions.

Exercises on Fractions

Pro 1: From the figure below, write the region shaded in the circle as a fraction of the whole circle

 i)



ii)





iii)



Ans : (i) `(2)/(10)` (ii) `(5)/(6)`       (iii) `(7)/(10)`

Pro 2: Identify the proper, improper and mixed fractions from the following

`(7)/(8)`  ,7 `(1)/(2)`   , `(16)/(2)` , `(1)/(8)` , `(4)/(3)`

Ans: Proper fractions - `(7)/(8)` ,`(1)/(8)`   - Numerator of these fractions are smaller then the denominator

Improper Fraction - `(16)/(2)` , `(4)/(3)`  - - Numerator of these fractions are bigger then the denominator

Mixed Fraction - 7 `(1)/(2)`   - This fraction has  a whole number and a proper fraction

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.

If you have problem on these topics circle Arc .

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

Algebra is widely used in day to day activities watch out for my forthcoming posts on Standard Deviation Calculator and relative standard deviation. I am sure they will be helpful.

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

Algebra is widely used in day to day activities watch out for my forthcoming posts on statistical graphs and free algebra help. I am sure they will be helpful.

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

Algebra is widely used in day to day activities watch out for my forthcoming posts on  list of irrational numbers. I am sure they will be helpful.

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


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