Definition Data Table

Let us see about definition data table. Generally, the data tables are made by the number of rows and the columns. In data table, the rows and columns are separated by the number of lines or line segments. In data table, each row and column has the data. Generally the data table is premeditated by the statistical graphs. The group of data can be creating the data table. Data tables are conniving by the graphs. The statistical graphs are depends on the data table. The data table rows are positioned as horizontal and the columns are positioned as vertical.

Examples of Data Table:

Let us see the example problems of data table.

Example 1:

Define the data table.

Name of pets	Number of students
Dogs	80%
Cats	20%
Fish	64%
Parrot	75%
Peacock	64%
Dove	29%
Sparrow	34%

Solution:

The definition of data table is described below. The data table exhibits the name of pets and the number of students. The data table is defined by the bar graph.  Name of pets is represented as x-axis and the number of students is represented as y-axis.



This is the definition of data table.

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Example 2:

Define the data table.

Activity	Number of students
Visit friends	20%
Talk on phone	15%
Play sports	46%
Earn money	50%
Use computers	61%
Preparing multimedia	31%
Playing cards	21%
Doing home work	11%

Solution:

The definition of the data table is defined below. The data table exhibits the activity and the number of students. The data table is defined by line graph.  Activity is represented as x-axis and the number of students is represented as y-axis.



This is the definition of data table.

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One more Example of Data Table:

Define the sample data table.

Year	Non-Employees rate
2000	12%
2001	45%
2002	31%
2003	61%
2004	55%
2005	72%
2006	81%
2007	74%
2008	94%
2009	75%
2010	95%

Solution:

The definition of the data table is described below. The data table exhibits the year and the non-employees rate. The data table is defined by the scatter plot graph.  The year is represented as x-axis and the non-employees rate is represented as y-axis.



This is the definition of data table.

Trigonometry Xi

In this section we will see about trigonometry xi. Eleventh average trigonometry is also recognized as the division of the main dealing with trigonometry functions, angle, etc. It gives the association and angles in detail with their problems. Sine, Cosine and Tangent are the trigonometric meaning concerned in trigonometric semi position identity. We have worked problems and practice problems along with solution in below. Let us see about the topic trigonometry xi.

Solved Problems for Trigonometry Xi:

Let us see about the topic trigonometry xi,

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Solved problem 1: Try to calculate the radius of the circle in which a central angle of 45 degree intercepts an arc of length 10 cm. (use π value as `22/7` )

Solution:

Given length = 10 cm and angle = 45 degree

θ = 45 degree = `(45Pi) / 180` = π/4

r = l/ θ

r = `(10 * 4) /Pi` = `(10 * 4 *7) / 22` = 12.72 cm

Therefore, the radius of the circle = 12.72 cm.

Answer: The radius of the circle = 12.72 cm.

Solved problem 2: If cos x = `1/5` , x lies in the first quadrant. Carry out step of the values of other five trigonometric functions.

Solution:

Given cos x = `1/5` , therefore, sec x = 5

We know that,

sin2 x + cos2 x = 1, that is sin2 x = 1 – cos2 x

sin2 x = 1 - `1/5` =`4/5`

sin x = ± `2/(sqrt 5)` (take square root on both sides)

x lies in 1st quadrant, sin x is negative.

Therefore, sin x = `2/(sqrt 5)` which also gives

cosec x = `2/(sqrt 5)`

Further, we have tan x = `(sin x)/(cos x)` = `(5(sqrt 5))/2` and cot x = `(cos x) /(sin x)` = `(2)/(5(sqrt 5))`

Answer: sec x = 5, sin x = `2/(sqrt 5),` cosec x = `2/(sqrt 5)` , tan x = `(5(sqrt 5))/2` , cot x =`2/(5(sqrt 5))`



Practice Problems for Trigonometry Xi:

Let us see about the topic trigonometry xi,

Practice problem 1: Determine the value of cos (370°).

Practice problem 2: Find the value of sin 5pi/3.

Solutions for prepare for trigonometry xi:

Solution 1: The value of cos (370°) is 0.98.

Solution 2: The value of sin 5pi/3 is -0.87.

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Solving Online Calculus Optimizing Problems

Study of rate of transformation is called calculus. Study of optimization calculus or mathematical encoding is disturbed to influence the excellent element from the group of elements. Optimization is a single technique to obtain a maximum or minimum value of a function. The smaller value of the function is known as minimum. The greater value of the function is known as maximum.

Online has emerged as one of the main key source for students to increase their knowledge topic wise.

Examples to Solving Online Calculus for Optimizing Problems:

Solving online calculus for optimizing example problems 1:

`y = 3x^2 - 5x` , solving for x and y for the optimizing calculus problems.

Solution:

Step 1: Equation is` y = 3x^2- 5x`

Step 2: Differentiate with respect to x

`dy / dx` = 6x - 5

Equate `dy / dx ` to 0.

`dy / d` x = 6x - 5 = 0

6x = 5

x = `(5) / 6` or   0.83

Step 3: Plug x = 0.83 in the given equation

y = `3 (0.83) ^2- 5(0.83)`

= 3(0.6889) - (4.15)

= 2.06 - 4.15

= -2.09

Therefore, x = 0.83 and y = -2.09

Step 4: From the given equation plot the graph and mark out the points in the graph.





Solving Online Calculus for Optimizing Example Problems 2:

`y = 5x^2 - 19x,` solving for x and y for the optimizing calculus problems.

Solution:

Step 1: Equation is `y = 5x^2- 19x`

Step 2: Differentiate with respect to x

`dy / dx ` = 5x - 19

Equate `dy / dx` to 0.

` dy / dx` = 10x - 19 = 0

10x = 19

x = `(19)/10`  or 1.9

Step 3: Plug x = 1.9 in the equation

y = `5(1.9) ^2- 19(1.9)`

= 5(3.61) - 36.1

= 18.05 - 18.05

= -18.05

Therefore, x = 1.9 and y = -18.05

Step 4: From the given equation plot the graph and mark out the points in the graph.



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Solving Online Calculus for Optimizing Example Problems 3:

`y = 4x^2- 7` , solving for x and y for the optimizing calculus problems.

Solution:

Step 1: The given equation is `y = 4x^2- 7`

Step 2: Differentiate with respect to x

`dy / dx ` = 8x

Step 3: Equate `dy / dx` = 0

8x = 0

x = 0

Step 4: Thus, `y = 4(0) ^2-7`

y = -7

So, x = 0 and y = -7.

Step 4: From the given equation plot the graph and mark out the points in the graph.

Same Perimeter Different Area

Perimeter is nothing but the path around the shape. And area is nothing but the space occupied by the 2 dimensional object. Here we are going to deal with the same perimeter and different area of the shape. Every shape is having different formulas for area. And for all shape if we want to find the perimeter we have to add the length of all sides. We will see some example problems for same perimeter and different area.

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Example Problems for same Perimeter and Different Area:

Example 1 for same perimeter and different area:



Find the area and perimeter of the following shapes and compare the area and perimeter.

Solution:

Shape 1:

The first shape is a rectangle. We know the area of the rectangle is l X w

Here length l = 3 cm and the width w = 4 cm

So the area of the rectangle = 3 X 4 = 12 cm2

Perimeter of the rectangle is = 2 (l + w) = 2 (3 + 4) = 2 X 7 =14 cm.

Shape 2:

The shape 2 is a triangle. To find the area of the triangle we have to use the following formula.

Area of the triangle = (`1 / 2` ) bh.

From the above b = 7 cm and h = 2 cm.

So the area =` (1 / 2) xx 7 xx 2 ` = 7 cm2

Perimeter of the triangle   = sum of all sides = 7 + 4 +3 = 14 cm.

From the above bath shape is having the same perimeter and different area.

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Example 2 for same Perimeter and Different Area:

Find the area and perimeter of the following shapes and compare the area and perimeter.



Solution:

Shape 1:

The first shape is a trapezoid. We know the area of the trapezoid is `(1/2) h (a + b)`

Here a and b are the base lengths. A = 6 cm. b = 4 cm and the height is 5 cm.

So the area of the rectangle = `(1/2) 5 (6 + 4) ` = 25 cm2

Perimeter of the rectangle is = sum of all sides = 6 + 5 + 4 + 3 = 18 cm.

Shape 2:

The second shape is a rectangle. We know the area of the rectangle is l X w

Here length l = 6 cm and the width w = 3 cm

So the area of the rectangle = 6 `xx` 3 = 18 cm2

Perimeter of the rectangle is = 2 (l + w) = 2 (6 + 3) = 2 X 9 =18 cm.

From the above bath shape is having the same perimeter and different area.

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Pictures of Division Sets

In this article we shall discuss  pictures of division sets. Here, division is also meant by fraction of a whole. A division can be creation over to a decimal through dividing the upper digit, or numerator, during the lower digit, or denominator. Division is instead of as ratios, and significance for fraction which is one of the important math processes. Thus the division `3/5` is also used to point out the ratio 3:5 and the division 3 ÷ 5 as well.

How to do Pictures of Division Sets:

The pictures of division sets are shown given below that,



 
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Example Problems Based on Learn Pictures of Division Sets:

The example problems based on learn about pictures of division sets are given below that,

Example 1:

How to learn about pictures of division for 277 divide by 8 sets?

Solution:

Step 1:

The given value is 277 divide by 8 sets.

Step 2:

Here, 277 divide by 8 sets is also denoted as 277/8.

Step 3:

Now, 277 divide by 8 sets is explain about using long division procedure pictures.

Here, using long division procedure pictures are shown given below that,



Step 4:

The final answer for pictures of division sets is 34.625.

Example 2:

How to learn about pictures of division for 197 divide by 6 sets?

Solution:

Step 1:

The given value is 197 divide by 6 sets.

Step 2:

Here, 197 divide by 6 sets is also denoted as 197/6.

Step 3:

Now, 197 divide by 6 sets is explain about using long division procedure pictures.

Here, using long division procedure pictures are shown given below that,



Step 4:

The final answer for pictures of division sets is 32.833







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Practice problems based on learn about pictures of division sets:

The practice problems based on learn about pictures of division sets are given below that,

Problem 1:

How to learn about pictures of division for 163 divide by 4 sets?

Answer: The final answer for pictures of division sets is 40.75

Problem 2:

How to learn about pictures of division for 52 divide by 4 sets?

Answer: The final answer for pictures of division sets is 13.

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.