Friday, November 30, 2012

First Decimal Place

Binary numerals are usually read digit-by-digit, in order to distinguish them from decimal numbers. For example, the binary numeral 100 is pronounced one zero zero, rather than one hundred, to make its binary nature explicit, Since the binary numeral 100 is equal to the decimal value four, it would be confusing to refer to the numeral as one hundred. (Source: Wikipedia)

First Decimal Place:

Let us take a number, 19.7.

1 indicates place value ten.

9 indicates place value one.

7 indicate 7/10(tenths decimal place)

Example problems for first decimal place:

Example 1:

Calculate the following 15.9+15.7-0.15+0.1

Solution:

Given that, 15.9+15.7-0.15+0.1

Numbers of digits behind decimal point are equal.

Step 1: Adding positive numbers,

= 15.9+15.7+0.1

= 31.7.

Step 2: Negative integer add to the primary step

= 31.7 – 0.15

= 31.55

The solution is 31.55

Example 2:

Calculate the following 18.2+0.1-18.7+1.8

Solution:

= 18.2+0.1-18.7+1.8

Numbers of digits behind decimal points are equal.

Step 1: Adding positive numbers,

= 18.2+0.1+1.8

= 20.1

Step 2: Negative number add to the first step

= 20.1– 18.7

= 1.4

The solution is 1.4

Addition for first decimal place:

Problem 1:

Add the following 1.3+1.5+1.7+1.9

Add the positive value,

=1.3+1.5+1.7+1.9

= 6.4

The solution is 6.4

Problem 2:

Add the following 1.6+1.5+1.7+1.7

Add the positive value,

=1.6+1.5+1.7+1.7

=6.5

The solution is 6.5

Example problem for first decimal place:

Subtraction for first decimal place:

Problem 1:

Find 9.3-9.5-9.8-9.7

= 9.3-9.5-9.8-9.7

Subtract the negative value,

= -19.7

The solution is -19.7

Problem 2:

Find 8.6-8.5-8.7-8.9

Subtract the negative value,

= 8.6-8.5-8.7-8.9

= -17.5

The solution is -17.5

Multiplication for first decimal place:

Problem 1:

3.5 * 3.5 * 3.5

Multiply the positive value,

= 3.5 * 3.5 * 3.5

= 12.25* 3.5

= 42.8

The solution is 42.8

Problem 2:

1.8 * 1.8 * 1.8

Multiply the positive terms,

= 1.8 * 1.8 * 1.8

= 3.24 * 1.8

= 5.8

The solution is 5.8

Division for first decimal place:

Problem 1:

`7.8/0.6`

Divide the positive numerator value,

= `7.8/0.6`

= 13.0

The solution is 13.0

Problem 2:

`9.7/8.8`

Divide the numerator value,

= `9.7/8.8`

=1.1

The solution is 1.1

Practice Problem for first Decimal Place:

Problem1: 7.2+8.3+8.2

Answer: 23.7

Problem2: 9.2+9.2+9.2

Answer: 27.6

Problem3: 15.3+15.2+15.2

Answer: 45.7

Thursday, November 22, 2012

Practice Model Definition


If any student wants to know about the definition for math practice model with examples, they can be referring the below examples for them. In this we can see some definitions for practice model in statistics with example problems. Let us see some of the definition in statistics such as mean, median and mode with example and practice problems with solve as follows.

Practice Model Definition - Definitions:

In this following we can see some definition for math practice model in statistics.

Practice Model Definition 1:

Mean:

It is a set of data which can find the average in dividing by sum of numbers with total numbers in the given data.

Practice Model Definition 2:

Median:

It is a set of numbers which can find the middle numbers and arrange the numbers in order to select the middle number.

Mode:

It is a set of numbers that can occur frequently in a set of data and no number can occur more than once.
Practice Model Definition - Examples and Practice Problems:

In this following we can see some definition for math practice model in statistics with example and practice problems.

Find the mean of weights for 6 peoples in kilograms are 15, 9, 10, 7, 16, and 23.

Solution:

                Sum of numbers
Mean = ------------------------
                  Total number

= `(15+9+10+7+16+23)/6`

= `80/6`

= 13.33

Find the median of 10, 24, 12, 21, 7 and 18.

Solution:

Arrange the data in ascending order as 7, 10, 12, 18, 21 and 24.

N = 6

Since n is even, median = `1 / 2` [`(nth)/2` item value + (`n / 2` + 1)th item value]

= `1 / 2` [6th item value + (`6/2` + 1)th item value]

= `1 / 2` [3rd item value + 4th item value]

= `1 / 2` [12 + 18]

= `1 / 2 xx 30`

= 15

Find the mode of 22, 17, 19, 15, and 22.

Solution:

22 are repeated twice.

Mode = 22

Practice Problems:

Find the mean of weights for 6 peoples in kilograms are 5, 9, 10, 7, 16, and 23.

Solution:

= 11.66

Find the median of 8, 25, 11, 21, 6 and 17.

Solution:

= 14

Find the mode of 20, 17, 11, 13, and 20.

Solution:

= 20



Monday, November 19, 2012

Integral of X Sinx

The term integral may also refer to the notion of anti-derivative, a function F whose derivative is the given function ƒ. In this case it is called an indefinite integral, while the integrals discussed in this article are termed definite integrals. Integrate the function f(x) can be written as`int ` f(x) dx. The integral is used to find the surface area, volume geometric solids by using stokes theorem, divergence theorem (like double integral, triple integral).
Source Wikipedia.

Integral Formulas:

1. `int` x n dx = `(x^n+1) / (n+1) ` + c

2. `int`sin x. dx = - cos x + c

3.` int ` cos x .dx = sin x. + c

4.  ` int` sec2x. dx = tan x + c

5. `int` cosec x.cot x. dx = -cosec x + c

6.`int dx / sqrt(a^2 - x^2) = sin^-1(x / a) + c`

Integral Problems:

Integral problem 1:

Find the integration of  x sin x with respect to x.

Solution:

Given term is x sin x

Integral of x sin x can be expressed as `int` x sin x dx

let u = x                         dv = sin x dx

`(du)/(dx)` = 1            and        v = - cos x

We know `int ` u dv  = uv - `int` v du

= x . (-cos x ) - ` int` (-cos x) 1.dx

= - x cos x +` int` cos x dx

= - x cos x + sin x + c

Answer:  `int` x sin x dx  = - x cos x + sin x + c

Integral problem 2:

Find the integration of  (2x + sin x) with respect to x.

Solution:

Given term is 2x + sin x

Integral of (2x + sin x )can be expressed as `int`(2x + sin x )  dx

`int`(2x + sin x )  dx =  `int` 2x dx +` int` (sin x).dx

= 2 `int` x dx +` int`sin x dx

= 2` (x^2/2)`+ (-cos x) + c

= x2 - cos x + c

Answer:  `int` (2x + sin x) dx  =   x2 - cos x + c

Integral problem 3:

Find the integration of  (sin x  + 3cosec2x)  with respect to x

Solution:

Given term is  (sin x  + 3cosec2x)

Integral of (sin x  + 3cosec2x) can be expressed as  `int ` (sin x + 3cosec2x) dx

`int ` (sin x + 3cosec2x) dx  =` int` (sin x dx) +` int` (3cosec2x) dx

= `int` (sin x dx) + 3` int ` (cosec2x) dx

=  (- cos x) + 3 (-cot x) + c

= - cos x - 3 cot x + c

Answer: `int `(sin x  + 3cosec2x) dx  = - (cos x + 3 cot x) + c

Integral problem 4:

Find the integration of  x2 sin x with respect to x.

Solution:

Given term is x2 sin x

Integral of x2 sin x can be expressed as `int`x2 sin x  dx

let u = x2                          dv = sin x dx

`(du)/(dx)` = 2x            and        v = - cos x

We know `int ` u dv  = uv - `int` v du

`int`x2 sin x  dx   = x2 . (-cos x ) - ` int` (-cos x) 2x.dx

= - x2 cos x +` int` 2x cos x dx

=  - x2 cos x + 2` int` x cos x dx + c

= - x2 cos x + 2 I1+c

Take I1 = ` int` x cos x dx

let u = x                       and     dv = cos x dx

`(du)/(dx)` = 1                                   v =  sin x

We know `int ` u dv  = uv - `int` v du

I1 =` int` x cos x dx = x sin x - `int` sin x dx

= x sin x - (-cos x) + c

I1 = x sin x + cos x + c

`int`x2 sin x  dx  =  - x2 cos x + 2 I1+ c

=  - x2 cos x + 2 (x sin x + cos x )+ c

= - x2 cos x + 2x sin x + 2cos x + c

= 2x sin x - (x2 - 2 )cos x + c

Answer:  `int`x2 sin x  dx   =  2x sin x - (x2 - 2 )cos x + c