Exponential Scientific Notation

This article is showing the exponential scientific notation. Scientific notation is nothing but transform very big numbers into simplest form of a number. That is called as scientific notation. Scientific notation is nothing but exponential of the base integer 10 have expressed in the number scientific notation. Example for scientific notation is 1860000000 this can be note down as 1.86 x 109. And the one more method in decimal is 0.0000081 = 8.1 x 10-6

Steps to find exponential scientific notation:

The steps to find exponential scientific notation are given below that:

The scientific notation has one number to the left of the decimal place.

We have to set the decimal point following the initial digit.

How many digits the decimal point was moved point out during the exponential of ten (10).


Example problems based on exponential scientific notation:

The example problems based on exponential scientific notation is given below that,

Example 1:

Change 82000 in exponential scientific notation.

Solution:

Step 1:

Specified that 82000.

Step 2:

Specified number is not a big number, but it is simple to convert scientific notation.

First note down 8.2 x 104 = 82000 [104 = 10000]

Step 3:

After that it can be write in scientific notation for 8.2 x 104.

Example 2:

Change 456000 in exponential scientific notation.

Solution:

Step 1:

Specified that 456000.

Step 2:

Specified number is a big number, but it is simple to convert scientific notation.

First note down 4.56 this is not equivalent to specified number.

Step 3:

Identify how many numbers available following the decimal point in previous step.

Present are 5 digits subsequent the decimal point.

4.56 × (100000)

Therefore the notation will be 105.

Step 4:

The scientific notation can be note down as 4.56 x 105

Example 3:

Change 987000 in exponential scientific notation.

Solution:

Step 1:

Specified that 987000.

Step 2:

Specified number is not a big number, but it is simple to convert scientific notation.

First note down the number as 9.87

Step 3:

There are 5 digit following the number is there so we can write 105.

Step 4:

The scientific notation can be note down as 9.87 x 105.
Practice problems based on exponential scientific notation:

The practice problems based on exponential scientific notation is given below that,

Problem 1:

Change 8960000 in exponential scientific notation.

Answer: The concluding answer for this problem is 8.96 x 106.

Problem 2:

Change 963000 in exponential scientific notation.

Answer: The concluding answer for this problem is 9.63 x 105.

Degree of Product Differentiation

The degree of the product differentiation represents the operations in the differentiation which deals with the product of the terms. For example if we have the function like xy then the degree of the function is 2 [x^1y^1 = ^2  ]. The homogeneous equation represents the degreee of the functions in the numerator and the denominator are same. In this article we are going to discuss about the degree of product differentiation for the students in detail with the homogeneous equations in different form with clear view.


Examples for the degree of product differentiation

Review on the degree of product differentiation for the differentiable function  dy/dx = [y^2-x^2]/[2xy]

Solution:

The function given for differentiation  is   dy/dx = [y^2-x^2]/[2xy]

Here it is present in the homogeneous form.

Since the numerator degree[ [y^2-x^2]] is two and the denominator [ [2x^1y^1]] degree is also two

Put y = vx and  dy/dx = v + xdv/dx in the equation given we get,

=>  v + x[dv]/dx   =    [(vx)^2-x^2]/[2x(vx)]

=>  v + x[dv]/dx   =    [v^2x^2-x^2]/[2vx^2]

=>  v + x[dv]/dx   =    x^2[v^2-1]/[2vx^2]

=>  v + x[dv]/dx   =    [v^2-1]/[2v]

=>   x[dv]/dx   =    [v^2-1]/[2v] - v

=>   x[dv]/dx   =    [v^2-1-2v^2]/[2v]

=>   x[dv]/dx   =    [-1-v^2]/[2v]

=>  [2v]/[-[1+v^2]] dv  =  dx/x

=>  [2v]/[1+v^2] dv  =  -1/x dx

Apply integration on both sides, we get

=>  int   [2v]/[1+v^2] dv  = int  -1/x dx

=>  int   [2v]/[1+v^2] dv  = -int  1/x dx

=>   log|1 + v^2|  =  - log |x| + log c

=>   log|1 + v^2|   + log |x| =  log c

=>   log|x[1 + v^2]|  =  log c

=>   x[1 + v^2]  =  +-   c

Put y = vx and v = y/x

=>   x[1 + [y/x]^2]  =  C_1

=>   x[1 + y^2/x^2]  =  C_1

=>   x[x^2 + y^2]/x^2  =  C_1

=>   [x^2 + y^2]/x  =  C_1

=>   x^2 + y^2  = xC_1           is the answer for the differentiation for the degree of product.

Problems for the degree of product differentiation

Review on the degree of product differentiation for the differentiable function dy/dx = [x^2y]/[x^3 + y^3 ]

Solution:

The answer is      [-x^3]/[3y^3] + log|y| =  C

Sectional Area of a Cylinder

A cylinder is one of the most basic curvilinear geometric shapes, the surface formed by the points at a fixed distance from a given straight line, the axis of the cylinder. The solid enclosed by this surface and by two planes perpendicular to the axis is also called a cylinder. The surface area and the volume of a cylinder have been known since deep antiquity. let us disuse about how to find the cross sectional area of cylinder. source-wikipedia


Cross section area of cylinder is nothing but the base area of cylinder. The base of cylinder is circle. So to find cross sectional area of cylinder, we use the area of circle formula.

Cross sectional area of cylinder = π r2 square unit.

r – Radius of circle

Example problems:

Ex:1 Find the cross sectional area of cylinder with radius 5cm.

Sol:

Given:

Radius (r) = 5 cm

Formula:

Cross sectional area of cylinder = π r2 square unit.

= π x 52

= 3.14 x 25

= 78.5

Cross sectional area of cylinder=78.5 cm2

Ex:2 Find the cross sectional area of cylinder with radius 3cm.

Sol:

Given:

Radius (r) = 3 cm

Formula:

Cross sectional area of cylinder = π r2 square unit.

= π x 32

= 3.14 x 9

= 28.26

Cross sectional area of cylinder= 28.26 cm2


Ex:3 Find the cross sectional area of cylinder with radius 5.5 m.

Sol:

Given:

Radius (r) = 5.5 m

Formula:

Cross sectional area of cylinder = π r2 square unit.

= π x 5.52

= 3.14 x 30.25

= 94.985

Cross sectional area of cylinder= 94.985 m2

I am planning to write more post on solve a system of linear equations and cbse books for class 9th. Keep checking my blog.


Ex:4 Find the cross sectional area of cylinder with radius 8cm.

Sol:

Given:

Radius (r) = 8 cm

Formula:

Cross sectional area of cylinder = π r2 square unit.

= π x 82

= 3.14 x 64

= 200.96

Cross sectional area of cylinder= 200.96 cm2

Sectional area of a cylinder- practice problem:

Find the cross sectional area of cylinder with radius 2.5cm.

Ans: cross section area=19.625 cm2

2.   Find the cross sectional area of cylinder with radius 3.8cm

Ans: cross section area = 45.3416 cm2

3.   Find the cross sectional area of cylinder with radius 7.7m

Ans: cross section area = 186.1706 m2

5 Properties of Parallelogram

Let us study about the 5 properties of parallelogram. Parallelogram is also a type of quadrilateral. Parallelogram is said to have two pairs of parallel sides.
Each pair has equal length. Also their diagonals are seemed to bisect each other in the center point.
Parallelograms have 5 important properties used in their construction point and evaluation point. All those 5 properties of parallelogram are discussed in detail with their examples below.




5 properties of parallelogram:

The 5 important properties of parallelogram are as follows:

a)      Parallelogram opposite sides are said to be parallel.

b)      Opposite sides are seemed to be congruent in parallelogram.

c)      Also parallelogram is said to have congruent opposite angles.

d)     Parallelograms two consecutive angles are supplementary angles.

e)      Parallelogram diagonals seemed to be bisecting each other right at their center point.

Parallelogram opposite sides are said to be parallel:

two pairs of parallel sides

This is the 1st property of parallelogram.
The above figure ABCD is seemed to be having two pairs of parallel sides as (AD, BC) and (DC, AB). Therefore it is a parallelogram.



Opposite sides are seemed to be congruent in parallelogram:

two pairs of congruent sides

This is the 2nd property of parallelogram.
The above figure MNOP is seemed to have the two pairs of congruent opposite sides as (MP, NO) and (PO, MN).



Parallelogram is said to have congruent opposite angles:

two pairs of congruent opposite angles

This is the 3rd property of parallelogram.
The above figure KLMN is seemed to have the two opposite pairs of congruent angles as (angle K, angle M) and (angle L, angle N).


Parallelograms two consecutive angles are supplementary angles:

two consecutive angles are supplementary

This is the 4th property of parallelogram.
The above figure ABCD is showing that all their two consecutive angles are supplementary angles as (A, B) or (A, C) or (C, D) or (B, D).




Parallelogram diagonals bisect each other right at their center point:

diagonals bisecting each other

This is the 5th property of parallelogram.
From the above figure ABCD we prove that the two diagonals of the parallelogram one which extends form A and other which extends from B will meet C and D respectively.
Also they bisect each other exactly at the center position of that parallelogram.

Mathematical Induction

The mathematical induction is one of the numerical proof naturally it is used to find the given statement is accurate for all normal numbers. It can be a deducted by proving that the 1st statement may be the countless succession of the statement is accurate and  we have to prove that if any other statement in the countless succession is real. In this article we shall discuss about the examples involved in induction proof.

Mathematical induction proof:

Principle of Mathematical Induction proof examples:

illustrate the  true for n = 1
suppose true for n = k
explain true for n = k + 1
Conclusion: Statement is true for all n >= 1


Examples in mathematical induction:

Examples in mathematical induction proof 1 :

Prove for n >= 1

1 * 1! + 2 * 2! + 3 * 3! + ….. + n * n! = (n+1)! – 1

This can be denoted by Sigma Notation

`sum_(k=1)^n` kk! = (n+ 1)! - 1

Proof:

Base case: n = 1

The left hand side is 1* 1!. The right hand side is 2! - 1. They are equal.


Examples in mathematical induction proof 2 :

Give the proof  when  amount of the n natural numbers by this formula

1+ 2 + 3 + … + n = n (n+ 1) / 2

Proof:

First, we have to denote the formula is real for n=k ,and it will be

1 + 2 + 3 + … + k = k (k + 1) / 2       (1)

This is the induction statement.  In order to prove  the method  is correct  means then

n = k + 1.

1+ 2 + 3 +… + (k + 1) = (k+1) (k + 2) / 2             (2)

Now add the ( k + 1) on both sides of the above equation in (2)

1 + 2 + 3 + … + k + (k + 1) = k (k + 1) / 2 + (k+ 1)

= k (k+ 1) + 2 (k+ 1) / 2

.

By adding  the  above  fractions we get ,

= (k + 1) (k+ 2) / 2

We can take  (k + 1) as a ordinary factor.

This is line (2), which is the first thing we need to explain

Next, we must explain that the formula is true for n = 1.  We have:

1 = ½· 1· 2

Thus the formula is true for n = 1. And  this fulfilled the both situation of the mathematical induction.

Percent in Math

The term percent in math is used to represent a value relative to hundred.  The same comparison can be expressed as a fraction if we compare a value with respect to another value. If the second value is 100, then we call it as percentage. For example:3/6 , Here we consider 3 relative to 6. If we consider relative to 100, it becomes 50 /100, we call this as 50 percent and is represented as 50 % .

If you need to convert a number into percentage, then multiply the number by 100 and put % symbol.

Now, Let us discuss about some example on percentages

Example problems to find percentage in math:

Percentage is mostly used in banking sector to find the interest.

The formula to find interest is,

Interest =Principle amount x time x rate of interest.



Example 1:

Suppose we need to find the interest will $2000 earn in one year at an annual interest rate of 5%

Solution:

5% = 5 /100

(5/100 ) x 2000 =100

So the answer is $100

Example 2:

Find the interest if amount if $1000 and month is 6 and rate of interest is 6%

Solution:

Interest =Amount x time x rate of interest.

Note : 6 month  = ½ of years.

Therefore,

6%  =>6/100  or 3/50

Substitute these values into interest formula. We get,

Interest = $1000 x ½ x 6/100

=$1000 x1/2  x 3/50

=$500  x 3/50

=$10 x 3

=$30

Example 3:

There are 45 boys  out of 90 students in the class room. Find the percentage of boys in the class room.

Solution:

Total students =90

Out of these 90 students 45 students are boys ,it is represented in terms of fraction.

= 45/90            [fraction form]

Convert the fraction into percentage,

So multiply the fraction number by 100,we get,

(45/90)  * 100

=50%

Example 4:

40 percent of X = 200 then find x?

Solution:

40 % * X =200

40 x */100  =200

40x  =200 *100

40x  = 20000

X =20000/40

X= 2000/4

X=500

Therefore, X value = 500

Verification:

40 percent of 500 =200

(40 x 500)/100  =200

20000/100   =200

200 =200

Hence Verified.


Practice problems to find percentage in math :

1)10 percent of p =100 find P?

Answer: 10

2) Find the interest will $5000 earn in one year at an annual interest rate of10%

Answer : 500

Combinatorial Learning Formula

Combination are selections ie. it involves only the selection of the required number of things out of the total number of things. Thus in combination order does not matter. For example, consider a set of three elements a,b,c and combination
made out of the set with


i) One at a time: a, b, c
ii) Two at a time: a,b, b,c, c,a
iii) Three at a time: a,b,c

The number of combinations of n things taken r, (r < n) is denoted by nCr or ([n],[r])

Learning the evaluation of combinations and the combinatorial formula

Learning to derive the formula for nCr:( Combinatorial formula)
No of combinations from ‘n’ things taken ‘r’ at a time = nCr
No of permutations from ‘n’ things taken ‘r’ at a time = nPr
Number of ways ‘r’ things that can be arranged among themselves = r!
Each combination having r things gives rise to r! permutations

Therefore  nPr = (nCr) r!


=> ((n - r )!) / (n!) = ( nCr) r!


nCr = r!(n - r )!

Learning observations related to combinatorial formula:

(i)   nCo = (n!)/(0! (n - 0 )!) = (n!)/(n!) = 1


(ii) nCn = (n!)/(n! (n -n)!) = (n!)/(n!0!) = 1


(iii) nCr = nCn-r


(iv) If nCx = nCy then x = y or x+y = n


(v) nCr =(^nPr)/(r!)

Learning examples on combinatorial formula

Learning to solve examples using the combinatorial formula :

Ex 1:Evaluate 8C3

Solution:

8C3  =  (8!)/(3! (8 - 3)!) = (8!)/(3! . 5!) = (8 . 7 . 6 . 5!)/((3 . 2 . 1) . 5!)  =  (8 . 7 . 6)/(3 . 2 . 1) = 8 . 7 = 48

OR another method

8C3 = (8 . 7 . 6)/(3 . 2 . 1) = 8 . 7 = 48

Ex 2:

Evaluate: 9C7

Solution:

9C7 = 9C2 = (9 . 8)/(2 . 1) = 9 . 4 = 36

Ex 3:    A committee of seven students is formed selecting from 6 boys and 5 girls such that majority are from boys. How many different committees can be formed?

Sol:

Number of students in the committee = 7
Number of boys = 6
Number of girls = 5
The selection can be done as follows


Boy (6)    Girl (5)
6              1
5              2
4              3


ie. (6B and 1G) or (5B and 2G) or (4B and 3G)

The possible ways are ([6],[6]) ([5],[1]) or ([6],[5]) ([5],[2 ]) or ([6],[4]) ([5],[3])

The total number of different committees formed

=  6C6  x  5c1 + 6C5  x  5C2  +  6C4  x  5C3

= 1 x 5 + 6 x 10 + 15 x 10 = 215

What is Continuous Integration

In what is continuous integration, Let f (x) be continuous and single valued for a <= x <= b. Divide the interval (a, b) of the x axis into n parts at the points (a -= X_0), X_1, X_2… ( X_n -= b) as in below figure.

Define Delta x_i = x_i - x_(i-1) and let xi_i be a value of x in the interval x_(i-1) < xi_i <= x_i. Form the sum of the areas of each approximately rectangular area to get

sum_(i=1)^(n) f(xi_i) Delta x_i

continuous integration

If we take the limit as n ->oo and Delta x_i -> 0, the approximation of the area under the curve as a rectangle becomes more accurate and we obtain the definition of the definite integral of f (x) between the limits a, b; thus

^lim_((n -> oo)_(Delta x_i -> 0)) sum_(i=1)^n f(xi_i) Delta x_i -= int_a^b f(x)dx

The points a, b are referred to as the lower and upper limits of integration, respectively. The definite integral

int_a^b f(x) dx

equals the area under the curve f(x) in the interval (a, b) for what is continuous integration.

Fundamental Theorem - what is continuous integration:

In what is continuous integration, the connection between differential and integral calculus is through the fundamental theorem of calculus. In particular, if f(x) is continuous in the interval a <= x <=b and G(x) is a function such that (dG)/(dx) = f(x) for all values of x in (a, b), then

int_a^b f(x)dx = G(b) - G(a)

The definite integral acts like an antiderivative of f(x), or the inverse operation to differentiation. The concept of inverse differentiation can be made more explicit by introducing the notation

[G(x)]_(a)^(b) = G(b) - G(a)

and using (dG)/(dx) = f(x) in the definite integral above equation to give

int_a^b ((dG)/(dx))dx = [G(x)]_a^b

Example - what is continuous integration:

In what is continuous integration, let f(y) = ln y, u(alpha) = a, and v(alpha) = alpha. In this case, the function f does not depend on alpha. Consequently, we substitute u, v, and f into find.

(d)/(d alpha) int_a^(alpha) In y dy = f(alpha)(d a)/(d alpha) = In alpha

because (da)/(d alpha) = 0 and f(alpha) = ln alpha. This result is verified by first evaluating the integral

int_a^b In y dy = [y In y –y ]_a^alpha = alpha In alpha - alpha - (a In a - a)

Taking the derivative of above equation with respect to alpha gives

(d)/(d alpha)[alpha In alpha - alpha - (a In a –a)] =(d)/(d alpha) [alpha In alpha - alpha]

Answer is     = In alpha

Arithmetic Adding Fractions

A fraction is a part of a whole. A fraction can be altered to a decimal by dividing the upper number, or numerator, by the lower number, or denominator. Fractions are to indicate ratios, and to represent division which is one of the basic arithmetic operations. Thus the fraction 3/4 is also used to signify the ratio 3:4 (three to four) and the division 3 ÷ 4 as well.(three divided by four). Here in this topic we see about adding fractions.

Arithmetic adding Fractions with same denominator - Example problems:

Arithmetic adding Fractions - Example 1:

Add: 2/7   +    3/7

Solution

Here the denominator is equal

So we keep the denominator same as 7 and add the numerator alone

So the solution is   (2+3)/7 =  5/7

Arithmetic adding Fractions - Example 2:

Add:  5/8 +  3/8

Solution

Here the denominator is equal

So we keep the denominator same as 8 and add the numerator alone

So the solution is ,

(5+3) / 8 = 8 / 8  = 1.

Arithmetic adding Fractions with different denominator - Example problems:

Arithmetic adding Fractions - Example 1:

10/8   +    12/9


Answer:

Here the denominators are different that are 8 and 9. So take LCD of 8 and 9 first.

LCD of 8 and 9 is 72.

Therefore,
10 x 9        12 x 8          90      96       186
------------  +   ------------  =  -------- +  ------ = ----------
8 x 9          9 x 8          72       72        72


Simplifying this we get the answer as 31/12 .
Arithmetic adding Fractions - Example 2:

16/3     +   10/10

Answer:

Here the denominators are different that are 3 and 10. So take LCD of 3 and 10 first.

LCD of 3 and 10 is 30.

Therefore,
16 x 10       10 x 3       160       30       190
------------  +   ------------  =  -------- +  ------ = ----------
3 x 10          10 x 3       30       30        30


Simplifying this we get the answer as 190/30 .

Learning Tangent Line

If a line and a curve intersect at a point only then the line said to be tangent to the curve at that point. And the point is called the point of contact.

Tangents at (x, y) to y = f(x)

Lt y=f(x) be a given curve and P(x, y) and Q(x + δx, y + δy) be two neighbouring points on it

Equation of the line PQ is

Y – y = y + δy – y/x + δx – x(X – x)

Or       Y – y = δy/δx(X – x)                                             …(1)

Equation of line PQ

Equation of the line PQ



The line (1) will be a tagent to the give curve at P if Q → P which in turn means that δx → 0 and we know that

Lim δy/δx = dy/dx

δx → 0

Therefore the equation of the tangent is

Y – y = (dy/dx) (X – x)

Geometrical Intrepetation - slope and tangent

Geometrical meaning of dy/dx

From equation (1) we observe that dy/dx represents the slope of the tangent to the given curve y = f(x) at any point (x, y)

...    dy/dx = tan Ψ

Where Ψ is the angle which the tangent to the curve makes with +ive direction of x-axis. In case we are to find the tangent at any point (x1, y1) then (dy/dx)(x1, y1) i.e. the value of dy/dx at (x1, y1) will represent the slope of the tangent and hence its equation in this case will be

y – y1 = (dy/dx)(x1, y1) (x – x1)

Algorithm to draw tangent of a circle

Algorithm to obtain the equations of tangents draw from a given point to a given circle.

Step i. Obtain the point, say (x1, y1)

Step ii. Write a line passing through (x1, y1) having slow m i.e., y – y1 = m (x – x1)

Step iii. Equate the length of the perpendicular from the centre of the circle to the line in step ll to the radius of  the circle

Step iv. Obtain the value of m from the equation in step iii

Step v. Substitute m in the equation in step ll

Normal to the tangent:

The normal to a curve at a given point is a straight line passing through the given point, perpendicular to the tangent at this point. From the definition of a normal it is clear that the slope of the normal m′ and that of the tangent m are connected by the equation m′ = –`1/m .`

i.e., m′ = –1/f ′(x1)

= − 1/  dx/dy    (x1, y1)

Hence the equation of a normal to a curve y = f(x) at a point P(x1,y1) is of the form y – y1= –1/f ′(x1)( x – x1).

The equation of the normal at (x1,y1) is

(i)                x = x1 if the tangent is horizontal

(ii)               y = y1 if the tangent is vertical and

(iii)              y – y1 = –1 / m (x – x1) otherwise.

Example problems:

Example 1 :  Find the equations of the tangents through (7, 1) to the circle x2 + y2 = 25

Solution :  The equation of any line through (7, 1) is

y – 1 = m (x – 7)

mx – y – 7m + 1 = 0             …(i)

The coordinates of the centre and radius of the given circle are (0, 0) and 5 respectively

The line (i) will touch the given circle if,

Length of the perpendicular from the centre = radius

| (1 - 7m)/√(m2 + (-1)2 ) | = 5

│(1 – 7m)/√(m2 + 1)│ = 5

(1 – 7m)2/(m2 + 1) = 25

24m2 – 14m – 24 = 0

12m2 – 7m – 12 = 0

(4m + 3) (3m – 4) = 0

m = –3/4, 4/3

Substituting the values of m in (i), we obtain

3/4x – y + 21/4 + 1 = 0 and 4/3x – y – 28/3 + 1 = 0

3x + 4y – 25 = 0 and 4x – 3y – 25 = 0 which are the required equations.


Example 2: Find the equation of the tangent and normal to the curve y = x3  at the point (1,1).

Solution:

We have y = x3 ; slope y′= 3x2.

At the point (1,1), x = 1 and m = 3(1)2 = 3.

Therefore equation of the tangent is y − y1 = m(x − x1)

y – 1 = 3(x – 1) or y = 3x – 2

The equation of the normal is y − y1 = −`1/m `  (x − x1)

y – 1 =–`1/3` (x – 1) or y = –`1/3` x +4/3 .

normal tangent line graph

Probability Simple Event

Probability is a method of expressing information or principle that an event will occur or has occurred. In mathematics the ideas has given a correct meaning in probability theory  is used widely in the areas of study as mathematics, statistics, science, and finance to depict conclusions about the possibility of potential events .An event is a group of possible outcomes.

A simple event in a probability is the event of a single outcome.

Simple Event

The simple event is as follows:

If we bring together a deck of 52 playing cards and no jokers, and pick a single card from the deck, then the sample space is a 52-element set, as each individual card is a possible outcome.

An event is any subset of the sample space, as well as any single-element set, the empty set and the sample space itself which is defined to have chance one.

Further events are correct subsets of the sample space that surround multiple elements. So, for example, potential events include:

Black and red at the same time without being a joker (0 elements),
The five of Diamond(1 element),
A Queen(4 elements),
A Face card(12 elements),
A Spade (13 elements),
A card(52 elements).

Seeing as all events are sets, they are generally written as sets (e.g. {1, 2, 3}), and symbolized graphically using Venn diagrams. Venn diagrams are mainly useful for symbolizing events because the probability of the event can be identified with the ratio of the area of the event and the area of the sample space.



Example:

The event that 1 roll a 7 with two dice is an event.

But the event that 1 roll snake-eyes (1 and 1) is a simple event.

A complementary event is just the opposite of an event.

The balancing event to rolling a 7 is the event that 1 do not roll a seven. It consists of all the outcomes not in the unique event.

Hypothesis Test Power

A statistical hypothesis test is a technique of making decisions using experimental information. In statistics, an effect is called statistically significant if it is doubtful to have happen by possibility. Critical examination of this category may be called tests of significance, and when such examination is available we may determine whether a next sample is or is not significantly dissimilar from the initial. Let us study about the topic hypothesis test power given below content with some example problems.


Example problems for hypothesis test power:

Example 1:

A sample of 900 members is establish to have a mean of 3.4 cm and standard deviation 2.61 cm. is the example in use from a great population of mean 3.25 cm and standard deviation 2.61 cm. If the population is usual and its mean is unknown, Get 95% self-confidence limits of exact mean.

Solution:

Given:

H0 : µ = 3.25
H1 : µ ? 3.25
L.O.S  a = 0.05

Test Statistic:

Z = (barx-mu)/(sigma)/(sqrt(n))

Given bar x = 3.4, µ=3.25, n=900, s=2.61

Z = (3.4- 3.25)/(2.61)/(sqrt(900))

= 1.724

Critical value:

At 5% level, the table  value of Z = 1.96

Conclusion:

Since |Z| <1 .96="" 5="" accepted="" at="" h0="" is="" level="" of="" p="" significance.="">
The sample is taken from population whose mean is 3.25 cm

Mean µ is unknown, 95% confidence limits of µ are = barx+-1.96 sigma/(sqrt(n))

=3.4+- 1.96*(2.61)/(sqrt(900))

= 3.4+- 0.17

=3.23,3.57

Example 2:-using hypothesis test power

A construct maintains that his artificial fishing line has a mean contravention strength of 8 kg and standard deviation 0.5 kg . Can we consider his state if a random sample of 50 lines yield a denote contravention power of 7.8kg?

Solution:

Given:

H0 : µ = 8
H1 : µ ? 8
L.O.S: a = 0.01

Test Stastistic

Z =(barx-mu)/((sigma)/(sqrt(n)))

Given:

barx = 7.8, mu = 8, n = 50, sigma = 0.5

Z = ((7.8)-8)/((0.5)/(sqrt(50)))

= - 2.828

|Z| = 2.828

Critical value:

At 1% level of significance the table value of Z = 2.58

Conclusion:

Since |Z|>2.58,H0 is rejected at 1% level.

The manufacture’s claim is not accepted.


Practice problem for hypothesis test power:

Problem 1: using hypothesis test power

On a test given 60 students at a huge number of several colleges, the mean position was 74.5 and S.D was 8.0. at one exacting college, where 200 students took test, the mean position was 75.9%. Discuss the meaning of this answer at 5% level.

Solution:

Z = 2.45, significant at 5% level.

Problem 2: using hypothesis test power

A random sample of 400 items is pinched from a normal population whose mean is 5 and whose variation is 4. if the sample mean is 4.45, can the sample be regard as truthfully random sample?

Solution:

Z= 5.5; H0 is rejected , No.

Standard Deviation

Standard Deviation is defined as "the measure of the variability". If the spread is more then the standard deviation will also be more. Standard deviation can also be stated as the "square root of the variance". Standard deviation has its application in computing the the annual returns of the banks

Arithmetic Mean:

A mathematical representation of the series of the number which is totalled as the sum of all the number in series and it is divided by the total number of values in the set.

Variability :

When the  standard deviation is raised to the power of two , the value which we get is the variance i.e Variance is equal to the square of the standard deviation. It is the measure of the degree of the spread among certain set of the values. Variance is also stated in the other way as follows"Variance is the measure of tendancy of individual values that vary from the arithmetic mean value

Formula to determine the Standard deviation :

[Formula of Standard deviation]

Where  ,     Σ = sum of

xi = Individual scores

m = arithmetic mean of all scores

n = Size of the number of scores

Variance :

Variance = σ2

Example

Example 1: To find the standard deviation of  2 , 4 , 6 , 8 , 10

Step-1 :Calculate the mean and the deviation

xi       m        xi-m         (xi-m)2

2        6        -4              16

4         6        -2              4

6        6          0              0

8        6          2              4

10       6         4             16

Step-2 :Find the sum of  (x-m)2

16 + 4 + 0 + 4 + 16 = 40

Step-3 :n = 5 , the total number of values , find n-1

5-1 = 4

Step-4 :Now find standard deviation

√40 / √4 = 3.162

3.162 is the standard deviation

To find the variance just square the standard deviation value

Step-5 : Variance = (3.162)2 = 9.998244

Example 2

Example 2 :To find the standard deviation of 3 , 6 , 9 , 12 , 15

Step-1 : Calculate the mean deviation

xi          m          xi-m           (xi-m)2

3           9           -6               36

6           9           -3                 9

9           9            0                 0

12          9            3                 9

15          9           6                36

Step-2 :

calculate the sum of (x-m)2

36+9+0+9+36 = 90

Step-3 :

n=5 ,the total number of values , find n-1

5-1 = 4

Step-4 :

Now let us find the standard deviation

√90/√4 = 4.743

The standard deviation is 4.743

Step-5 :

Variance = (4.743)2

= 22.496

Non Terminating Decimals

In this page we are going to discuss about Non terminating decimals.Below we will get introduction to Non terminating decimals. A non terminating decimal is one of the type of decimals which are divided into two types. They are,

Terminating Decimals
Non terminating Decimals

Non terminating Decimals is the continuous decimal number. It has been going on till the value infinitive. It is the very big value sometimes these non terminating decimals are rounded to its nearest values. But the Terminating decimal number is the fixed value of the decimal value.

Calculating Non Terminating Decimals

Below you can see the steps of calculating non terminating decimals:

Divide 5/3

Step 1: This is the type of an improper fraction. because here numerator takes the large value than the denominator.

Step 2: One time 3 is 3. So the remainder is 2 and the quotient value is 1

Step 3: Now 2 is not divided by 3

Step 4: So we can add zero to the remainder 2 and put a point the quotient value 1

Step 5: Now the value is 20. Again we can do the same procedure

Step 6: Now 6 times 3 is 18 again the remainder is 2 and the quotient value is 6

Step 7: Again we can add zero to 2 then the value is 20

Step 8: Again 6 times 3 is 18. Again the remainder is 2 and the quotient value is 6

Step 9: Now the decimal value is 1.66 …….its going on

Step 10: This decimal part 1.66……is known as the non terminating decimal.

Step 11: It can be rounded as 1.67

Step 12: Because it’s decimal part value is greater than or equal to 5. So the value of the last decimal value is consider as one and that value is added to the previous decimal value.

Step 13: Otherwise the last decimal part value is less than five. So the value of the last decimal value is simply leave. And the previous decimal value is considered as the last decimal digit value.

Pictorial Representation of this problem is shown below:

Non- terminating decimals

Practice problems

Here are some practice problems on non terminating Decimals:

Identify which of the fractions gives the non terminating decimals:

(4)/(2)
(2)/(3)
(6)/(3)

Answer:

Terminating decimals (4)/(2) =2 and (6)/(3) =2

Non terminating decimal (2)/(3) = 0.6666.....