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.