Average Velocity

Before going to the concept of average velocity,you should know about the concept of velocity.So, here is the introduction to velocity.

Velocity - the rate of change of position of an object. Velocity is a vector physical quantity because it requires both magnitude and direction to define. Speed is an absolute scalar value (magnitude) of velocity. Velocity is measured in meter/second (m/s).

In mathematics, velocity is the ratio of  displacement and  during the time taken to the interval. As it is associated to how fast waves travel between layers of the earth, it also tells us of how compact they are in between.

For example, 10 meter per second is a scalar value of velocity, but 10 m/s east is a vector. The rate of change in the velocity is called as acceleration.

Average velocity:

The Average velocity topic is dealt under mathematics. Students get to learn how to calculate the average velocity when the rest of the values are given. Students get to learn to practice various word problems on the concept of average velocity and also they get ample support from the online tutors regarding their homework problems and examination preparation.They can get all the answers for homework problems.

The “Rate of change(derivative) of position at which an object ” is known as velocity . It is a vector physical quantity; both magnitude and direction are essential to define it. The scalar complete value (magnitude) of velocity is speed, a quantity that is measured in meters per second (m/s or ms-1) when using the SI (metric) system.

The average velocity v of an object affecting through a displacement  (?x) during a time interval (?t) is described by the formula.Here is the Formula for average velocity:

V = Final displacement / total time taken

=  ?x / ?t

Students can get more detailed explanation on the topic on the Physics help page.

Finding average velocity

Example 1: If an object is thrown from the ground at an initial velocity of 10 meter per second, after t seconds the height of the object “h” in meter is given by,   h = t2 + 10t.  Find the time taken to the object to reach a height of 200 meter.

Solution:

Given, Height, h = 200 meter

h = t2 + 10t --> (1)

t =? (At 200 meter)
Plug h = 200 in equation (1)

200 = t2 + 10t

t2 + 10t – 200 = 0

By solving the quadratic equation,

t2 + 20t -10t -200 = 0

t(t+20) – 10(t+20) = 0

(t+20) (t -10) = 0

t = -20 or t = 10

We have two values for time “t”. Since time cannot be negative, any problem that gives a negative answer for one of its answers is always false, so we just go for the positive value.
So the final answer is 10 seconds.

So, the object reaches the height (200 meters) in 10 seconds.

Students can get more solved examples and problems to practice upon on the mastering physics page.

Calculate average velocity

Here is one more example of how to Calculate average velocity.

Example :  A particle moving along the x axis is located at 17.2 m at 1.79 s and at 4.48 m at 4.28 s. what is the average velocity in the particular time interval?

Solution:

Distance traveled by particle beside x-axis = Final location  - Initial location

= 17.20 - 4.48

= 12.72 meters

Time taken to cover up this distance = Final time - Initial time

= 4.28 - 1.79

= 2.49 seconds

Average velocity    = (Distance traveled)/(Time taken)

= 12.72 / 2.48

= 5.1084 (approximately)

Average velocity of the particle is    5.1084 m/s

Bijective Function Examples

In mathematics, a bijection, or a bijective function, is a function f from a set X to a set Y with the property that, for every y in Y, there is exactly one x in X such that f(x) = y and no unmapped element exists in either X or Y.

A one - one onto function is said to be bijective or a one-to-one correspondence.

A few examples of bijective function is given below which helps you for learning bijective function.

(Source: Wikipedia)

Examples of bijective function:

Example 1:

Show that the function f : R → R : f(x) = 3 - 4x  is one-one onto and hence bijective .

Solution:

We have

f(x1) = f(x2)

3 - 4x1 = 3 - 4x2

x1 = x2

Therefore, the function f is one-one.

Now, let y = 3 - 4x. Then, x = (3 - y)/4

Thus, for each y ε R (codomain of f), there exists x = (3 - y)/4 ε R

such that f(x) = f((3 - y)/4)

= {3 - 4 (3 - y)/4 }

= 3 - (3 - y)

= y

This shows that every element in codomain of f has its pre-image in dom(f).

Therefore, the function f is onto.

Hence, the given function is bijective.

Example 2:

Let A = R - {3} and B = R - {1}. Let f : A → B : f(x) = (x - 2)/(x - 3) for all values of x ε A.

Show that f is one-one and onto.

Solution:

f is one-one, since

f(x1) = f(x2)

(x1 - 2)/(x1 - 3) = (x2 - 2)/(x2 - 3)

(x1 - 2)(x2 - 3) = (x2 - 2)(x1 - 3)

x1x2 - 3x1 - 2x2 + 6 = x1x2 - 2x1 - 3x2 + 6

x1 = x2

Let y ε B such that y =  (x - 2)/(x - 3) .

Then, (x - 3)y = (x - 2)

x = (3y - 2)/(y - 1)

Clearly, x is defined when y ≠ 1.

Also, x = 3 will give us 1 = 0, which is false.

Therefore,

x ≠ 3.

And, f(x) = ((3y - 2)/(y - 1) - 2)/((3y - 2)/(y - 1)- 3) = y

Thus, for each y ε B, there exists x ε A such that f(x) = y.

Therefore, f is onto.

Hence, the given function is one-one onto.

These examples of bijective function help you to solve the following practice problems.

My forthcoming post is on icse board question papers and Nonlinear Partial Differential Equations will give you more understanding about Algebra.

Practice problems of bijective function:

Following examples of bijective function is given for your practice which helps you to learn more about bijective function.

1) Show that the function f : R → R : f(x) = x3  is one-one and onto.

2) Let R0 be the set of all non zero real numbers. Show that f : R0 → R0 : f(x) = 1/x is a one-one onto function.

Probability Distribution

In probability theory and the statistics, a probability distribution identifies either the probability of each value of a random variable (when the variable is discrete), or the probability of the value falling within a particular interval (when the variable is continuous). The probability distribution describes the range of the possible values that a random variable can get and the probability that the value of the random variable is within any subset of that range.

I like to share this Non Central T Distribution with you all through my article.

The Normal distribution is often called as the "bell curve", when the random variable takes the values in the set of real numbers. Let us see some sample problems on probability distribution statistics.

Examples

Given below are some of the examples on Probability Distribution Statistics.

Example 1:

A continuous random variable X has probability distribution function f(x) = 3x2, 0 ≤ x ≤ 1, Find a and b such that. (i) P(X ≤ a) = P(X > a) and (ii) P(X > b) = 0.05. Calculate probability distribution for this function.

Solution:

(i) Since the total probability is 1, [Given that P(X ≤ a) = P (X > a)

P(X ≤ a) + P(X > a) = 1

i.e., P(X ≤ a) + P(X ≤ a) = 1

⇒ P(X ≤ a) =1/2⇒ `int`3x2dx = 12

i.e.,[3x3/3]a0=1/2 ⇒ a3 =1/2

i.e., a = 1/213

(ii) P(X > b) = 0.05

∴ `int` f(x) dx = 0.05 ∴ `int` b1 3x2  dx = 0.05

[3x3]31b= 0.05 ⇒ 1 − b3 = 0.05

b3 = 1 − 0.05 = 0.95 =95

100 ⇒ b = 19/2013

Example 2:  A random variables X has probability mass function as in the probability distribution tables given below
X
0 1 2 3 4 5 6
P(X = x) k 3k 5k 7k 9k 11k 13k



(1) Find k.

(2) Evaluate P(X < 4), P(X ≥ 5) and P(3< X ≤ 6)

Solution:

(1) Since P(X = x) is a probability mass function `sum_(n=0)^6` P(X = x) = 1

ie.,P(X=0) + P(X = 1) +P(X = 2) +P(X = 3) +P(X = 4) +P(X = 5)+P(X = 6) = 1.

⇒ k + 3k + 5k + 7k + 9k + 11k + 13k = 1 ⇒ 49 k = 1 ⇒ k =1/49

(2) P(X < 4) = P(X = 0) + P(X = 1 ) + P(X = 2) + P(X = 3) =1/49 +3/49 +5/49 +7/49 =16/49

P(X ≥ 5) = P(X = 5) + P(X = 6) =11/49 +13/49 =24/49

P(3 < X ≤ 6) = P(X = 4) + P(X = 5) + P(X = 6) =9/49 +11/49 +13/49 =33/49

∴ The smallest value of x for which P(X ≤ x) > 1/2 is 4.

Example 3:

In a class, the average number of marks obtained by student in Physics is 0.52, chemistry is 0.48 and in both Physics and chemistry they obtained 0.37. Find the total average obtained in either Physics or chemistry.

Solution:

P(A) = Average number of marks in Physics = 0.52

P(B) = Average number of marks in Chemistry = 0.48

P(A and B) = Average number of marks in both physics and chemistry = 0.37

P(A or B)    = P(A) + P(B) – P(A-B)

= 0.52 + 0.48 – 0.37

= 0.63

Calculating Interquartile Range

Calculating interquartile range is referred as a measure of variability, spread or dispersion, H-spread. It is the difference sandwiched between the 75th percentile (often called Q3 or third quartile) and the 25th percentile (often-called Q1 or first quartile). The primary step is to find the interquartile range is to arrange the given set of numbers in ascending order. The standard formula to find the interquartile range is given as

Interquartile ranges = Quart3 – Quart1.    ----- Standard formula

Whereas Q1 = first quartile.

Q3 = third quartile.

Example Problems for calculating interquartile range:

Ex : Determine the interquartile range of following set of numbers

34, 15, 8, 26, 22, 9, 19

Sol :  The following are the steps to find the interquartile range of a set of numbers.

Step 1:  Arranging of numbers

The initial step is to modify the given data in order, from smallest to biggest.

9, 8, 15, 19, 22, 26, 34

Step 2:    Calculating 1st quartile Q1.

The next step is to find the lower median (1st quartile Q!).

This is the middle of the lower three numbers.

1st quartile Q1 is 8.

Step 3:    Calculating 3rd quartile Q3.

Now find the upper median (The 3rd quartile Q2).

This is the middle of the upper three numbers.

The 3rd quartile Q3 is 26.

Step 4:     Calculating interquartile range.

The formula used to find the interquartile range is

Interquartile range = Q3 – Q1.

Q1 = first quartile.

Q3 = third quartile.

Here

Q1 the first quartile = 8

Q3 the third quartile = 26

Interquartile range = Q3 – Q1. ----- Standard formula

Plug in the Q1 and Q3 values in the standard formula Q3 – Q1.

Interquartile range = 26 – 8.

Interquartile ranges = 18.

Practice Problems for calculating interquartile:

Pro 1:  Find the interquartile range of following set of numbers

42, 81,56,21,63,12,5

Ans : Interquartile range = 51.

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Pro 2:  Find the interquartile range of following set of numbers

56, 23, 14, 25, 86, 45, 63, 15, 49, 18, 16

Ans:   Interquartile range = 40.

Pro 3:   Find the interquartile range of following set of numbers

12, 8, 5, 22, 15, 45, 2

Ans :   Interquartile range = 17

Permutations in Math

In mathematics, the notion of permutation is used with several slightly different meanings, all related to the act of permuting (rearranging in an ordered fashion) objects or values. Informally, a permutation of a set of values is an arrangement of those values into a particular order. Thus there are six permutations of the set {1,2,3}, namely [1,2,3], [1,3,2], [2,1,3], [2,3,1], [3,1,2], and [3,2,1]. (Source – Wikipedia.)

Example problems for permutations in math:

Example 1

1) How many permutations of 5 apparatus are probable from 2 apparatus. what is a permutation in given math problem?

Solution:

By means of the formula we can evaluate the answer,

P (n, r) = n! / (n – r)!

Here, n = 5, r = 2

P (5, 2) = 5! / (5–2)!

P (5, 2) = 5! / 3!

=   `(5xx4xx3xx2xx1)/ (3xx2xx1)`

Now 3, 2, 1 gets crossed out

= 5 x 4

= 20

Answer is: 20

Example 2

How many permutations of 9 files are possible from 5 books. what is a permutation in given math problem?

Solution:

Formula for calculate permutations,

P (n, r) =n! / (n – r)!

Now, n = 9

r = 5

P (9, 5) = 9! / (9 – 5)!

P (9, 5) = 9! / 4!

= `(9 *8 * 7 * 6 *5 * 4 *3 *2 *1)/ (5 * 4 *3 *2 *1)`

Here 5, 4, 3, 2, 1 gets crossed out

= 9x 8 x 7x 6

= 3,024

The answer is: 3, 024

Example 3

How many ways can 6 graduates from group of 15 are lined up for a function. what is a permutation in given math problem?

Solution:

There are 15P6 possible permutations from a group of 15.

15P6 = (15!)/(15-6!)

= (15!)/ (9!)

= 15x 14x 13x 12x 11x 10

= 3, 603, 600 different lineups.

The answer is: 3, 603, 600 different lineups.

Example 4

How many ways can 3 kids from group of 13 are lined up for a photograph. what is a permutation in given math problem?

Solution:

There are 13P3 possible permutations of 3 students from a group of 13.

13P3 = (13!)/ (13- 3!)

= (13!)/ (10!)

= 13 x 12 x 11

= 1716 different lineups.

The answer is: 1716 different lineups.

My forthcoming post is on Divide Fractions by Whole Numbers and cbse sample papers for class 9 science will give you more understanding about Algebra.

Practice problems for what is a permutations in math:

Practice problem -1

1) How many permutations of 9 components are possible from 2 elements. what is a permutation in given math problem?

Ans: 72

Practice problem -2

2) How many ways can 3 girls from group of 14 are lined up for a photograph?

Ans: 2184

Practice problem -3

3) How many permutations of 17 books are possible from 4 books?

Ans: 57, 120

Practice problem -4

4) How many 4-digit numbers can be ordered from the digits 1, 2, 3, 4, and 5, if each digit is unique. what is a permutation in given math problem?

Ans: 120.

Non linear Differential Equations

An ordinary differential equation is an differential equation in For instance (i)dy/dx = x + 5 (ii) (y′)2 + (y′)3 + 3y = x2 which  is a single independent variable enters either explicitly or implicitly are all ordinary non linear differential equations And now we see about the non linear differential equations.

Non linear differential equations in First order

Order and degree of a non linear differential equations:

Definition of non linear differential equations :

The order of an differential equation is the order of the highest order derivative occurring in it. The degree of the differential equation is the highest order derivative which occurs in it, after the differential equation has been made free from radicals and fractions as far as the derivatives are concerned.The degree of a non linear differential equation does not require variables like r, s, t … to be free from radicals and fractions.

Find the order and degree of the following differential equations:

ii) y = 4dy/dx + 3xdx/dy

y = 4dy/dx + 3xdx/dy  y = 4(dy/dx) + 3x 1/(dy/dx )

Making the above equation free from fractions involving

dy/dx we get y .dy/dx = 4(dy/dx)2 + 3x 2Highest order = 1

Degree of Highest order = 2

(order, degree) = (1, 2)

Problem for non linear differential equations:

Formation of non linear differential equations :

Form the differential equation from the following equations.

y = e2x (A + Bx)

ye−2x = A+ Bx … (1)

Since the above equation contains two arbitrary constants, differentiating

twice, we get y′e−2x − 2y e−2x = B

{y′′e−2x − 2y′ e−2x} − 2{y′e−2x − 2y e−2x} = 0

e−2x {y′′ − 4y′ + 4y} = 0 [‡ e−2x ≠ 0]

y′′ − 4y′ + 4y = 0 is the differential equation.

Between, if you have problem on these topics Formula for Conditional Probability, please browse expert math related websites for more help on cbse previous year question papers.

Non linear Differential equations of first order and also first degree :

Solve: 3ex tan y dx + (1 + ex) sec2y dy = 0

`=>` 3 log (1 + ex) + log tan y = log c

`=>` log [tan y (1 + ex)3] = log c

`=>`   (1 + ex)3

required solution to the problem is tan y = c

Learning Sample Space

Percentage is a number expressed  as a fraction of  100. The word percent is short form of the Latin word "Percentum" meaning out of hundred.  It means ratio of some number to hundred. We use the sign" %" to denote percentage.  It is basically a fraction, with 100 as the denominator and the number as the numerator.

For example 25% :

25% = (25)/(100)

In everyday life, we relate percentage to profit, loss, discount,rate of interest in banks, sales tax, income tax.

(i) Convert a ratio into percent , we write it as a fraction and multiply it with 100.

Example:  To express 20 : 50 as percent , we write (20)/(50) x 100 = 40%

(ii) To convert decimal into percent  we also multiply with 100

Example:   0.175 to percent   0.175 x 100 = 17.5%

(iii) To convert percent to fraction or decimal.  We drop the sign % and divide the remaining number by 100

Example:   45% = (45)/(100)

(iv) To find the certain percent of a given quantity, we multiply the given percent with the given quantity

Example:  30 % of 120   = (30)/(100) x 120 = 36

Some examples on solve finding percentages

1) what is 25 % of 200 ?

Solution:   (25)/(100)  x 200   = 0.25 x 200 = 50

2) What is 45% of 20 ?

Solution:   (45)/(100) x 20 = 0.45 x 20 = 9

3) We have to pay sales tax of 2% on product for $6.25. What is the amount we pay in all for that product.

Solution:  2% = (2)/(100) = 0.02

0.02 x 6.25 = 0.125

Total amount paid is $ ( 6.75 +0.125) = $  6.875

4) A shoe shop marks 40% discount on each pair of shoes , what is the cost price of a pair of shoes that cost $ 67.

Solution:     40 % = (40)/(100) = 0.4

0.4 x 67 = 26.8,

Selling price = cost price - discount = 67 - 26.8 = $ 40.2

5) 43 % of a number is 53.75, find the number .

Solution:   Let the number be x

43% = (43)/(100) = 0.43

0.43 x  = 53.75   divide by 0.43

x = (53.75)/(0.43) = 125

Word problems on solve finding percentages

1) Mary had 24 pages to write .  By the evening, she completed 25% of her work .  How many pages did she have left ?

Solution:    Mary completed 25% means  (25)/(100) x 24  = 6 pages
Number of pages left is 24 - 6 = 18 pages

2) John's income is 20% more than that of Tom .  How much percent is the income of Tom less than that of Jim.

Solution:   Let Tom's income is  $ 100

John's income is 20% more means 20 + 100 = $ 120

Tom's income is 100

Jim's income is 120

Tom's income is $20

% is  (20)/(120) x 100 = (100)/(6) % = (50)/(3) % = 16 (2)/(3) %

3)A 90 kg solution has 10% salt .  How much water must be evaporated to leave the solution  with 20% salt.

Solution:    Let amount of water evaporated is x
90 kg solution has 10  % salt means   (10)/(100) x 90  = 9 kg
After evaporation total quantity left is ( 90 -x )
In this solution salt is 20% means   (20)/(100)  (90-x) 20/(100) = (1)/(5) (90-x)
Quantity of salt remains same in both of these , as only water evaporated
(1)/(5) (90 - x) = 9
90 - x = 9 . 5,   90 -x = 45 ,  90 -45 = x ,  x = 45
45 Kg water evaporated.