Learning Compound Interest

Compound interest is paid on the principal and also for the interest accumulated in the past years. Compound interest emerges when interest is added to the principal, so the interest that has been added also earns interest. This adding of interest with the principal is said to be compounding. Compounding can be done for a time like yearly, quarterly, monthly, daily etc.


Learning Compound Interest Formula:


Learning The basic formula for Compound Interest is:

FV = PV (1+r)n

PV is the present value

r is the annual rate of interest (percentage)

n is the number of years the amount is deposit or borrowed for.

FV = Future Value is the amount of money accumulate after n years, including interest.

Learning Frequent Compounding of Interest:

If interest is paid more frequently, Here are a few examples of the formula:

Annually = P × (1 + r) = (annual compounding)

Quarterly = P (1 + r/4)4 = (quarterly compounding)

Monthly = P (1 + r/12)12 = (monthly compounding)

Learning Find the Present Value when know a Future Value, the Interest Rate and number of Periods.

PV = FV / (1+r)n

Learning Find the Interest Rate when know the Present Value, Future Value and number of Periods.

r = ( FV / PV )1/n – 1

Learning Find the number of Periods when know the Present Value, Future Value and Interest Rate

n = ln(FV / PV) / ln(1 + r)

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Example of compound interest


Ex 1: I have $1000.00 to invest for 3 years at rate of 6% compound interest.

Sol:  Here p=$1000, n=3, r=5

A = 1000 (1 + 0.06)3 = $1191.02.

1000.00 is worth $1191.02

Ex 2:  How many years to turns 1000into10,000 at 5% interest?

Sol :    n = ln( 10000 / 1,000 ) / ln( 1 + 0.05 ) = ln(10)/ln(1.05) = 2.3026/0.04879 = 47.19

Ex 3 :  I have $2000.00 to invest for 3 years at rate of 7% compound interest.

Here p=$2000, n=3, r=7

A = 2000 (1 + 0.07)3 = $2450.09.

2000.00 is worth $2450.09.

Angle of Elevation Learning

Introduction:

Assume that you are walking on a road. Normally your eyes will be viewing objects at a horizontal level at a distance of your height from the ground. Suppose you see a tall building in the vicinity and you want to see the top of the building. What you do is, rotate your line of sight from horizon to spot the top of thr building. The angle by which you do that is called the Angle of Elevation.

The angle of elevation and depression are a wonderful concepts in the subject of trigonometry.


Angle of elevation and depression- Definition


Let AB be an object. The angle θ that is required to see the point A from point O is called the angle of elevation of the object at point O.

The angle of elevation of an object depends on the point from where it is measured. The more you are closer to the object, the more is the angle of elevation and vice versa.

Hence, it is a must that the angle of elevation of an object should always be accompanied by the information from where it is measured.

Angle of Depression is the angle by which you tilt your eye down from horizontal line to see an object which is below you.


Angle of elevation – Practical applications

The concept of angle of elevation and depression is used very widely in estimations of heights and depths. In fact, in trigonometry, a separate topic heights and distances deals with angle of elevation and angle of depression. The angle of depression is the angle made below the horizon for objects at lower levels.



Suppose the height of the building shown in the above picture (h) is to be estimated. It may not be practicable to measure that directly. But it is always possible to measure the angle of elevation θ (equipment for such equipment is available) and also the horizontal distance between the building and the point of angle measurement (d).

The basic trigonometric identity, tan θ =`(h)/(d)`, will easily give you the estimate.


Problems based on Angle of elevation and angle of depression


Let us do a  problems relating to angle of elevation and angle of depression to understand the concept properly.

Angle of Elevation Problems :

1)  A balloon is flying high. A boy looks at it. His eyes make an angle of 30 º from the ground to the balloon.

This is angle of elevation. The distance between the boy and  the flying balloon is 10 m. Find the height of the balloon above the Ground.

Here we use sine 30º formula =  opposite side / hypotenuse   = height / hypotenuse

Sine 30º= 1/2     hypotenuse = 10 m

so height = sine 30º x 10  =    `(1)/(2)` x 10m =  5 meter

So the ballon flies at a height of 5 meter above the ground

Problems based on Angle of depression

Problem 2)From the top of a light house, the angle of depression of 2 ships on the opposite side of the light house  are

observed as  45º and 3 angle 0º. The height of the light house is 200 m. Find the distance between the ships.


Let  M be  a ship and N be another ship.   Let AB be the light house

CD is a line drawn horizontally through B.

From B  the angle of depression is 45º and 30º.

CD is parallel to MN

So angle CBM= angle BMA and angle DBN = angle BNA.

That makes angle BMA=45º and angle BNA= 30 degrees(Interior alternate angles  are equal)

In the right triangle BAN ,  tan 30º = AB/AN

Tan 30º= `(1)/(sqrt(3))`        So `(1)/(sqrt(3))`  = `(200)/(AN)`

So AN =  200√3 = 200 x 1.732 = 346.4 m

Now let us measure AM

Tan 45º = 1   Tan 45º = `(AB)/(AM)`  so AM x 1 = AB   AM= AB   AB= 200m so AM = 200 m

Distance between Ship M and the light house is 200m

Distance between ship N and the light house is 346.4 m

Add them both.

Distance between ship M and N is 200+346.4 = 546.4 meters

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Probability Problems with Dice Eighth Grade

Probability is the chance that something will happen - how likely it is that some event will happen.If a trial will produce N commonly exclusive and equally probable outcomes out of which n outcomes are positive to the amount of event A, then the probability of A is mentioned by P(A) and is defined as the ratio n/ N. Thus the probability of A is given by

                                           Event (P)   = Number of Successful outcomes
                                                                   Number of total outcomes

probability problem with dice for 8th grade:

Problem:

A fair die is thrown. Find the probabilities that the face on the die Is 

 (1) Maximum (2) Prime (3) Multiple of 3 (4) Multiple of 7

Solution:
            In the above probability problem there are 6 possible outcomes when a die is tossed. We assumed that all the 6 faces are equally likely. The classical definition of probability is to be applied here
            

The sample space is ,S={1,2,3,4,5,6}  => n(S)=6

(1) Let A be the event that the face is maximum
            Thus,
                       A=(6),        n(A)=1
            Therefore, P(A) = n(A) / n(S)

                                  = 1/6

(2) Let B be the event that the face is prime

            Thus,
                       B=(2,3,5)        n(B)=3   
            Therefore, P(B) = n(B) / n(S)

                                  = 3/6

                                  => 1/2

(3) Let C be the event that the face is multiple of 3
            Thus,          C=(3,6)    n(C)=2
          Therefore, P(C) = n(C) / n(S)

                                = 2/6

                                =>1/3

(4) Let D be the event that the face is multiple of 7
            Thus,          D  =0,           n(D)=0
           Therefore,P(D)  = n(D)/n(S)

                                  =0/6

                                  =0   (not possible)

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Dice problem for 8th grade:

Example:

Solve the problem for probability that when 2 standard 6-sided dice are rolled, the sum of the numbers on the top faces is 4.

Solution:

In the above problem, A dice contains 6 faces, so the possible out comes from the  both dices are      

                            = 6 * 6 
                            = 36 total number of possible outcomes 
Here the given condition is “Some of the numbers on the top of the face is 4”.
So the possible out comes are 
                   Dice 1             Dice 2
                      1         +         3           = 4
                      2         +         2           = 4
                      3         +         1           = 4
the total number of successful outcomes are 3

Final solution 
                          P (sum of 4)     = 3 / 36

                                                 = 1 / 12.

Volume of a Cylinder

Volume of a cylinder is a measurement of the occupied units of a cylinder. The volume of a cylinder is represented by cubic units like cubic centimeter, cubic millimeter and so on. Volume of a cylinder is the number of units used to fill a cube.

A cylinder has two parallel faces in its structure. The height of a cylinder is perpendicular to the two bases. The cylinder has a round base and a perpendicular height.



The volume of a cylinder formula depends on the area of the cylinder. The formula for volume of a cylinder can be written as,

Volume of cylinder = pi * r2 * h.


Formula for Surface Area of a Cylinder


The formula for volume of a cylinder depends on the area of the cylinder. The area of a cylinder can be written as, A = pi * r2. Here, pi is the constant value and r is the radius of the cylinder. To find the volume of the cylinder, we have to multiply the height of the cylinder to the area of the cylinder. The volume of cylinder formula can be written as,

Volume of cylinder = pi * r2 * h.

Here, pi is the constant variable, r is the radius of the cylinder and h is the height of the cylinder. The constant variable pi is equal to (22 / 7) or 3.14.


How to Find The Volume of a Cylinder


Given below are some of the examples to find the volume of a cylinder.

Example 1:

Calculate volume of a cylinder having the radius of cylinder is 3cm and height is 4cm.

Solution:

The given radius of the cylinder is 3cm and height of the cylinder is 4cm

Formula:

Volume of a cylinder = pi * r2 * Height

Volume of cylinder = 3.14 * 32 * 4 cm3

= 113.04 cm3

Answer:

The volume of cylinder is 113. 04 cm3

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Example 2:

The radius and height of the cylinder is 5 cm, 21 cm. Find the volume of cylinder.

Solution:

Given radius r = 5 cm, height = 21 cm.

Formula:

Volume of cylinder = pi * r2 * h

= 3.14 * 52 * 21 cm3

= 1648.5 cm3

Answer:

The volume of given cylinder is 1648.5 cm3

Example 3:

Find the volume of cylinder for the given base and height. base = 6 cm and height = 4 cm

Solution:

Given base = 6 cm and height = 4 cm

Formula:

Volume of cylinder = pi * r2 * h

= 3. 14 * 62 * 4 cm3

= 452.16 cm3

Answer:

The volume of cylinder is 452.16 cm3

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Probability Mass Function Examples

Introduction:

In probability, a probability mass function (pmf) is a function that gives the probability that a discrete random variable and accurately equals some value. Probability mass function examples differ from portable document format defined only for continuous random variables are not probabilities as such examples. The integral over a range of possible values [a, b] gives the probability of the random variable. Some examples for probability mass function are below. I like to share this Probability Set with you all through my article.

Example 1:

At the same time two coins are tossed, what is the probability result of getting (i) accurately one head (ii) no less than one head (iii) more or less one head.

Solution:

The sample space is S = {HH, HT, TH, TT}, n(S) = 4

Let A be the event of getting one head, B be the event of getting at least one

Head and C be the event of getting almost 1head.

∴ A = {HT, TH}, n(A) = 2

B = {HT, TH, HH}, n(B) = 3

C = {HT, TH, TT}, n(C) = 3

(i)               P(A) = n(A) / n(S) = 2/4 = 1/2

(ii)              P(B) =n(B)/n(S) = 3/4

(iii)             P(C) = n(C)/n(S) = 3/4


Example 2:


Two cards have chosen without replacement from 54 cards X measures the number of heart cards drawn y measures the number of clubs drawn. Find probability of joint probability mass function?

Solution:

A deck involves 52 cards.

Sample space = {all ordered pair (p, q)} = {(0, 0), (0, 1), (0, 2), (1, 0), (1, 1), (2, 0)}

Here, we have to find out the probability at each element of the sample space, and then we have to joint probability mass function.

X (0, 0) = (26 / 52) * (25 / 51) = 650 / 2652

X  (0,1) = (26 / 52) * (13 / 51) + (13 / 52) * (26 / 51) = 676 / 2652

X  (0,2) = (13/52) * (12/51) = 156 / 2652

X (1, 0) = (13/52) * (26/51) = 676/2652

X (1, 1) = (13/52) * (13/51) + (13/52) * (13/51) = 338/2652

X (2, 0) = (13/52) * (12/51) = 156/2652

X (s, t) = 0 if (s, t) is not in the sample space.

Free Calculus Tutoring

Introduction:

British mathematician, Isaac Newton and the German man Gottfried Leibnitz, invented the calculus. Calculus has  two classifications: Differentiation calculus, Integration calculus. Calculus is the study of rates of change, area, or volume. In symbol, we require to find f(x) where, d/dx f(x) = g(x). Integration is one of most important study of calculus in mathematics. We require to find f(x) = ∫ g(x) dx.In the trigonometric calculus the derivative of a constant is zero. There is no exact value for the integral.


Examples for free calculus tutoring:

Differential free calculus tutoring:

Free calculus tutoring problem 1:

Find the first and second derivative of f(x) = 8x4 + 7x3 - 6x2 - 5x +3

Solution;

Steps to solve:

• First differentiate f(x) = 8x4 + 7x3 - 6x2 - 5x +3                                  
• we know, d / dx (x n ) = n xn-1
• when we differentiate the first term we get  (8×4) x (4-1)
• The above equation can above simplified as 32 x 3
• Like wise we can differentiate  and simplify all the terms in the equation
• Finally after the first derivative we will get  32 x 3 + 21 x2 -12x -5
• Now we will go for Second derivative
• Second derivative of  f(x) = 32 x 3 + 21 x2 -12x -5
• Differentiate all terms   (32 × 3)x2 + (21 × 2) x - (12 × 1) x0
• End of the second derivative we will get   96x2 + 42 x - 12

First derivative:

f' = df / dx

= (8 × 4) x (4-1) + (7 × 3)x(3-1) - (6 × 2)x(2-1) - (5 × 1)x (1-1)

= 32 x 3 + 21 x2 – 12x -5

Second derivative:

f '' = df ' / dx

= 32 x 3 + 21 x2 – 12x -5

= 96x2 + 42 x - 12

Answer for the given free calculus tutoring problem is:

`d^2/dx^2 ` (8x4 + 7x3 - 6x2 - 5x +3) = 96x2 + 42 x - 12



Free calculus tutoring problem  2:

Find the first and second derivative of sin 2x.

Solution;

Steps to solve:

• First differentiate f(x) = sin 2x                                                       
• (we know, d/dx (sin x) = cos x)
• when we differentiate sin x  = cos x So, we get  sin 2x = cos 2x
• Then we differentiate 2x  So, we get  2
• The differentiation of sin 2x = 2 cos 2x
• Finally after the first derivative we will get  2 cos 2x
• Now we will go for Second derivative
• we know,  d/dx (cos x) = −sin x
• The second derivative 2 cos 2x = 2 ×2 (- sin 2x)
• So, The second derivative of  sin 2x = -4 sin 2x

First derivative:

f' = df / dx

= `d/dx` (sin 2x)

= 2 cos 2x

Second derivative:

f '' = df ' / dx

= `d /dx` (2 cos 2x)

= -4 sin 2x

Answer for the given free calculus tutoring problem is:

`d^2/dx^2 `(sin 2x) = -4 sin 2x

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Integral free calculus tutoring:


Free calculus tutoring problem 1:

Integrate  `int ` (-2x2 + 8x3 + 2x4 + x5) dx

Solution;

Steps to solve:

• First seperate each term ` int` (- 2x2 + 8x3 + 2x4 + x5) dx
• So, we get `int` (- 2x2 +8x3 + 2x4 + x5) dx =  `int ` - 2x2 dx + ` int ` 8x3 dx + `int` 2x4 dx + `int`   x5  dx
• Integrate first term ` int ` - 2x2 dx =  -2 (1/3) x3                                                                  
•  [ we know,   `int` xn dx = 1/ n+1 (xn+1)]
• Integrate all term and we get   (- 2/3) x3 + (8/4) x4 + (2/5) x5+(1/6) x6
• Finally after the integration we will get  (- 2/3) x3 + 2 x4 + (2/5) x5+(1/6) x6

` int ` (-2x2 + 8x3 + 2x4 + x5) dx =` int` -2x2 dx + `int` 8x3 dx + ` int ` 2x4 dx +` int` x5  dx

= -2 `int ` x2 dx +  8  `int` x3 dx + 2` int`x4 dx + `int` x5  dx

= -2 (1/3) x3 + 8(1/4) x4 + 2(1/5) x5+(1/6) x6

= (-2/3) x3 + (8/4) x4 + (2/5) x5+(1/6) x6

= (-2/3) x3 + 2 x4 + (2/5) x5+(1/6) x6

Answer for the given free calculus tutoring problem is:

` int` (-2x2 + 8x3 + 2x4 + x5) dx = (-2/3) x3 + 2 x4 + (2/5) x5+(1/6) x6

How to solve linear programming problems

Linear Programming is one of the operations research techniques. It is one of the best mathematical techniques for finding the limited use of resources of a concern in a best way. Complex problems can be modeled using linear functions in a presentable way by the management. The linear programming technique is used in solving a wide range of operations management problems.

Definition of linear programming problems:

Linear Programming is defined as a technique which allocates the available resources in an optimum manner for achieving the company’s objective which is for maximizing the overall profit or to minimize the overall cost under conditions of certainty.

Linear Programming can be applied to areas which are given below:

Allocation of resources to various activities of the concern, for example: man power, machine etc.
Production scheduling.
The common characteristics in the above mentioned areas are to allocate limited resources to the activities of the concern.

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How to solve Linear Programming Problems: Mathematical Formulation

Linear Programming can be used in a variety of situations. In most of the business or economic situations, the resources will be limited, the problem there will be to make use of the available resources in such a way as to maximize the production or to maximize the profit or to minimize the expenditure. This can be formulated as linear programming models.

Mathematical Formulation of the problem:

How to solve linear programming problems?? here are the steps which you need to follow:

Step 1:

Write down the decision variables of the problem.

Step 2:

Formulate the objective function to be optimized as a linear function of the decision variables.

Step 3:

Formulate the other conditions of the problem as Linear equations or In equations in terms of the decision variables.

Step 4:

Add the non negativity constraint from the consideration that negative values of the decision variables do not have any valid physical interpretation.

The objective function, the set of constraints, and the non negative constraints together form an LPP.


Steps to solve linear programming problems using Graphical Method:


When a LPP has only two variables in the objective function and constraints, it can be easily solved using the graphical method. The given information of a LPP can be plotted on the graph and the optimal solution can be obtained from the graph.

The steps to solve an Linear Programming Problem using Graphical method is given below:

Step 1:

Identify the decision variables, the objective function and the restrictions for the given Linear Programming Problem (LPP).

Step 2:

Write the Mathematical Formulation of the problem.

Step 3:

Plot the points on the graph representing all the constraints of the problem. Find the feasible region or solution space. The intersection of all the regions represented by the constraints of the problem is called the feasible region and is restricted to the first quadrant only.

Step 4:

The Feasible region obtained in the step 3 may be bounded or un bounded. Determine the Co-ordinates (x, y) values of all the corner points of the feasible region.

Step 5:

Find the value of the objective function at each corner points (solution) determined in step 3.

Step 6:

Select a point from all the corner points that optimizes (Maximizes or Minimizes) the values of the objective function. It gives the Optimum Feasible Solution.

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Some Exceptional Cases of Linear Programming Problem:


There may be an LPP for which no solution exists or for which the only solution obtained is an unbounded one. The exceptional cases arise in the application of graphical method are

• Alternative Optima
• Unbounded Solution
• Infeasible Solution or Non existing Solution

Alternative Optima:

When the objective function is parallel to the binding constraint, the objective function will assume the same optimal value at more than one solution point, because of this reason, they are called as Alternative Optima.

Unbounded Solution:

When the values of the decision variables may be increased in definitely without violating any of the constraints, the feasible region is unbounded. In such cases, the value of the objective function may increase (for maximisation) or decrease (for minimisation) in definitely. Thus, both the solution space and the objective function value are unbounded.

Infeasible Solution:

When the constraints are not satisfied simultaneously, the LPP has no feasible solution. This solution can never occur, if all the constraints are of less than or equal to type.




Example for some exceptional cases:


The general form of the LPP is used to develop the procedure for solving a common programming problem.

A standard LPP Some exceptional cases is of the form
Max (or min) Z = c1x1 + c2x2 + … +cnxn
x1, x2, ....xn these are called decision variable.

Ex: Show graphically that the model

Maximize Z = -5y

Subject to

x+y<span style="font-family: Serif;">?</span> 1

0.5x-5y<span style="font-family: Serif;">?</span> -10

x<span style="font-family: Serif;">?</span> 0

y<span style="font-family: Serif;">?</span> 0 has no feasible solution.

Sol:

Draw the graphs x + y = 1

- 0.5 -5y = - 10

Shade the half planes of the constraints x + y 1 …(1)

-0.5x - 5y -10 …(2)



Points are (0,1)(0,2)(1,0)(20,0)

Note that the origin (0, 0) does not satisfy the in 2nd equation hence the required region is the upper half plane.

From the graph, that the intersection of the constraints is empty. Therefore the given problem has no feasible solution. So, the some exceptional cases of given LPP has no solution.

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