Weight of the Cylinder

Cylinder is the three dimensional figure. Weight of the cylinder is same as the volume of the cylinder. The formula to find the weight of the cylinder is pi * radius2 * height. Height is the total height of the cylinder and radius is the radius of the circular face which is at the bottom and top of the cylinder.

Example diagram and formula - Weight of the cylinder:

Formula - Weight of the cylinder:

Weight of the cylinder = pi * r2 * h, where r`=>` radius of the cylinder and h`=>` height of the cylinder.

Example problems – Weight of the cylinder:

Calculate the weight of the cylinder whose radius is 10cm and height is 12 cm.

Solution:

Given, radius of the cylinder = 10cm and height of the cylinder= 12cm.


The formula to find the weight of the cylinder is pi * r2 * h

= 3.14 * 102 * 12

= 3.14 * 100 * 12

= 3768cm3.

The weight of the cylinder is 3768cm3 .

Calculate radius of the cylinder whose weight is 3000cm3 and height is 30cm.

Solution:

Given weight of the cylinder= 3000cm3 and height of the cylinder = 30cm


Weight of the cylinder = pi * r2 * h

3000   = 3.14 * r2 * 30

3000   = 94.2 * r2.

Divide the above expression by 94.2 on both sides,

3000/94.2 = 94.2 * r2/94.2

31.85        = r2

r               = 5.6cm.

Radius  of the cylinder    = 5.6cm.

Calculate the weight of the cylinder whose radius is 5cm and height is 2 cm.

Solution:

Given, radius of the cylinder= 5cm and height of the cylinder= 2cm.


The formula to find the weight of the cylinder is pi * r2 * h

= 3.14 * 52 * 2

= 3.14 * 25 * 2

= 157cm3.

The weight of the cylinder = 157cm3 .

Calculate radius of the cylinder whose weight is 300cm3 and height is 3cm.

Solution:

Given weight of the cylinder= 300cm3 and height of the cylinder= 3cm


Weight of the cylinder = pi * r2 * h

300   = 3.14 * r2 * 3

300   = 9.42 * r2.

Divide the above expression by 9.42 on both sides,

300/9.42 = 9.42 * r2/9.42

31.85      = r2

r             = 5.6cm.

Radius of the cylinder   = 5.6cm.

Percent Loss Formula

Loss is defined as the difference between the cost price and the selling price. Through the loss we can calculate loss percentage. If they give loss percentage we also find the cost price and the selling price. Let us see the formula and examples for loss and loss percent in this article.

Formula for Loss Percentage

Variation between the cost price and the selling price is called as loss, If the cost price is greater than the selling price means we obtain loss.

Loss = cost price- selling price

Selling price is indicated by S.P and cost price is indicated by C.P.

Based on the cost price we determine the loss.

Formula for loss Percent: Loss Percentage formula = (lossxx 100)/(C.P)

Sometimes they give loss percentage and ask cost price or selling price.

Example on Loss Percent

Example 1: Ragu buys pineapple at 5 for 50Rs ech and sold them at 3 for 25 each. Find his loss and loss percent?

Solution: Given

Number of pineapple bought = lest common divisor of 5,3 =15
Investment or cost price = (50 / 5 ) *15 =150
Selling price = (25 /3)  *15 = 125, here cost price is greater then selling price,we can say it is loss.
Loss amount =cost price – selling price
Loss =150 – 125
Loss amount =25
Loss percentage formula = (lossxx 100)/(C.P)
Loss percentage = (25 xx 100)/(150)
Loss percentage = 16.6%
Loss and loss percentage of Ragu were 25 Rs and 16.6%

Example 2: Cost price =50 Rs, Selling price =45 Rs then find the loss and loss percentage?

Solution: Given

Cost price =50
Selling price =45
Loss = cost price - selling price
Loss =50- 45
Loss =5 Rs.
Loss percentage formula = (lossxx 100)/(C.P)
Loss percentage = (5xx 100)/(50)
Loss percentage = 10%

Example 3: Deepen loss 80 cents on $80 .What is the loss percentage of Deepen?

Solution: Given

Loss 80 paisa on 80 rupees
Here Cost price = $80
Loss =80 cents
Loss percentage formula = (lossxx 100)/(C.P)
Loss percentage = (0.8 xx 100)/(80)
Loss percentage = 1 %

Learn Online Linear Equations

A linear equation is an algebraic equation in which each term is either a constant or the product of a constant and (the first power of) a single variable. Linear equations can have one or more variables. Linear equations occur with great regularity in applied mathematics. Source - Wikipedia

Let us Learn linear equation in online. In online we can understand very easily about linear equation.

Learning Concepts of Linear equations in online:

Learning Concepts of Linear equations are as follows:

• In the graph showed, we would like to get the equation of the graphed line. To complete this we start linear equation. We have innocently been functioning with linear equation, so this is NOT a fresh concept for us Most probable you have been working with the standard form of a line, which is Ax + By = C. This is a linear equation with two unfamiliar variables, x and y.

• The ordinary form of the line in the graph is 2x + y = 5, where A =2, B = 1, and C = 5.

• We want to restructure this equation and set it into the y-intercept form which is y = mx + b, in which m is the slope and b is the y-intercept. In this shape, it is very simple to graph. Just explain the standard equation for “y” and you have the y- intercept form.

Example problems for learning linear equations in online:

Example problems for learning linear equations are as follows:

Example 1:

x - 4 = 10

Add 4 to both sides of the equation:

x = 14

The answer is x = 14

Make sure the resolution by substituting 14 in the original equation for x. If the left side of the equation generation the right side of the equation after the substitution, you have found the correct answer.

Example 2:

2x - 4 = 10

Add 4 to both sides of the equation:

2x = 14

Divide both sides by 2:

X = 7

The answer is x = 7.

Algebra is widely used in day to day activities watch out for my forthcoming posts on Unit of Length and Mean Value Theorem Examples. I am sure they will be helpful.

Check the solution by substituting 7 in the original equation for x. If the left side of the equation generation the right side of the equation after the substitution, you have found the correct answer.

2(7) - 4 = 14 - 4 = 10.

Multiplication Rule

The multiplication rule is a product used to determine the probability that two events, A and B, both occur.

The multiplication rule follows from the description of conditional probability.

The result is often written as follow, using set notation:

P( A ∩ B ) = P(A | B ) . P(B)

Or

P(A ∩ B) = P(B | A) . P(A)

Where:

P(A) = probability that event A occurs

P(B) = probability that event B occurs

P(A ∩ B) = probability that event A and event B occur

P(A | B) = the conditional probability that event A occur given that event B has occurred already

P(B | A) = the conditional probability that event B occur given that event A has occurred already

For independent an event, that is events which have no influence on one another, the rule simplifies to:

P(A nn B) = P(A). P(B)

That is, the probability of the joint events A and B is equivalent to the product of the individual probabilities for the two events.

Multiplication Rule of Probability

The addition rule helped us resolve problems when we performed one task and wanted to know the probability of two things happening during that task. This topic deals with the multiplication rule. The multiplication rule also deal with two events, but in these problems the events occur as a result of more than one task (rolling one die then another, drawing two cards, spinning a spinner twice, pulling two marbles out of a bag, etc).

When ask to find the probability of A and B, we would like to find out the probability of events A and B happening.

The Multiplication Rule:

Consider events A and B. P(AB)= P(A) P(B).

What The Rule Means:
Expect we roll one die followed by another and want to find the probability of rolling a 4 on the first die and rolling an even number on the second die. Notice in this problem we are not trade with the sum of both dice. We are only commerce with the probability of 4 on one die only and then, as a separate event, the probability of an even number on one dies only.
P(4) = (1)/(6)
P(even) = (3)/(6)
So P(4even) = ((1)/(6) )((3)/(6) ) = (3)/(36)  = (1)/(12)
While the rule can be applied in any case of dependence or independence of events, we should note here that rolling a 4 on one die followed by rolling an even number on the second die are independent events. Each die is treated as a split thing and what happens on the first die does not influence or effect what happens on the second die. This is our basic description of independent events: the outcome of one event does not influence or affect the outcome of another event.

Let's Practice the Multiplication Rule

Assume you have a box with 3 blue marbles, 2 red marbles, and 4 yellow marbles. You are going to draw out one marble, record its color, put it back in the box and draw another marble. What is the probability of pull out a red marble followed by a blue marble?


The multiplication rule says we need to find P(red) P(blue).

P(red) = (2)/(9)
P(blue) =(3)/(9)

P(redblue) = ((2)/(9) )((3)/(9) ) = (6)/(81)  = (2)/(27)
The events in this example were independent. Once the first marble was pull out and its color recorded, it was returned to the box. Therefore, the probability for the second marble was not affected by what happened on the first marble.

Some students find it supportive to simplify before multiplying, but the final answer must always be simplified.


Think the same box of marbles as in the previous example. However in this case, we are going to draw out the first marble, leave it out, and then pull out another marble. What is the probability of pull out a red marble followed by a blue marble?


We can motionless use the multiplication rule which says we need to find P(red) P(blue). But be alert that in this case when we go to pull out the second marble, there will only be 8 marbles left in the bag.

P(red) = (2)/(9)
P(blue) = (3)/(8)

P(redblue) = ((2)/(9) )((3)/(8) ) = (6)/(72)  = (1)/(12)

The events in this example were dependent. When the first marble was pull out and kept out, it affected the probability of the second event. This is what is meant by dependent events.


Assume you are going to draw two cards from a standard deck. What is the probability that the first card is a champion and the second card is a jack (just one of several ways to get “blackjack” or 21).

Using the multiplication rule we get

P(ace) P(jack) = ((4)/(52) )((4)/(51))  = (16)/(2652)  =  (4)/(663)

Notice that this will be the similar probability even if the question had asked for the probability of a jack followed by an ace.

Preparation for Sample Size

Our critical characteristic of preparation of sample size plan is deciding how big your sample size should be. If you amplify your preparation about sample size you add to the exactitude of your estimates, which means that, for any given approximation / size of result, the better preparation about the sample size is  the more “ statistically significant” the outcome will be. In other words, if an examination is too small then it will not sense fallout that is in actuality significant.

Defintion of preparation for sample size

The preparation about sampling is that a division of statistical performs troubled with the assortment of an impartial or casual division of personality explanation within a inhabitants of persons intended to give up some information regarding the population of distress, mainly for the purposes of construction predictions based on arithmetic supposition. Sampling is a central part of data collection.

Sample size process:

Step 1: Essential the population of worry.

Step 2: Indicate a sampling outside, a spot of material or trial likely to estimate.

Step 3: Specifying a sampling system for selecting items or actions from the enclose.

Step 4: Decisive the sample size.

Step 5: Implementing the sampling map.

Step 6: Gather the study sample data and information.

Step 7: Reviewing the example process.

Formula of preparation for sample size:

The formula for sample size = n = t2 * p (1-p) / m2.

Description:

N represents the required sample size.

T represents the confidence level. (Regular value 1.96).

P represents the estimated prevalence of malnutrition in the area

M represents margin of error. (Steady value 0.05).

Example for the preparation of sample size:

Ex:1 Find  the sample size when the population is 15.

Sol:

The sample size = t2 * p(1-p) / m2.

Where, the t is constant value of 1.96,

the m is constant value of 0.05.

Size = (1.96)2 * (15)*(1-15) / (0.05)2.

= 3.8416 *15 *14 / 0.0025.

= 322694.4

So, the sample size is 322694.4

Ex:2 Find  the sample size when the population is 30.

Sol:

The sample size = t2 * p(1-p) / m2.

Where, the t is a constant value of 1.96,

The m is a constant value of 0.05.

Size = (1.96)2*(30)*(1-30) / (0.05)2.

= 3.8416*30*29 / 0.0025.

= 1336876.8

So, the sample size is 1336876.8

Learn Common Ratio

In mathematics, a geometric series is a series with a permanent ratio between successive terms. For example, the series 1/2+1/4+1/8+1/16.....is geometric, so each term except the first can be obtained by multiplying the previous term by 1/2.Geometric series are one of the easiest examples of infinite series with finite sums. Basically, geometric series played an important role in the early development of calculus, and they continue to be central in the study of convergence of series. Geometric series(GS) are used throughout mathematics, and they have important applications in physics, engineering, biology, economics, computer science, queueing theory, and finance.

Common ratio:

The terms of a geometric series form a geometric progression(GP), meaning that the ratio of successive terms in the series is constant. Below table shows several geometric series with different common ratios:

Common ratio


Example

10


4 + 40 + 400 + 4000 + 40,000 + ···

1/3


9 + 3 + 1 + 1/3 + 1/9 + ···

1/10


7 + 0.7 + 0.07 + 0.007 + 0.0007 + ···

1


3 + 3 + 3 + 3 + 3 + ···

−1/2


1 − 1/2 + 1/4 − 1/8 + 1/16 − 1/32 + ···

–1


3 − 3 + 3 − 3 + 3 − ···

The behavior of the terms depends on the common ratio r:

If r is between -1 and +1 the terms of the series become smaller and smaller, approaching zero in the limit. The series converges to a sum, as in the case, where r is a half, and the series has the sum one.

If r is greater than 1 or less than -1 the terms of the series become larger and larger. The total sum of the terms also gets larger and larger, and the series has no sum. (The series diverges.)

If r is equal to 1, all of the terms of the series are the same. The series diverges.
If r is minus one the terms take two values alternately (e.g. 2, −2, 2, −2, 2,... ). The sum of the terms oscillates between two values (e.g. 2, 0, 2, 0, 2,... ). This is a different types of divergence and again the series has no sum.

Examples:

Consider the sum of the following geometric series:

s = 1+2/3+4/9+8/27+......

This series has common ratio 2/3. If we multiply through by this common ratio, then the initial one becomes a 2/3, the 2/3 becomes a 4/9, and so on:

2/3 s = 2/3 + 4/9 + 8/27 + 16/81 + .....

Mean Value Theorem

The Mean Value Theorem has very important consequences in differential calculus.

THEOREM: Let the function f such that

i.           continuous in the closed interval [a,b]
ii.            derivable in open interval (a,b)

Then there exists at least one c with  a < c < b such that

calculus formula

The result in the theorem can be expressed as a statement about graph of f: if A(a , f(a)) and B(b , f(b)), are the end points on the graph, then there is at least one point C between A and B,such that the tangent is drawn from C is parallel to the chord AB.

Mean value theorem graph

Mean value theorem is also known as Lagrange’s Mean Value Theorem or First Mean Value Theorem or Law of Mean.

Applications of mean value theorem

1. Let the function be f such that

(i)                  Continuous in interval [a,b]

(ii)                Derivable in interval (a,b)

(iii)              f'(x) = 0 `AA` x  `epsi` (a,b) , then f(x) is constant in [a , b].

2.Let f and g be a functions such that

(i)             f and g are continuous in interval [a,b]

(ii)            f and g are derivable in interval(a,b)

(iii)          f'(x) = g'(x) `AA` x  `epsi`  (a,b) , then f(x) - g(x) is constant in [a,b]

3.Let the function be f such that

(i)             Continuous in interval[a,b]

(ii)            Derivable in interval(a,b)

(iii)          f'(x) > 0 `AA` x  `epsi`  (a,b), then f(x) is strictly increasing function in [a,b]

4.Let the function be f such that

(i)             Continuous in interval[a,b]

(ii)            Derivable in interval (a,b)

(iii)          f'(x) < 0 `AA` x  `epsi`    (a,b), thenf(x)  is strictly decreasing function in[a,b]

Special case of Mean Value Theorem is when f(a) = f(b).Then there exists at least one c with  a < c < b such that f'(c)= 0 . This case is known as Rolle’s Theorem.

Cauchy’s mean value theorem in calculus

Let f and g be functions such that

i.            both are continuous in closed interval [a,b]
ii.            both are derivable in open interval (a,b)
iii.             g'(x) `!=` 0 for any x `epsi` (a,b)  then there exists at least one number c `epsi` (a,b) such that

`(f'(c))/(g'(c))`  =  `(f(b) - f(a))/(g(b) - g(a))`

Mean value theorem example



Verify Rolle's theorem for the function

f (x) = x2 - 8x + 12 on (2, 6)

Since a polynomial function is continuous and differentiable everywhere f (x) is differentiable and continuous (i) and (ii) conditions of Rolle's theorem is satisfied.

f (2) = 22 - 8 (2) + 12 = 0

f (6) = 36 - 48 + 12 = 0

Therefore (iii) condition is satisfied.

Rolle's theorem is applicable for the given function f (x).

\ There must exist c  (2, 6) such that f '(c) = 0

f '(x) = 2x - 8

`=>`  c = 4 `in`(2,6)

Rolle's theorem is verified.