Quick Math Answers for Algebra

Introduction to quick math answers for algebra:

Algebra is the branch of mathematics concerning the study of the rules of operations and relations, and the constructions and concepts arising from them, including terms, polynomials, equations and algebraic structures. The part of algebra called elementary algebra is often part of the curriculum in secondary education and introduces the concept of variables representing numbers. (Source: From Wikipedia). In this article, we are going to see some of the math algebra problems quick help.

Quick Math Algebra Problems:

Example problem 1:

Simplify the expression: 14b + 6 - 21a + 15b + 13 + 15a

Solution:

14b + 6 - 21a + 15b + 13 + 15a

Combine the like terms in the given expression

-21a + 15a + 14b + 15b + 6 + 13

Add the like terms in the expression

(-21 + 15)a + (14 + 15)b + (6 + 13)

-6a + 29b + 19

So, the answer is -6a + 29b + 19.

Example problem 2:

Solve for the value of c: 9c - 2 = 79

Solution:

9c - 2 = 79

Add 2 on both sides of the equation

9c - 2 + 2 = 79 + 2

9c = 81

Divide by 9 on both sides of the equation

`(9c) / 9 = 81 / 9`

c = 9

So, c = 9 is the solution of the given equation.

Few more Math Algebra Problems Quickly:

Example problem 3:

Find the x and y intercept of the equation: 12x + 6y = 48.

Solution:

12x + 6y = 48

To find the x intercept value, plug y = 0 in the given equation

12x + 6(0) = 48

12x = 48

x = 4

So, the x intercept is (4, 0).

To find the y intercept value, plug x = 0 in the given equation

12(0) + 6y = 48

6y = 48

y = 8

So, the y intercept is (0, 8).

Example problem 4:

Solve the inequality: 20y - 43 < 17

Solution:

20y – 43 < 17

Add 43 on both side of the inequality

20y- 43 + 43 < 17 + 43

20y < 60

Divide by 20 on both sides of the equation

`(20y) / 20 < 60 / 20`

y < 3

So, the solution is (-infinity, 3).

Practice Math Algebra Problems Free Help Quickly:

1)      Simplify the expression: -15x + 16y + 13x - 12y. (Answer: -2x + 4y)

2)      Solve for the variable z:  10z + 34 = 24 + 8z (Answer: z = -5).

3)      Solve for the variable x:  13x - 68 = 10 (Answer: x = 6).

Sum of Two Irrational Numbers

Introduction to sum of two irrational numbers:

Let us study about sum of two irrational numbers. Irrational numbers are defined as the numbers which cannot be written in the form of simple ratios of two integers.
The irrational numbers are commonly said to have infinite values. These irrational numbers can get added with each other by approximating or rounding those infinite values to some nearest values.
Some examples for sum of two irrational numbers are discussed below.

Two Irrational Numbers:

Two irrational numbers – example 1:

Add the following two irrational numbers 3.4256… and 2.3227…


Solution:

The two given irrational numbers are 3.4256… and 2.3227…
To add the two given irrational numbers follow the steps below given steps:
3.426 + 2.323 (by rounding the values with three decimal points)
Therefore the added value of the two given irrational number is found to be as ‘5.749’


Two irrational numbers – example 2:

Add the following two irrational numbers 0.436… and 12.987…


Solution:

The two given irrational numbers are 0.436… and 12.987…
To add the two given irrational numbers follow the steps below given steps:
0.44 + 12.99 (by rounding the values with two decimal points)
Therefore the added value of the two given irrational number is found to be as ’13.43’
Looking out for more help on Solve Equations in algebra by visiting listed websites.

Two irrational numbers – example 3:

Add the following two irrational numbers `sqrt(3) and sqrt(12)`


Solution:

The two given irrational numbers are `sqrt(3) and sqrt(12)`
To add the two given irrational numbers follow the steps below given steps:
`sqrt(3) + sqrt(12) `
`sqrt(15)` = 3.872983…
3.873 (round the values with three decimal points)
Therefore the added value of the two given irrational number is found to be as ‘3.873’


Two irrational numbers – example 4:

Add the following two irrational numbers 6.2328… and `sqrt(8)`


Solution:

The two given irrational numbers are 6.2328… and `sqrt(8)`
To add the two given irrational numbers follow the steps below given steps:
6.2328… + `sqrt(8)`
6.2328… + 2.8284…
6.233 + 2.828 (by rounding the values with three decimal points)
Therefore the added value of the two given irrational number is found to be as ‘9.061’
Two irrational numbers – exercises:

Add the following two irrational numbers `sqrt(2) and sqrt(6).` (Answer: 3.863)
Add the following two irrational numbers 0.2028… and 9.45454... (Answer: 9.658)

Second Grade Math

Introduction to second grade math:

In the second grade math the students learn about the counting and number patterns, comparing and ordering, place values, estimation and rounding, names of numbers, logical reasoning, addition one digit and two digit, subtraction one digit and two digit, addition three digits, subtraction three digits, probability and statistics, multiplication and division, and mixed operations.

Second Grade Math Counting and Place Values

Second grade math to study counting and number patterns:

Example problems:

What is the missing number in the given sequence?

10, 20, 30, ____, 50, 60.

Answer: The missing number in this sequence is 40.

Explanation:

Note that the numbers, each number 10 more than the previous number.

Problem 2:

What is the missing numbers in the following sequence?

510, 511, 512, _____, 514, 515, 516, _____, 518, 519.

Answer:

The missing numbers in the following sequence is 513 and 517.

Explanation:

513 is comes between the numbers 512 and 514.

517 is comes between the numbers 516 and 518.

Second grade math to study place values:

In this the students can place the values in the form of ones, tens, and hundreds.

Example problems:

Example 1:

How many hundreds, tens, and ones present in the value 234?

Answer: There are 2 hundreds, 30 tens, and 4 ones present in the above value.

Example 2:

Write the value in the given term.

3 thousands + 6 hundreds + 5 tens + 0 ones = _______.

The correct answer is: 3,650.

Second Grade Math Geometry and Addition

Second grade math to study geometry:

Example 1:

Identify the following figure.

a)      Rectangle

b)      Square

c)       Circle

d)      Triangle

Answer: option b.

Addition:

Example for one digit addition:

5 + 6 = 11.

8 +9 = 17.

9 +5 = 14.

Example for two digit addition:

Example 1:

Add the given two digit number 35 and 23.

35

23

______

58

______

Example 2:

Add the given two digit number 67 and 42

67

42

_____

109

_____

Example for 3 digit addition:

Example 1:

Add the given three digit number 237, 429, and 320.

Write the numbers in vertical form.

237

429

320

______

986

______

Add the following three digit numbers 345 and 123

345

123

______

468

______

Connected Math Variables and Patterns

Introduction about connected math variables and patterns:

Variables are one of the branches in algebra. In math, variables are very necessary concept. It does not change the meaning of expressions. We are mostly use variables to stand for the algebra expression like a,b,c and d. In math, generally variables can be defined using alphabets. Patterns are generally the collection of numbers which are listed in an order under a certain conditions. There are three different types of general patterns like Arithmetic Pattern, Geometric Pattern and Alphabetic Pattern. Here we are going to study about connected math variable and patterns.

Examples of Connected Math Variables:

Example 1:

(6m^2+9) + (-5m^2+7).

Solution:

Step 1: Here we are going to add the two terms

Step 2:  Its also like a normal addition.

Step 3: It can be written as 6m^2-5m^2+9+7.

Step 4: Now we can easily add the terms.

Step 5: Therefore, the answer is m^2+16.

Example 2:

Add (8n3+12) +( -15n3-6)

Solution:

Step 1: Here we are going to add the two binomials.

Step 2:  Its also like a mathematical addition.

Step 3: It can be written as 8n^3-15n^3+12-6.

Step 4: Now we can easily add the terms.

Step 5: Therefore, the answer is 7n^3+6.

These are the examples of connected math variables. Having problem with trig identities solver keep reading my upcoming posts, i will try to help you.

Examples of Patterns:

Example 1:

Compute the missing terms from the pattern given below.

2, 4, 8, 16, 32, 64, ___, ____

Solution:

Step 1: The first term of the pattern is 2.

Step 2: The second term of the pattern = 2 `xx` 2 = 4.

Step 3: The third term is 4 `xx` 2 = 8.

Step 4: The fourth term is 8 `xx` 2 = 16.

Step 5: Similarly, the missing terms can be determined as follows.

Step 6: The seventh term will be 64 `xx` 2 = 128.

Step 7: The eighth term is 128 `xx` 2 = 256(So, the pattern is we need to multiply 2 with the previous term).

Step 8: Therefore, the pattern is 2, 4, 8, 16, 32, 64, 128 and 256.

Example 2:

Compute the next two terms in the pattern given below.

1, 3, 5, 7, 9, 11, ___, ____

Solution:

Step 1:The first term of the pattern is 1

Step 2:The second term of the pattern = 1 + 2 = 3

Step 3:  The third term is 3 + 2 = 5

Step 4: fourth term is 5 + 2 = 7

Step 5: Similarly, the missing terms can be determined as follows.

Step 6: The seventh term will be 11 + 2 = 13

Step 7: The eighth term is 13 + 2 = 15

Step 8: This is an odd sequence of number.

Step 9: So the exact pattern is 1, 3, 5, 7, 9, 11, 13 and 15.

These are the example problems of connected math variables and patterns.

Negative Whole Numbers

Introduction :
Negative numbers are defined as the numbers that less than zero. It is always an opposite of the positive numbers. A negative whole number means negative numbers which consist of all negative integer numbers. We can add, subtract, multiply and divide by using the negative whole numbers. For example 0, 1, 2, 3… and -1, -2-, 3… etc

Examples of Negative Whole Numbers:

Negative whole numbers example 1:

Simplify (-120) + (-250)

Solution:

Given problem is (-120) + (-250)

We can add the given negative whole numbers by using the following methods:

Positive * negative = negative

In the given problem, first term as -120 and then the second term as + (-250) = -250

Negative + negative = negative

Therefore, -120 + -250= -370

So we get the final result as negative number.

Answer: -370

Negative whole numbers example 2:

Simplify (-200) - (-500)

Solution:

Given problem is (-200) - (-500)

We can subtract the negative whole numbers by using the following methods:

Negative * negative =positive

In the given problem, first term as -200 and then the second term as - (-500) = + 500

But here negative + positive = positive. Because of positive number (500) is larger than negative value (200).

That is, -200 + 500= 300

Here we get the final answer is positive. Understanding Is pi a Rational Number? is always challenging for me but thanks to all math help websites to help me out.

Answer: 300

Negative whole numbers example 3:

Simplify (-400) * (-100)

Solution:

Given problem is (-400) * (-100)

We can multiply the negative whole numbers by using the following methods:

Negative * negative = positive

But here we can just multiply the whole numbers. Then we get

-400 * (-100) = 40000

Answer: 40000

Negative whole numbers example 4:

Simplify (-1200) / (-300)

Solution:

Given problem is (-1200) / (-300)

Here we are dividing the above whole numbers.

Negative / negative = positive

Therefore we get the positive answer.

That is, -1200/-300

Here minus sign are canceled on both numerator and denominator.

Then we get,

1200/300

1200 is 4 times of 300

1200/300= 4

Therefore we get the final answer is 4.

Answer= 4

Practice Problems of Negative Whole Numbers:

Simplify the following problems:

(-100) + ( -50)
(- 120) – ( -60)
( -30) * (-40)
(-400) / (-50)
Answer keys:

-150
-60
1200
8

Trig Word Problems

Introduction to trig word problems:

Greek Mathematician Ptolemy, Father of trigonometry proved the equation sin2A+cos2A=1 using geometry involving a relationship between the chords of a circle. Trigonometry was mainly concerned with establishing the relations between sides and angles of a triangle. The trigonometry consists of angles, quadrants, ratios and Identities, Compound Angles, and trigonometrical Equations. The trigonometry word example problems and practice problems are given below.

Example Problems for Trig Word Problems:

Trig word problems - Example: 1

The top of a tower was seen from the top and the bottom of a building of height 10 m at angles of elevation 45° and 60°. Determine the height of the tower.

Solution:

 

Let AB be the tower CD be the building of height 10 m.

Let DE be perpendicular to AB from D.

At C, the angle of elevation of A is 60°.

That is ?BCA = 60°

At D, the angle of elevation of A is 45°.

That is ?EDA = 45°

BE = CD = 10 m (since BEDC is a rectangle)

Let AE be x in metres, then AB is = x +10 m

In a right angled triangle ABC

tan60° = AB/BC (or) v3 = (x + 10m)/BC

After solve this, we get

ED = BC = (x + 10m)/v3

In a right angled triangle AED

tan45° = AE/ED   (or)

v3 x = x + 10 m

v3 x – x = 10 m

x( v3 –1) = 10 m

After solving this, we get

= 5 (1.732+1)m = 5 (2.732) m = 13.66 m

Height of the tower AB = x + 10 m = 13.66 m + 10 m = 23.66 m

Trig word problems - Example: 2

Determine the length of the chord of a circle of radius 10 cm subtending at the centre the angle of 144°.

Solution:

Let AB be a chord of a circle with centre 0 of radius 10 cm. Draw OC?AB. Then C is the mid point of AB

?AOB = 144°

?COB = 72°

In right angled triangle OCB

OB/BC = sin 72°

BC = 10 sin 72° cm

= 10 × 0.9511 cm

= 9.511 cm

Length of chord AB = 2 × BC

= 2 × 9.511 cm =  19.022cm . Is this topic Translation Math hard for you? Watch out for my coming posts.

Practice for Trigonometry Word Problems:

1. A flag staff stands on the top of 6 m high tower. From a point on the ground the angle of elevation of given top of the flag staff is 60° and from the same point the angle of elevation of the top of the tower is 45°. Determine the height of the flag staff.

Answer: 4.392

2. A ladder placed that against the wall such that, the ladder are reaches the top of the wall then height 6 m and the ladder is inclined at an angle of 60°. Determine how far the ladder is from the foot of the wall.

Answer: 3.464

Random Prime Number

Introduction :

Prime number is an important topic in mathematics.Prime number is defined as the number which is  divisible by  1 and the number by itself. In this article we shall discuss  random prime number with suitable example problem. In a set of random data we need to identify the prime number.

Random Prime Number Examples:

Prime numbers between 1 and 100:

The following numbers are the least prime numbers

2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97.

Random prime number example 1:

Solution:

Find is 4 a prime number?

Step 1:

The number of 4 is

Step 2:

Now We should identify 4 is prime number or not.

Step 3:

Factors 4 is  1  2  4

Step 4:

So  4 has three factors. So it is not a prime number.

Random prime number example 2:

Solution:

Find  is 14 a  prime number in the random set of data 1,2.4,7,13,23,14,15

Step  1:

The number of 14 is

Step 2:

Now We should identify 14 is prime number or not.

Step 3:

factors of 14 is  1  2  7  14

Step 4:

So  14 has four factors. So it is not a prime number.


Few more Random Examples:

Random prime number example 3:

Solution:

Find  is 7 a prime number in the random of data 1,2.4,7,13,23,14,15

Step  1:

The number of 7 is

Step 2:

Now We should identify 7 is prime number or not.

Step 3:

Factors of 7 is 1  7

Step 4:

So  7  has  two factors. So it is  a prime number.Having problem with how to add percentages keep reading my upcoming posts, i will try to help you.

Random prime number example 4:

Solution:

Find  is 13 a  prime number in random of data 1,2.4,7,13,23,14,15

Solution:

Step  1:

The number of 13 is

Step 2:

Now We should identify 13 is prime number or not.

Step 3:

Factors of 13 is 1  13

Step 4:

So  13  has  two factors. So it is  a prime number.

Solving Box and Whisker Plots Practice

Introduction To solving box and whisker plots practice:

Box-and-whisker diagram is also called as box plot or Whisker plot.
It is a suitable way of graphically give a picture of groups of numerical data with the help of their five-number summaries.
Box and Whisker plots are generally used in the display of statistical analyses of a group of numbers.
The five number summary is a different name for the visual illustration of the box and whisker plot.
The five number summary consist of:

1. The 2nd quartile.
2. The 1st quartile.
3. The 3rd quartile.
4. The Largest value in a data set.
5. The minimum value in a data set.

Diagram of Box and Wisker Plot :-



Now Lets see an example problem that helps you to understand the topic Solving box and whisker plots practice .

Solved Example Problem on Solving Box and Whisker Plots Practice

Solve and draw  the box plot for the following set of numbers

35, 24, 53, 57, 14, 78, 95

Solution:-

To draw the box plot for the given set of the numbers we have to arrange the given set of numbers in increasing order.

35, 24, 53, 57, 14, 78, 95

A box plot is entirely based on medians. The basic step is to find the median of the given set of the numbers.

Practice Help Step 1:-

We can find the median of set of numbers by using the formula` (n+1)/2`

14, 24, 35, 53, 57, 78, 95

Here the numbers in the set is odd (n=7) so we use this formula.

`(n+1)/2 = (7+ 1) /2 `

By Solving the above step we get  4

So the number in the fourth position is the median.

Here the number is 53

So the median is 53.

Practice Help Step 2:-

The median of lower numbers is called as lower quartile. (1st quartile).

14, 24, 35

Here the numbers in the set is odd (n=3) so we use this formula.

`(n+1)/2 = (3+ 1) /2 `
By solving the above we get as 2

Medial of lower set of numbers is 2

So the lower quartile is 24

Practice Help Step 3:-

The median of upper numbers is called as lower quartile. (3rd quartile).

57, 78, 95

Here the numbers in the set is odd (n=3) so we use this formula.

`(n+1)/2 = (3+ 1) /2 `   By solving the above we get as 2
Medial of upper set of numbers is 78

So the upper quartile is 78

Practice Help Step 4:-

The sample maximum is the number with the largest value in the data set.

Here it is 95

The sample maximum is 95

Practice Help Step 5:-

The sample minimum is the number with the smallest value in the data set.Having problem with 9th class cbse question papers keep reading my upcoming posts, i will try to help you.

Here it is 14

The sample minimum is 14

The 5 Number Summary is

The solved box plot gives the following information

Median – 53
Lower Quartile -24
Upper Quartile - 78
The sample maximum - 95
The sample minimum - 14

Sampling Distribution of Proportion

Introduction to Sampling and Sampling Distributions:

Often in food stalls, the shoppers often taste a small piece of an item and based on their experience with the small piece they decide to buy or not to buy. This is an example of sampling. This is done in many factories as well. Consider a tyre manufacturing factory. The quality inspection engineers pull out a few manufactured tyres and test the tyres thoroughly and in that process these tyres are destroyed.

Note in the previous two cases why was the sampling done? If the shoppers in the food stall had tasted the entire quantity available, there would have been nothing left for sale. Similarly if the quality inspector had tested all the tyres, all the tyres would have been destroyed and there would have been nothing available to sell. So sampling for inspection was unavoidable in these cases.

Consider another case where you want to study about the average income of graduates in a particular region. If we try to get information from all the graduates in the region, the process will be time consuming and would be elaborate. However, if we collect data from a representative group of people, the same can be done quickly and easily

In the above example we deduce information about a bigger set based on a sample. The bigger set is called the population. We infer the data of the population based on the sample data. In the above cases, the total amount of the food in the food stall, total tyres produced by the factory, total graduates in the region are the population. The total tyres produced by

If the above samples are taken repeatedly and the mean of each sample is plotted is plotted against its probability then the distribution obtained is called sampling distribution.

Sampling Distribution of Proportions:

A sampling distribution described above can be partially described by the mean and the standard deviation. If we collect samples repeatedly from a population and calculate the mean, there will be difference in means for different samples. The means will be different. This is because each sample has different sampling elements from the population. This variation is called the standard deviation of the distribution of sample means or simply as standard error of the means. Similarly, the standard deviation of each sample might vary and this variation in the standard deviation is called the standard deviation of the sampling distribution or simply as standard errors.

The aim of taking samples is to estimate the population characteristics like mean and standard deviations. If we take samples repeatedly then the values of mean and standard deviation for each sample will be different. If so, which one will we consider as an accurate representation of the population data?  The standard error tells us how reliable it will be if we predict the population values from the sample statistics. There is an important theorem correlating the characteristics of the sample and the population and this is called the central limit theorem

Central Limit Theorem says the mean of the sampling distribution will be equal to the population mean regardless of the sample size. The sampling distribution of the means will approach normality as the sample size increases. This theorem helps to make inference about the population without knowing much about the population or the distribution of population.Understanding need help with math problems is always challenging for me but thanks to all math help websites to help me out.

This means the sampling distribution means will be the population means and the standard deviation of the population is accurately estimated if we use a large sample size. This is illustrated in the figure below. As we can see form the graph, as the sample size increase the mean is the same but the standard deviation decreases.



The process of evaluating the population data from the sample is called estimation. Central limit theorem is the basis on which the theories of estimations have been propounded. The theory of estimation helps us to find the population values of mean and standard deviation form the sample values.

Statisticians often use a sample to estimate the proportion of occurrences in a population. For example if we want to measure the unemployment rate or the proportion of unemployed people in the population. When we make try to estimate the proportion of a population from the sample we call it as a estimate of the population proportional and the sampling distribution is the sampling distribution of the proportion.

Exercises on Sampling Distribution:

Q:1 A machine is supposed to fill on an average 125 grams of a liquid in a bottle with a standard deviation of 20 grams. A random quality inspection shows a mean of 130. The inspector concludes that the sample is wrong. Is he correct?

Ans: No. The sample data need not accurately represent the mean. So just because the sample value was 130 grams does not mean it is a wrong sample.

Q:2 A sample of patients with tooth diseases wanted to be carried out regarding their brushing and eating habits. The statistician approaches a group of dentists and asks them to submit data. Each dentist takes data from 50 patients and submits the average value to the statistician. The statistician draws inference based on this. Was this a sampling of the patients?

Ans: No. The statistician used the data from the dentists, which was a mean of the 50 patients and not individual patient. So the statistician had plotted a sampling distribution and not a sample value.

Prob 3: What sample size will give you mean value equal to the population mean and without any error?

Ans: The sample size should be the same as population. This means that we need to carry out the data collection for the entire population.

One Variable Statistics

Introduction :

The statistics includes the variables as mean, median, mode, range, variance and standard deviation. Statistics is that one of the important branch of applied mathematics which deals with the scientific analysis of data. In this article we shall discuss the concept of mean, median, mode, range and standard deviations as well as example and find the one variable value in statistics.

Examples – One Variable Statistics:

In this example we will declare the variable as x and find that value with help of example.

Now we will solve the example problems for one variable in statistics.

14, 12, 16,18,52,20.

Now locate the mean, median, mode and range for solve variables in statistics.

Mean:

Here we will declare the mean as `stackrel(-)(x)` variable.

Mean is the sum of values divided by the number of given data in the given numbers.

`stackrel(-)(x)` = Mean = `("Sum of given values")/("Total number of given values")`

The given numbers are,

14, 12, 16,18,52,20.

Mean = `stackrel(-)(x)` = `132/6` .

`stackrel(-)(x)`= 22.

Median:

In this example we will declare median as x.

Median means center values.

The given values are 80, 74,96,32,65.

Now sort the numbers in ascending orders.

32, 65,74,80,96.

So, the x = median = 74.

Mode:

In this example we will declare mode as x.

The given values are

88, 22, 36, 50, 40, 88.

In the given values most repeated value is known as mode.

In the given values 88 is repeated.

x = 88.

Range:

In this example we will declare the variable as x.

The range is defined as difference among smallest value and maximum value.

88, 22, 36, 50, 40, 88.

In this given numbers highest value is 88.

Then smallest value is 22.

So the range is,

= 88 - 22

x = 66.

These are mode, mean, range and median examples in one variable statistics.

Standard Deviation – One Variable Statistics:

Let us we will solve standard deviation example problems for one variable in statistics.

In this example the standard deviation declared as `sigma`

Example 1:

The given values are 100, 58,96,84,60.

Solution:

The given numbers are 100, 58,96,84,60.

Find the approximating mean value for the given data.

Mean (M) = 100 + 58 + 96 + 84 + 60.

= `398/5`

= 79.6.

Find

`sigma`   = `(1555.2)/(5-1)`

`sigma` = 388.8

Formula of the standard deviation =`sqrt((sum_(i=1)^n (x-m)^2) / (n-1))`

The answer for standard deviation is 19.72

That’s all about one variable statistics.

True Mean Statistics

Introduction to true mean statistics:

True mean is the average rate for the given set of data. It is one of the most frequently used terms in the statistics. Mean presents the average for the set of numbers. The set of numbers contains the positive numbers and also the negative numbers. It is also called as the mean statistics. The mean value is calculated by adding the set of numbers and divides that ensuing number by the total number of values in the set of numbers.

Formula Used in True Mean Statistics:

The given set of values is x1, x2, x3 . . . . . . . . xn, the set of values contains n terms.

Mean= `(x1+x2+x3+. . . . . .+x n)/n`

Examples for True Mean Statistics:

Example 1 for true mean statistics:

Determine the net mean for the set of values 4, 5, 13, 17, 14, 18, 19, 21.

Solution:

The given set of values is 4, 5, 13, 17, 14, 18, 19, 21.

The formula for mean = `(x1+x2+x3+. . . . . .+x n)/n`

Step 1: Mean= `(4+5+13+17+14+18+19+21)/8`

Step 2: Mean= `111/8`

Step 3: Mean=13.875

The mean value for the set of values 4, 5, 13, 17, 14, 18, 19, 21 is 13.875. 

Example 2 for true mean statistics:

Compute the net mean for the set of values 1, 3, 4, 5, 7, 9, 4, 6, 8.

Solution:

The given set of values is 1, 3, 4, 5, 7, 9, 4, 6, 8.

The formula for mean = `(x1+x2+x3+. . . . . .+x n)/n`

Step 1: Mean = `(1+3+4+5+7+9+4+6+8)/9`

Step 2: Mean = `47/9`

Step 3: Mean = 5.2

The mean value for the set of values 1, 3, 4, 5, 7, 9, 4, 6, 8 is 5.2. 

Example 3 for true mean statistics:

Calculate the net mean for the set of values 10, 14, 17, 21, 18, 19, 22, 20, 15, 13.

Solution:

The given set of values is 10, 14, 17, 21, 18, 19, 22, 20, 15, 13.

The formula for mean = `(x1+x2+x3+. . . . . .+x n)/n`

Step 1: Mean= `(10+14+17+21+18+19+22+20+15+13)/10`

Step 2: Mean= `169/10`

Step 3: Mean=16.9

The mean value for the set of values 10, 14, 17, 21, 18, 19, 22, 20, 15, 13 is 16.9.

Understanding free math problem solver online is always challenging for me but thanks to all math help websites to help me out.

Practice Problem for True Mean Statistics:

Compute the true mean for the values 12, 18, 25, 14, 27, 16, 15, 28.
Answer: 19.375.

Calculate the true mean for the data 5, 14, 16, 9, 13, 10, 17, 19, 14, 21.
Answer: 13.8.

Study Online Harmonic Progression Help

Introduction to study online harmonic progression help:

A harmonic progression is a series of numbers that are all obtained by taking the inverse of the arithmetic progression. Hence it is necessary to find the arithmetic progression for calculating the harmonic progression. Let us solve some of the example explaining arithmetic progression.When student have doubt in study, they come to online for their help. The tutors will help the students to clarify the doubts in study through online white board.

Formula for Study Online Harmonic Progression Help:

The general form of the harmonic progression is given by,

`a, a/(1+d), a/(1+2d),a/(1+3d), a/(1+4d), ....`

These terms are all the inverse of the arithmetic progression.

nth term of harmonic progression  is given  by,

`T_n =1/( a + (n - 1)d)`

Where

a is the first term

d is the common difference between the successive numbers.

The harmonic mean of two numbers a and b are,

` (2ab)/(a+b)`

The harmonic mean of three numbers a, b and c are,

`(3abc)/(ab+bc+ac)`         

Similarly, the harmonic mean of two numbers a and b are,

`n/(1/a + 1/b + 1/c +. . . 1/N)`

Example Problems for Study Online Harmonic Progression Help:

Example 1:

Find the harmonic progression up to fourth term with difference is 4 and the series starts from 3.

Solution:

Step 1: The given terms are

a = 3;

d = 4;

Step 2:  Find Harmonic progression.

Step 3: Formula for harmonic progression is given by,

`1/a, 1/(a+d), 1/(a+2d),1/(a+3d), 1/(a+4d),...`

Step 4: on applying the all values we get,

= `1/3, 1/(3 + 4), 1/(3+2xx4), 1/(3+3xx4)` ,…

= `1/3, 1/7, 1/11, 1/15` ,…

= 0.333, 0.143, 0.091, 0.067,...

This is the harmonic progression series.

Example 2:

Find the harmonic progression up to fourth term with difference is 4 and the series starts from 2.

Solution:

Step 1: The given terms are

a = 2;

d = 4;

Step 2:  Find Harmonic progression.

Step 3: Formula for harmonic progression is given by,

`1/a, 1/(a+d), 1/(a+2d),1/(a+3d), 1/(a+4d),...`

Step 4: on applying the all values we get,

=` 1/2, 1/(2 + 4), 1/(2+2xx4), 1/(2+3xx4),...`

= `1/2, 1/6, 1/10, 1/14,...`

= 0.5, 0.167, 0.1, 0.017,...

This is the harmonic progression series.

Example 3:

Find the harmonic progression up to fourth term with difference is 4 and the series starts from 1.

Solution:

Step 1: The given terms are

a = 4;

d = 2;

Step 2:  Find Harmonic progression.

Step 3: Formula for harmonic progression is given by,

`1/a, 1/(a+d), 1/(a+2d),1/(a+3d), 1/(a+4d),...`

Step 4: on applying the all values we get,

= `1/1, 1/(1 + 4), 1/(1+2xx4), 1/(1+3xx4),...`

= `1/1, 1/5, 1/9, 1/13,...`

= 1, 0.2, 0.111, 0.077...

This is the harmonic progression series.

Adding and Subtracting Real Numbers

Introduction :

Let us discuss about the adding and subtracting real numbers. The real number is declare the positive number, negative number, whole number, decimal number, large number, small number. The real number do not used the imaginary number. The example of the real number is 8, -4, 2.36, 789 and etc. The general notation of the real number is R. Next we see the adding and subtracting of the real numbers.

Adding and Subtracting Real Number

Adding real numbers

The adding the real number is used the two operands and one operator. The operands are positive or negative number. The operator of adding real number is ‘+’. The example is 4.83 +1 = 5.83.

The 4.83 and 1 is the operands of adding real number.
The + is the operator of adding real number.
Subtracting real numbers

The subtracting the real number is used the two operands and one operator. The operands are positive or decimal number. The operator of subtracting real number is ‘-’. The example is 3.14 - 2.

The 3.14 and 2 is the operands of subtracting real number.
The - is the operator of subtracting real number.


Example of Adding and Subtracting Real Numbers

The example of the adding real number is

Problem 1:

Find the 88.3 + (-41)

Solution

= 88 .3 + (-41)

= 47.3

Answer of the adding real number is 47.3

Problem 2:

Find the (-52) + (-38)

Solution

= (-52) + (-38)

= -90

Answer of the adding real number is -90

Example problem of subtracting real number

Problem 1:

Find the (-96) - 197

Solution

= (-96) - 197

= -293

Answer of the subtracting real number is 101

Problem 2:

Find the 124 – (-87)

Solution

= 124 – (-87)

= 211

Answer of the subtracting real number is 211

Volume Cubic Feet

Introduction to volume cubic feet

Volume is how much three-dimensional space a substance (solid, liquid, gas, or plasma) or shape occupies or contains, often quantified numerically using the units, the cubic meter or cubic feet. The volume of a container is generally understood to be the capacity of the container, i. e. the amount of fluid (gas or liquid) that the container could hold, rather than the amount of space the container itself displaces. (Source: From Wikipedia).

Formulas to Find the Volume of Shapes in Cubic Feet

The volume of a cube = a3 cubic units
Here, a is the side of the cube

The volume of a sphere = `4/3 pi r^3` cubic units
Here, r is the radius of the sphere

The volume of a cylinder = `pi r^2 h` cubic units
Here r and h are radius and height of the cylinder respectively

The volume of a cone = `1/3 pi r^2 h` cubic units
Here r and h are radius and height of the cone respectively.

I like to share this hard math problems for college with you all through my article.

Example Problems to Find the Volume of Shapes in Cubic Feet

Example 1

Find the volume on a cube with side length of 5 feet.

Solution

Volume of a cube = a3 cubic units

= 53

= 5 * 5 * 5

= 125

So the volume of the cube is 125 cubic feet

Example 2

The radius of a sphere is 3 feet. Find it's volume.

Solution

The volume of a sphere = `4/3 pi r^3` cubic units

= `4/3` * `pi`* `3^3`

= `4/3` * 3.14 * 3 * 3 * 3

= 4 * 3.14 * 3 * 3

= 113.04

The volume of the sphere is 113.04 cubic feet.

Example 3

If perpendicular height and the base radius of a cylinder are 10 feet and 6 feet respectively, find it's volume.

Solution

The volume of a cylinder = `pi r^2 h`

= 3.14 * 6 * 6 * 10

= 1130.4

The volume of a cylinder is 1130.4 cubic feet.

Example 4

If perpendicular height and the base radius of a cone are 10 feet and 6 feet respectively, find it's volume in cubic feet.

Solution

The volume of a cone = `1/3 pi r^2 h` cubic units

= `1/3` * 3.14 * 6 * 6 * 10

= 3.14 * 2 * 6 * 10

= 376.8

So, the volume of the cone is 376.8 cubic feet.

Negative Prime Numbers

Introduction :

Let us study about the negative prime numbers. The prime numbers are commonly termed to be as the numbers which gets divisible by ‘1’ multiplication table and its own multiplication table alone.
Therefore the prime numbers are said to have only two factors. The negative prime numbers are said to be as the normal prime numbers along with the negative sign.
Some of the examples for negative prime numbers are discussed in detail as below.


Negative Prime Numbers:

Negative prime numbers example 1:

From the following series of numbers, find the numbers that belongs to negative prime numbers ones?
1, -2, 7, -5, -7,-9, -21


Solution:

The number in the first position is ‘1’ which is not a prime number since the numbers greater than ‘1’ is considerable in the prime numbers family and also it is not with the negative sign for the negative prime numbers category.
The number in the second position is ‘-2’ which is said to be an even prime number. Since it is with the negative sign it is a negative prime number.
The number in the third position is ‘7’ which is said to be a prime number but it is not a negative prime number since there is no negative sign in it.
The number in the fourth position is ‘-5’ which is said to be a prime number. Since it is with the negative sign it is a negative prime number.
The number in the fifth position is ‘-7’ which is said to be a prime number. Since it is with the negative sign it is a negative prime number.
The number in the sixth position is ‘-9’ which is said to be a prime number. Since it is with the negative sign it is a negative prime number.
The number in the seventh position is ‘-21’ which is said to be a non prime number. Since it is with the negative sign it is a negative non prime number.


Negative prime numbers example 2:

From the following series of numbers, find the numbers that belongs to negative prime numbers ones?
2, -3, 5, -9, 13, -17, -27

Solution:

The number in the first position is ‘2’ which is said to be an even prime number but it is not a negative prime number since there is no negative sign in it.
The number in the second position is ‘-3’ which is said to be a prime number. Since it is with the negative sign it is a negative prime number.
The number in the third position is ‘5’ which is said to be a prime number but it is not a negative prime number since there is no negative sign in it.
The number in the fourth position is ‘-9’ which is said to be a non prime number. Since it is with the negative sign it is a negative non prime number.
The number in the fifth position is ‘13’ which is said to be a prime number but it is not a negative prime number since there is no negative sign in it.
The number in the sixth position is ‘-17’ which is said to be a prime number. Since it is with the negative sign it is a negative prime number.
The number in the seventh position is ‘-27’ which is said to be a prime number. Since it is with the negative sign it is a negative prime number.


Negative prime numbers exercises:

From the following series of numbers find the numbers that belongs to negative prime numbers ones?
-17, 19, -29, 31, 41, -43, 58. (Answer: -17, -29 and -43)

From the following series of numbers find the numbers that belongs to negative prime numbers ones?
-2, 5, -101, 102, -107, 109, 131. (Answer: -2, -101 and -107)

Polynomial and Rational Inequalities

Introduction to polynomial and rational inequalities:

Polynomial and rational inequalities mean we are going to learn about both inequalities separately. Polynomials are nothing but the sum of more than one number of monomials. Polynomial inequalities mean it is having a simple difference which is less than and greater than symbol. Likewise rational inequalities mean we have to solve the inequalities which are having the rational form. It will be like `(p(x)) / (Q(x))` . We will see some example problems for solving polynomial and rational inequalities.

Example Problems for Polynomial Inequalities:

Solve the polynomial function inequalities 5x2 - 15x + 10 = 0

Solution:

The given polynomial inequality is 5x2 - 15x + 10 = 0

To solve this inequality first we have to factor the given inequality.      

So 5x2 - 15x + 10 = 0

5x2 - 5x - 10x + 10 = 0

5x (x - 1) - 10 (x - 1) = 0

(5x - 10) (x - 1) = 0

5x - 10 = 0 and x - 1 = 0

From this x = 2 and x = 1

So the solution of the inequality lies between 1 and 2.

From the we can learn how to solve the polynomial inequalities. Here we solved the quadratic inequality.

My forthcoming post is on Solving Decimal Equations, Linear Combinations will give you more understanding about Algebra.

Example Problems for Rational Inequalities:

Find the solution of the x from the rational inequalities `(3x + 6) / (2x + 9)` = 1

Solution:

Given inequality is `(3x + 6) / (2x + 9)` = 1

If we want to solve we have to multiply by 2x + 9 on both sides

So we get,

3x + 6 = 2x + 9

Now we have to add -6 on both sides of the function so we get

3x + 6 - 6 = 2x + 9 - 6

From this 3x = 2x + 3

Add –2x on both sides

3x – 2x = 2x + 3 – 2x

x = 3         

So x is always greater than 3.

These are some of the example problems for polynomial and rational inequalities. From the above we can learn how to solve the polynomial and rational inequalities.

Properties of Rational Exponents

Introduction for exponents:

The exponents are which integer is placed in the power of base numbers. It can be easily represent as, that is a small number to the right side and above of base number. It is called as exponents. In this exponents have many types. Rational exponents are one of the types of exponents. These rational exponents have some of important rules and laws. In this rules and laws are called as properties. In article we are going to explain about the important properties of exponents.

Explanatory for Rational Exponents:

Rational exponents are nothing but, it is one type of exponents. These exponents are in fraction form. Otherwise the power values raised to fraction. It is known as rational exponents.

Types of rational exponents properties:

There are seven important rules or properties are there, but here we have to given five important properties only. There are given below,

Properties of multiplying powers,
Power of quotient properties,
Power of product properties,
Power of power properties, and
Power of zero properties.
These all are the important properties of the rational exponents.

Between, if you have problem on these topics Descriptive Statistic, please browse expert math related websites for more help on Relation and Function.

Properties Explanations and Examples:

1. Properties of multiplying powers:

In this term the base numbers have same values and exponents only different. In this multiplying term, we will add the power values and take the base value as same.

It can be denoted as, xm * xn = xm+n.

2. Power of quotient rules:

In this term the base numbers have two different numbers and only one power for both base numbers. Then we will separate and write it as, (x/y) m = xm / ym.

3. Power of product properties:

This term will be having two different bases, but only one power value. And it can be denoted as, (xy) m = xm * ym.

4. Power of power properties:

In this power of power properties has only one base number, and having the one power value. Then whole value has one more power value. It is denoted as, (xm) n = xmn.

5. Power of zero properties:

This power of zero properties has any one base number with the power of zero values only. It can be getting constant answer is 1. It can be denoted as x0 = 1.

Example Problems in Properties of Rational Exponents:

1. Simplify: `(9)^ (2/3) * (9)^ (3/2)`

Solution:

Given: `(9)^ (2/3) * (9)^ (3/2)`

Use the property of multiplying powers, and write the expression,

= `(9)^ (2/3) * (9)^ (3/2)` = `9^ ((2/3) + (3/2))`

The LCD of exponents is `6` . So,

=` (9)^ ((4/6) + (9/6))`

= `(9)^ (13/6)`

Simplify the radicand, and we get

= `(3^2)^ (13/6)` .

Here we use the power of power property, and simplify, and we get

= `((3)^ (2)) ^ (13/6)`

= `(3) ^ (13/3)` .

Answer is `(3) ^ (13/3)` .



2. Factor the number into the square of another number and then simplify `196^-(1/2)`

Solution:

Given:` 196^-(1/2)`

Take the square root value,

=` (14^2) ^ - (1/2)`

Here we use the power of power property, and simplify

= `(14 ^2)^ (-(1/2))`

= `(14) ^ (-1)`

= `(14/1) ^ -1`

Here we use the power of quotient property,

= `(1/14) ^1`

= `1/14.`

Answer is `1/14` .

Those above explanations and examples problems makes this properties of exponents will clear.

Quotient Rule Algebra

Introduction to quotient rule in algebra:

The integer part is divided into the two integer is called as the quotient. The quotient is the theoretical branches of mathematics. The quotient is used to the sets, spaces, or algebraic structures of these elements. These elements are the some equivalence relation. Consequence of the division is the quotient. The quotient is classified into number of the times the divisor divide into the dividend.

Quotient Rule Algebra:

The quotient rule algebra is the technique of the decision the derivative purpose that is the quotient.

Production rule:

The quotient rule algebra is a process of sentence the derivative of a function which is the quotient of two further functions for which derivatives are present.

The quotient rule algebra points cannot be in the where the similar to the numerator or denominator is not differentiable

Multiplying two power of the same base is the production rule, we can insert the element.

Exercise 1:

`A^(a).A^(b)=A^(a+b)`

`11^(2).11^(3)=11.11.11.11.11`

`11^(2+3)=11^(5)`

Power rule:

The power rule means power to power, simply multiply the exponents.

Example:

`x^(ab)=x^(ab)`

`(5^(2))^(3)=5^(2.3)=5^(6)` 

Quotient rule:

The Quotient rule algebra can divide the two powers with the similar base by subtracting the exponents.

Example:

`x^(a)-:x^(b)=x^(a+b)`

x=0

`1^(5)-:1^(2)=(1.1.1.1.1)/(1.1)`

`1^(5-2)=1^(3)`

Zero rule:

Zero rules some non zero number is move up to the power of zero equivalent to 1.

Example:

x°=1

x`!=0`

Negative exponents:

This is the last rule of the quotient rule. Some the non zero of the element is move up to the negative power equal. Reciprocal is moved to the opposite positive power.

Example :

`11^(-2)=(1)/(11^(2))`

` = (1)/(121)`

Algebra is widely used in day to day activities watch out for my forthcoming posts on Addition Property of Equality Definition and How do you Simplify Expressions. I am sure they will be helpful.

Example Problems:

Problem 1:

`((x+3)-(x-2))/((x+3)^(2))`

Solution:

` = (x+3-x+2)/(x^(2)+6x+9)`

` = (5)/(x^(2)+6x+9)`

Problem 2:

`x^(2)-5x+6-:x-2`

Solution:

`= x^(2)-2x-3x+6`

`= x(x-2)-3(x-2)`

=  `((x-3)(x-2))/(x-2)`

`= (x-3)`

Differentiation of Function Using the Quotient Rule:

The following formula is the differential of function using the quotient rule,

D = `(f(x))/(g(x))=(g(x)f'(x)-f(x)g'(x))/{g(x)}^(2)`

Quotient rule algebra is represented by the formal rule for differentiating problems; the information is classified by another. In the calculus, the finding the value of the function is the quotient is the quotient rule algebra.

Solving Quadratic Equations

Introduction :

An equation in the form of ax2 + bx + c = 0, where a, b, c ? R and a ?0 is called a quadratic equation. It is an equation with degree 2. For example, 5x2 + 6x + 7 = 0, 9x2 – 4 = 0, 2x2 – 3x =0 are quadratic equations. If p(x) = 0 is a quadratic equations, then the zero of the polynomial  p(x) are called the roots of equation p(x) = 0. Find the roots of a quadratic equation is called as solve the quadratic equations.

Solution of Quadratic Equations Example

To solve quadratic equation by using factorization method using at what time the quadratic equation is expressible as the product of two linear equations. Quadratic equation of the form ax2+bx+c=0 here a,b,c,d are called constants can be used to reduced the form of a quadratic equations.

Example 1:

Solving quadratic equations X2+9x-10

Solution:

Step 1:

List out the factors

1x-10, -1x10

Step 2:

Find the factor the sum value is 9

-1+10=9

Step 3:

Check using distributive property

(X-1)(x+10)=x2+10x-x-10

X2+9x-10

Step 4: back to original equation

X2+9x-10 we get two values for x.

X-1=0=>x=1

x+10=0=>x=-10

Answer:  x=1, x=-10

Example 2:

Solving the quadratic equations 2x2 + 3x – 5 = 0

Solution: 2x2 + 3x – 5 = 2x2 + 5x – 2x – 5

= x (2x + 5) – 1(2x + 5)

= (x – 1) (2x + 5)

Since 2x2 + 3x – 5 = 0 we get (x – 1) (2x + 5) = 0

=>x – 1 = 0 or 2x + 5 = 0

=>x = 1, –5/2

The solution set = 1, -5/2

Example 3:

Solving the quadratic equations 64x2 – 36 = 0

Solution: 64x2 – 36 = 0 or (8x + 6) (8x – 6) = 0

Or x = –6/8 = –3/4

Or x = 6/8 = 3/4

The solution set =-3/4, 3 /4

Example 4:

Solving the quadratic equations 3x2 – 4x = 0

Solution: 3x2 – 4x = x (3x – 4)

Since 3x2 – 4x = 0, we get x (3x – 4) = 0

Or x = 0

Or 3 x –4 = 0

=> 3x = 4

=> x = 4/3

Quadratic Equation Practice Problem

1) Solving the quadratic equations x2+4x-5=0

Answer:

X=1, x=-5

2) Solving the quadratic equations x2-5x-6

Answer:

x=-1, x=6

Solving a Literal Equation

Introduction for solving a literal equation:
The literal equations are nothing but the equations where it can be called as the formulas for finding the recipes of the given equation variables. From a given formula, some other variables can be solved can be called as the literal equations. This can be applicable for many of the algebraic and geometric formulas. Now we are going to see about solving a literal equation.

About Solving a Literal Equation:

Now we are going to see solving a literal equations. The solving literal equations can be found for the thing which is additional information in that equation.

For example for solving a literal equation:

The area of triangle can be given as `1/2` * base * height

Generally this formula is used to find the area of the triangle. But from this we can find the base or the height,

Base = `(2 * area)/(height)`

Height = `(2 * area)/(base)`

The above two equations can be called as the literal equations.

Algebra is widely used in day to day activities watch out for my forthcoming posts on math algebra solver and factoring algebraic expressions. I am sure they will be helpful.

Problems for Solving a Literal Equation:

Example 1:

Find the base length of the given triangle where the area is about 15 cm2 and the height is about 5 cm.

Solution:

Now we can use the literal equation for the triangle as follows,

Base = `(2 * area)/(height)`

Base = `(2 * 15)/5`

Base = `30/5`

Base = 6 cm.

Example 2:

Solve the perimeter of the rectangle P = 2L +2W for W.

Solution:

Let us take the given equation

P = 2L +2W

Now take the L term to the left hand side we get as,

P – 2L = 2W

Now divide by 2 on either side of equal sign

` (P - (2 * L))/2 ` = W

Example 3:

Solve for the literal equation for the area of parallelogram where the area of parallelogram is 20 cm2 and the base of the parallelogram is about 10 cm. Determine the height of parallelogram.

Solution:

The literal equation or the formula for the parallelogram can be given as follows,

Area = Base * height

Height = `(area)/ (base)`

Height = `20/10`

Height = 2 cm.