Statistics Practice Problems

Introduction to Statistics Practice Problems Tutoring

The tutoring is a fresh way for the student to collect the information from the online. Generally, the tutoring contains chat, teleconferencing, web conferencing and another easy way of the information from back and front. In statistics practice problems tutoring, the tutoring giving the information about mean, median, mode, and range with some of the practice problems.


Example Problems - Statistics Practice Problems Tutoring:


Example 1:

Find out the mean of the following given sequence in statistics?

3,8,17,26,36,39,40.

Solution:

The given numbers are 3,8,17,26,36,39,40.

Mean:

Mean is the average of the given number. First find the total value of the given numbers.

Sum of the given numbers are = 3+8+17+26+36+39+40.

= 169.

The total value is divided by 7 (7 is the total given numbers) = `(169)/(7)`

= 24.1.

Example 2:

What is the median for the following sequence. 12,18,33,37,39.

Solution:

Center element of the given series is said to be median.

The number series is 12,18,33,37,39.

The middle element of the above series is 33.

So the value of median is 33.

Example 3:

Solve the range of the following in basic statistics?

11, 17, 19, 24, 28.

Solution:

The given numbers are 11,17,19, 24, 28.

Minus the lowest value from the peak value of the series.

Range = 28 - 11

= 17.

These examples are the statistics problems given by tutoring.

I have recently faced lot of problem while learning Free Bar Graphs, But thank to online resources of math which helped me to learn myself easily on net.

Practice Problems - statistics practice problems tutoring:


Practice Problem 1

Find the mean for the below number sequence.

12,17,19,26.

Answer

=18.5

Practice Problem 2

Find the range for the below number sequence.

15,17,21,28.

Answer:

= 13.

Practice problem 3

What is the mode for the following numbers? 13,18,13,19.

Answer:

13 is the mode.

Triangles in Math

Introduction of Triangles in Math:

Triangles are closed loop that consists of three sides and three angles. The total interior angle of triangles is 180°. The triangles consist of vertex by joining the two sides of the triangles. The amplitude of the triangle is the distance between vertexes of triangles in to the opposite side of that same vertex of triangles. The length of the altitude is the height of the triangles .


Classification of Triangles in Math:


Classification of Triangles:

There are different types of triangles; they are acute angle triangles, obtuse angle triangles, right angle triangles, Isosceles triangles, equilateral triangles and scalene triangles.

Classification of triangles is based on sides and angles.

Classification based on the sides of the triangles:

Isosceles Triangles – Two sides are equal in this triangle.

Equilateral Triangles – All the three sides of the triangle is equal.

Scalene Triangles – All the three sides are different in length of the triangles.

Classification based on the angle of the triangles:

Acute angle Triangles – The three angles of the triangle is less than 90°.

Obtuse angle Triangles – One of the angles in triangle is greater than 90° (less than 180°).

Right angle Triangles – One of the sides of the triangle is 90°.


Formula and Example Problems – Triangles in Math:


Basic Formula – Triangles in Math:

Area of the triangles in math – ½ b(h)

Where,

b = Base length of the triangles.

h = Height length of the triangles.

Perimeter of the triangle – x + y + z

Where, x, y, and z are the sides of the triangles.

Example Problems – Triangles in Math:

Example 1:

Find the area and perimeter of the equilateral triangle, whose base is 5 cm and height, is 7 cm.

Solution:

Given: base (b) = 5 cm

Height (h) = 6 cm

Formula to find area and perimeter:

Area = 0.5 (b) (h)

Perimeter = a + b + c, where a, b, c are the sides of the equilateral triangle,

Here equilateral triangle have all three sides are equal. Therefore perimeter = 3 a

Area = 0.5 (5 (6)) = 15 cm^2

Perimeter = 3 (5) = 15 cm

Example 2:

Find the perimeter of the triangle whose sides are 4cm, 6cm and 7cm.

Solution:

Given: The sides of the triangle are, a = 4cm

b = 6cm and c = 7cm.

Formula:  Perimeter = a + b + c = 4 + 6 + 7 = 17cm.

Answer: perimeter of the triangle  = 17cm.

Grade 8 Math Inequalities

Introduction to grade 8 math inequalities:
In mathematics, an inequalities is a statement about the relative size or order of two objects or about whether they are the same or not.

The notation a < b means that a is less than b.
The notation a > b means that a is greater than b.
The notation a ? b means that a is not equal to b.
In this article we shall discuss about math grade 8 inequalities. (Source: Wikipedia)

Grade 8 math inequalities example problem


Here we are going to discuss some 8th grade inequalities problems with detailed solutions.

Example:

Solving the inequalities 5x – 12 > 4x +12

Solution:

The given inequalities is

5x – 12 > 4x +12

Adding the 12 on both side of equation

5x -12+12> 4x +12+12

5x>4x+24

Subtract 4x on both side of the inequality equation

x>24

Example:

Solving the following inequality equation -5< 4(x+2)-5<19 br="">
Solution:

The given inequality is

-5< 4(x+2)-5<19 br="">
Multiplying the factor for given inequality equation

-5<4x br="">
-5<4x br="">
Subtracting three on both sides of the inequality equation

-5-3<4x br="">
-8<4x br="">
Divide by 4 for all terms in inequality

-2
Conclusion:

The solution includes all real number value the interval is (-2, 4)

Example:

Solving the following inequality equation -4< 2(x+6)-2<12 br="">
Solution:

Given inequality is

-4< 2(x+6)-2<12 br="">
Multiplying factor values for given inequality equation

-4<2x br="">
-4<2x br="">
Subtracting 10 on both sides of equation

-4-10<2x br="">
-14<2x br="">
Divide by 2 for all terms in equation

-7
Conclusion:

The solution includes all real number value the interval is (-7, 1)

Example:

Solving the inequality 6x – 4 > 3x +14

Solution:

The given inequality is

6x – 4 > 3x +14

Adding the value four on both side of equation

6x -4+4> 3x +14+4

6x>3x+18

Subtract value 3x on both side of the equation

3x>18

Simplifying the value x

x>18/3

x>6

Conclusion:

The solution includes all real number value the interval is (6, infinity)

Example:

Solving the inequality 8x – 2 > 3x +13

Solution:

The given inequality is

8x – 2 > 3x +13

Adding the value two on both side of equation

8x -2+2> 3x +13+2

8x>3x+15

Subtract value 3x on both side of the equation

5x>15

Simplifying the value x

x>15/5

x>3

Conclusion:

The solution includes all real number value the interval is (3, infinity)

I have recently faced lot of problem while learning Properties of Real Numbers Chart, But thank to online resources of math which helped me to learn myself easily on net.

Grade 8 math inequalities practice problem


Problem:

Solving the inequality -4< 4(x+2)-4<16 br="">
Answer:

The solution includes the interval is (-2, 3)

Problem:

Solving the inequality equation -2< 4(x+6)-3<10 br="">
Answer:

The solution includes all real number value the interval is (-6, -2.75)

Multiplying Integers Practice

Introduction to multiplying integers practice:

Multiplying integers practice  is one of the important topics in mathematics. A multiplying is of the mathematical operation of scaling a number by another number. It is the basic operations in elementary arithmetic. Integer is the set of numbers in which positive whole numbers, negative whole numbers and zero. It has a complete unit or entity. But it has a no fractional part.

Positive whole number = {1, 2, 3, 4, …..}

Negative whole number = {-1, -2, -3, -4, …..}

Example for integers:

25, 1897, -665, 0, etc.,


Rules for multiplying integers practice:


Different rules for multiplying integers practice are,

Rule 1:

Positive number × Positive number = Positive number

Rule 2:

Positive number × - Negative number = - Negative number

Rule 3:

- Negative number × Positive number = - Negative number

Rule 4:

- Negative number × - Negative number = Positive number

Example problems for rules for multiplying integers practice:

Using rule 1:

12 × 12 = 144

Using rule 2:

15 × - 8 = - 120

Using rule 3:

-10 × 5 = - 50

Using rule 4:

- 14 × - 20 = 280


Example problems for multiplying integers practice:


Example 1:

Multiply the two integer numbers

12 × 3

Solution:

Given

12 × 3

Both are two positive integers so, the result is also a positive numbers

Here we add 12 into 3 times, then we get

12 + 12 + 12

36

It is the simplest method of multiplying integers.

Example 2:

15 × 10

Solution :

Given

15 × - 10

In the given integer numbers, positive and negative numbers so result of this given integer is also a negative numbers.

If we multiply by 10, we have add to zero to the result

- 150

Solution to the given two integer is - 150.

Example 3:

- 8 × - 7

Solution:

Given

- 8 × - 7

Both are negative numbers so the result is also a negative numbers

If we multiply the - 8 by -7 we get

56

Solution to the given integers is 56.

Having problem with math online tutoring keep reading my upcoming posts, i will try to help you.

Practice problems for multiplying integers:


Some practice problems for multiplying integers are

1). 15 × 20

Solution = 300

2). 8 × - 30

Solution = - 240

3). 2 × 3

Solution = 6

4). - 6 × 8

Solution = 48

Geometric Proofs Practice

Introduction for geometric proofs practice:

The geometric proofs practices are generally involved in solving the proof for some problems that involving the statements from different theorems that are earlierly solved. we have to analyze the question first and then we have to choose what kind of statements are used for the problems given. After those steps are taken, we have to match it for the statements with the problems. The proof represents the way that how to we prove the given problem or some postulates.

(Source from Wikipedia)

I like to share this Calculate Area of Polygon with you all through my article.

Examples to explain "geometric proofs practice"


We have to prove the straight line's equation is 3x + 2y = 7 and which is passing through the point (1,2) and which making the intercepts on the axes of the co-ordinate which are in the ratio 2 : 3.
Geometry proof:

The intercept form is    `x/a` + `y/b`  = 1   --------- (1)

The intercepts are in the ratio 2 : 3 ? a = 2k, b = 3k.

(1) becomes `x/(2k)` + `y/(3k)` = 1

i.e. 3x + 2y = 6k

We know that the point (1, 2) lies on the straight line given above, 3 + 4 = 6k

i.e. 6k = 7

Hence the required straight line's equation is 3x + 2y = 7

Thus we have proved the proof for finding the straight line's equation.

Practice:

In this problem as we are seen that it have specific points to prove the problems such that it having ratio on the concern and the point which the line passing.

Problems to explain "geometric proofs practice"


We have to prove that the point's co-ordinates are (8, 9) and (- 42, - 41) and given that it is reside on the straight line y = x + 1 which are at a distance of 5 units from the straight line 4x - 3y + 20 = 0
Geometry proof:

Let (x1, y1) be a point on y = x + 1

? y1 = x1 + 1 … (1)

The length of the perpendicular from (x1, y1) to the straight line

4x - 3y + 20 = 0 is `|(4x1 - 3y1 + 20)/sqrt(4^2 + (-3)^2)|`   = `+-`    `((4x1 - 3y1 + 20)/5)`

But the length of the perpendicular is given as 5.

?  `+-`    `((4x1 - 3y1 + 20)/5)`  = 5

4x1- 3y1 + 20 = `+-`  25

Considering the positive sign, 4x1- 3y1 + 20 = `+` 25

?                                   4x1 - 3y1 = 5 … (2)

Considering the positive sign, 4x1- 3y1 + 20 = `-` 25

?                                   4x1 - 3y1 =`-` 45 … (3)

Solving (1) and (2), we get x1 = 8, y1 = 9

Solving (1) and (3), we get x1 = - 42, y1 = - 41.

? The required points co-ordinates are (8, 9) and (- 42, - 41).

Thus we have proved the proof with the given points.

Practice:

In this problem as we are seen that it have specific points to prove the problems such that it having the perpendicular distance from the two straight lines.

Practice problems on geometric proof:

We have to prove the straight line's equation is x - `sqrt(3)` y + 12 = 0, if the perpendicular comes from the origin, which makes an 120° angle with x-axis and the distance from perpendicular comes from the origin is 6 units.
We have to show that the straight lines 132x +13 y - 9 = 0 and 132x + 13y - 10 = 0 are parallel.

Division Practice problems

Introduction to division practice problems:

In mathematics, especially in elementary arithmetic, division (÷) is the arithmetic operation that is the inverse of multiplication.

The following division methods are all based on the form Q = N / D where

• Q = Quotient

• N = Numerator (dividend)

• D = Denominator (divisor).

Specifically, if c times b equals a, written:

c x b = a

Where b is not zero, then a divided by b equals c, written:

a / b = c

In the above expression, a is called the dividend, b the divisor and c the quotient. (Source: Wikipedia)

I like to share this Difference Quotient Examples with you all through my article.

Example problems on division practice


Ex:1  Jack built a tower of blocks forty-five inches high. Each block in the tower is five inches tall. How many blocks were used to build the tower?

Sol: Jack built a tower of blocks forty-five inches high.

Each block in the tower is five inches tall.

So, total blocks = 45 / 5

Therefore total blocks used to build the tower = 9 blocks

Ex:2 The school's Internet connection transferred thirty six megabytes of data in six seconds. How many megabytes can it transfer in just one second?

Sol:

The school's Internet connection transferred thirty six megabytes Data in six seconds.

So, therefore total megabytes can it transfer in just one second

= 36 / 6 = 6 megabytes.

Ex:3 There are eight soft drink machines in the university. They hold sixty - four cases of soda altogether. How many cases does each machine hold?

Sol:

There are eight soft drink machines in the university.

They hold sixty - four cases of soda altogether.

Total cases of soda machine hold = 64 / 8

= 8 Cases of soda

Ex:4 Zachary used five thousand, five hundred forty-four chips to make a big batch of giant chocolate chip cookies. Each cookie got about eighteen chips. How many cookies did Zachary make?

Sol:

Zachary used five thousand, five hundred forty-four chips to make a big batch of giant chocolate chip cookies

Each cookie got about eighteen chips

So, total cookies Zachary make = 5544 / 18

= 308 Cookies

Ex:5 Twenty-five busses brought a total of seven hundred fifty passengers to the city. About how many passengers were on each bus?

Sol:

Twenty-five busses brought a total of seven hundred fifty passengers to the city.

So, total passengers were on each bus = 750 / 25

= 30 Passengers 

Understanding solve the system of linear equations by graphing is always challenging for me but thanks to all math help websites to help me out.

Division practice problems:


Ex:1 There are thirty apartments in that building. The building has five stories. How many apartments does each story of the building have?

Ans: 6 apartments

Ex:2 Jenna will mail out seven copies of her resume on special paper. She needs twenty-eight sheets of the paper. How many pages long is her resume?

Ans: 4 pages

Ex:3 Jeremy has twenty-four balloons. He wants to give each of his four friends an equal number. How many balloons should each friend be given?

Ans: 6 balloons

Practice Act Question

Act Test preparation:

Act test is one of test which is used to get admissions in mid east countries. Depending on the test marks students get admissions in mid east colleges. Many students around the globe are writing this test hence this test mark is now accepted in US countries also. Act test is some what different from sat test. Act test consists of science reasoning questions but sat test does not. The grammatical talent of the student is checked in the act test while it is not in the sat test. Both the sat and act test are skill oriented test.  The act test is conducted for nearly 4 hrs and optional of 30 minutes.

Please express your views of this topic Significant Figures Practice by commenting on blog.

Act practice problems:

Act practice question 1:

Anand's new job comes with a salary increase of 8%. If he currently makes $74,000 per year, which of the following is the amount he will earn per year at his new job?

A. $79,000
B. $72,000
C. $79,200
D. $75,550
E. $79,920

Solution:

Increase percent in salary = 8 %

8% of $74,000                     = (8 x 74,000) / 100

= 8 x 740

= $5920

The total amount her earned is = $74,000 + $5920

= $79,920

The amount he earned is $79,920

Answer: E



Act practice question 2:

If Suresh biked 10 miles in 4 hours and arun biked three times as much in half the time, what was arun's average rate of speed?



A. 10 mph
B. 12 mph
C. 15 mph
D. 23 mph
E. 27 mph

Solution:

Let the speed of suresh per hour = x

Speed of arun                              = 3x

Arun biked suresh distance in ½ hour.

Total distance covered by arun   = 3*10

= 30 miles

Arun covered the distance in half the time of suresh,

= 30/2

Average speed of arun                  = 15 miles/hr

The answer is 15 miles/hr

Answer: C


Practice act questions:


Practice question 1:

In an urn there are 15 balls: 8 balls are black, 4 are red and 3 are orange. Then 1 black and 1 red ball are taken from the urn and put away. What is the probability that a red ball is selected at random from the urn?

A) 3/13
B) 5/15
C) 6/15
D) 7/13
E) 4/13

Answer: A

Practice question 2:

A group of 7 friends are having dinner together. Each person eats at least 3/4 of a cake. What is the smallest number of whole cakes needed for dinner?

A) 7
B) 5
C) 6
D) 28
E) 21

Answer: C

Reasoning Practice Test

Introduction:

Let us learn about the reasoning practice test. Reasoning test is a test which is conducted to get placed in international colleges, universities and in companies. Reasoning test  practice consists of quantitative aptitude, verbal reasoning and analytical writing. Quantitative aptitude involves math problem solving questions on various topics. Practice of reasoning test questions helps the students to get skilled in aptitude questions. It also helps the students to get prepared for SAT and ACT test.

I like to share this Quantitative Data Collection Methods with you all through my article.

Reasoning practice test:


Example 1:

A trader mixes 30 kg of rice at Rs. 18 per kg with 28 kg of rice of other variety at Rs. 34 per kg and sells the mixture at Rs. 32 per kg. What is the profit percentage?

Solution:

Total price of 58kg mixture rice = ( 30 x 18 + 28 x 34)

= 540 + 952

= Rs 1492.

Sell price of 58 kg rice = (58 x 32)

= Rs 1852

Total gain amount        = Rs 1852 - Rs 1492

= Rs 360

Total gain percentage = (360/1492)*100

= 0.24 * 100

= 24%

The profit gain is 24%



Example 2:

Vinay has 100 kg of wheat, part of wheat which he sells at 8% profit and the rest of wheat at 18% profit. He gains 14% on the whole. What is the quantity sold at 18% profit?

Solution:

From the given data, form two equations,

Part of wheat sold at 8%    = X

Part of wheat sold at 18%  = Y

X + Y = 100  equation 1

0.08X + 0.18Y = 100(0.14)

0.08X + 0.18Y = 14 equation 2

Multiply equation 1 with 0.18

0.18X + 0.18Y = 18  equation 1

0.08X + 0.18Y = 14  equation 2     (subtract)

0.10X                = 4

Divide 0.10 on both sides,

0.10X/0.10 = 4/0.10

X = 40 kg

Total quantity            = 100

Remaining quantity = 100 - 40

Y  = 60 kg

The quantity which is sold at  18% profit is 60 kg.

Practice questions:


Practice problem 1:

The ratio between the speeds of two trains is 6 : 8. If the second train runs 300 km in 3 hours, then the speed of the first train is:

A) 60 km/hr

B) 65 km/hr

C) 75km/hr

D) 73 km/hr

Answer: Option C



Practice problem 2:

544, 509, 474, 439, ... Find the next number in the given series.

A) 420

B) 404

C) 454

D) 528

Answer: Option B



Practice problem 3:

201, 202, 204, 207, ... Find the next number in the given series.

A) 211

B) 212

C) 225

D) 230

Answer: Option A

Practice 9th Grade Algebra Midterm

Introduction to practice algebra problem for 9th grade midterm:

Algebra which is fetched by a finite number which is used to define the following operations like exponents, adding the values, subtracting the values, multiplying and dividing as well.
Basically algebra midterm defines about the number systems , and here in 9th grade we can see about the following methods. They are,
Numbers

Having problem with Two Digit Subtraction with Regrouping keep reading my upcoming posts, i will try to help you.

In number systems, we can see the operations of fractions,exponents,basic arithmetic operations and place values.
Measurement:

Measurement is a process of measuring the object or shapes values in the required units.
Geometry:

Geometry is a process,in which we can declare the different shapes of polygons like regular shapes and irregular shapes.
Algebra / pattern and Probability.

Algebra pattern is a method of arranging the algebraic values in the proper order and it performs a prominent role in algebra terms.

practice problem for 9th grade algebra midterm


Example 1:

Solve the  practice algebraic problem 5(-5x - 2) - (-x - 6) = -6(4x - 5) -36

Solution:

Given the equation

5(-5x - 2) - (-x - 6) = -6(4x - 5) - 36

Multiply factors.
-25x - 10 + x + 6 = -24x + 30-36

Grouping the terms.

-24x - 4 = -24x - 4

Add 24x +4 to both sides, the above equation becomes

0 = 0

All real values are solution to this equation.

Example 2:

Simplify the  practice algebraic expression    3(a -4) + 6b - 3(a -b -2) + 8

Solution:

Given the algebraic expression

3(a -4) + 6b - 3(a -b -2) + 8

Multiply factors.

= 3a - 12 + 6b -3a + 3b + 6 + 8

Grouping the above terms.

= 9b + 2.

I have recently faced lot of problem while learning Statistics Calculator, But thank to online resources of math which helped me to learn myself easily on net.

Practice algebra problem for 9th grade midterm:


Try to solve this practice 9th grade algebra midterm problem:  a+ b = 3, b + c = –5, c + a = 2.

(solution: a = 5, b = -2, c = -3)

Trigonometry Practice Questions

Introduction to trigonometry practice questions:

The branch of mathematics which deals with the relations of the sides and angles of triangles, which the methods of deducing from certain given parts other required parts, and also of the general relations which exist between the trigonometrical functions of arcs or angles with the trigonometric functions.


Solved problems on trigonometry practice questions


1. Solve the trigonometric equation and find the (6 tan^2 x - 2) (2 tan^2 x - 6) = 0

2. Solve the trigonometric equation and find the interval [0 , 2Pi).

-2 sec^2 x + 4 = -2sec x

3. Solve the trigonometric equation and find the interval [0 , 2Pi).

2sin (x) cos (-x) = 2 sin (-x) sin (x)

4. Two tennis players stand on opposite sides of a net. When the ball is directly above the net, at a height of 7m above the ground, the angle of elevation from player A's position to the ball is 41±  and the angle of elevation from player B's position to the ball is 53±. What is the distance between player A and player B?

5. A pilot flying at an altitude of 1200 feet’s is starting his approach. He is 2400 feet along his flight path from the runway. What is his horizontal distance x from the runway and what is his angle of approach µ?

6. The distance between two points on the map is 5 cm correct to the nearest centimeter.

(a) Write down the

(i) Least upper bound of the measurement

(ii) Greatest lower bound of the measurements

(b) The scale of the map is 1 to 20 000. Work out the actual distance in real life (in kilometers) between the upper and lower bounds.

Having problem with Binomial Distribution Calculator keep reading my upcoming posts, i will try to help you.

7. The temperature from the factory furnace a varies inversely as the square of the distance from the furnace.

The temperature 2 meters from the furnace is 50 degrees Celsius.

Calculate the temperature 3.5 meters from the furnace. Give your answer to 2 decimal places.

Fourth Grade Free Math

Introduction of fourth grade free math:-

In Internet or website is the best place for study free fourth grade math. The students learn number of skills in math website and to work practice problems and homework problems. It is more helpful for students to improve our practice skills. In year 4 is based on fourth grade. The fourth grade student’s does on math regular basis. In number of  websites have a special tutoring service to provide math worksheets, practice problems and homework problems. The parents to teach our children’s it is also help our children’s homework problem.

Lessons in fourth grade free math:-


In the following topics involves the fourth grade free math

Number sense
Addition
Subtraction
Multiplication
Division
Algebra

Number sense

In fourth grade free math number sense is nothing but converting the numbers into wordings or converting the wordings into numbers.

For example,

2250 = Two thousand and fifty

Addition

In fourth grade free math adding the two are more value is called the addition by using the (+) operator sign.

For example,

114 + 22 = 136

Subtraction

In fourth grade free math subtract the two are more value is called the subtraction by using the (-) operator sign.

For example,

57 - 41 = 16

Multiplication

In fourth grade free math multiply the two are more value is called the multiplication by using the (x) operator sign.

For example,

134 x 13 =1742

Division

In fourth grade free math divide the two are more value is called the Division by using the (`-:` ) operator sign.

For example,

235 `-:` 21 = 11.19

Algebra

In fourth grade free math algebra is nothing but the study of operations and relations.

Please express your views of this topic Prime Factorization Calculator by commenting on blog.

Practice problems for fourth grade free math:-


Problem 1:-

Solve the number sense problem 8739.

Answer: Eight thousand seven hundred and thirty nine.

Problem 2:-

Solve the addition operation 326+45.

Answer: 371

Problem 3:-

Solve the subtraction operation 122 - 13.

Answer: 109

Problem 4:-

Solve the multiplication operation 212 x 46.

Answer: 9752.

Problem 5:-

Solve the division operation 50 `-:` 4.

Answer: 12.5

Problem 6:-

Find algebra without identity 15 x 14.

Answer: 200

Practice Algebra Division

Introduction of practice algebra division:

Algebra is a branch of mathematics, which is used to make mathematical problems of real-world events and switch problems that we cannot explain using arithmetic.

Algebra is used the symbols for addition, subtraction, multiplication and division and it includes constants, operating symbols and variables

Division is one of the arithmetic operation. Manually division is defined as the reverse operation of multiplication. Variables and constants are combined or grouped and to make algebraic expressions .it contains variables, expressions, terms, polynomials, and equation .

Main goal of division is minimizing the value of dividend or series of subtraction from dividend. The symbol for division is “/”.

Ex :     x ^2+6x+`7/x` +8


Operations of Algebra division:


The Algebra division is defined as repeated subtraction of divisor from the dividend.

The Form of Manual division method,

a / b = c

Where, a is called as  dividend. b is called as divisor.c is called as quotient.

Example:

45 / 5 = 9

example of algebra division using polynomial:

x ^2+12x+27 /  x+9

Here, x ^2+12x+27 is dividend

x+9 is divisor

x+3 is quotient of (x ^2+12x+27) / ( x+9)

I like to share this Degree of Polynomial with you all through my article.

Example problems:


Division of: (x ^2+4x+6)/(x+5)

Sol:   (x ^2+4x+6) is dividend and (x+5) is the divisor.

This is the given standard form of algebra division. In first we can check the order of terms on both dividend and divisor. If it was not in degree of  order, we can arrange the terms.

Like x ^2+x+x3 mean we can change x^3+x ^2+x

After rearrange the terms we can  divide the first term of the dividend   by the first term of the divisor, it mean x ^2/x=x .It gives the first terms of quotient.

After we got the first terms of quotient, and then multiply the first term with quotient after  then subtract this product from the dividend.

The product is x ^2+5,and the subtracted value is –x+6 (remainder)

Again we can  divide the first term of the dividend   by the first term of the divisor, it mean   -x/x =-1 .

It gives the second terms  after then multiply the first term with quotient after  then subtract this product from the dividend .

Now got the remainder is 11

Final answer is x-1

Algebra 2 Practice Tests

Introduction to online algebra 2 practice tests:

There are many online tests available for algebra 2. The algebra 2 defined as solving equation like graphs, function, quadratic equations, slope of a line and trigonometric Identities. In algebra we can refer about the quadratic equations and slope. Quadratic equation in which from the second power is the highest degree to which one denotes degree of variable formation is called as quadratic equation. Algebra deals with unknown values called variables, unlike arithmetic which is based entirely on known number values.

Is this topic help on algebra 2 hard for you? Watch out for my coming posts.

Algebra 2 practice test equation of a line:


If the slope m of a line plus a point (x1, y1) on the line are both known, then the equation of the line can be found using the point-slope formula:

(y- y1) = m (x-x1)

Online Algebra 2 practice test problem:

Find the equation of the line in slope- intercept form passes through (-4, 5) with a slope 1/2.

Solution :

slope formula for given points (y - y1) = m(x - x1)

It goes through the point (3, 2) and has a slope of 1/2. So appropriate the information to the point-slope formula gives:

substitute the given values into the formula

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

(y - 2) = 1/2(x + 3) multiply that (1/2) inside the brackets.

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

y = x/2 +3/2 + 2

y = x/2 + 7/2

The final equation is in the slope-intercept form, which is y = mx + b, where m is the slope and b is the y-intercept.

Please express your views of this topic Definition of Right Triangle by commenting on blog.

Example for online algebra 2 practice test:


(x + 2)(x + 3) = 0

x + 2 = 0 or x + 3 = 0

x = -2 or x = -3

Therefore, the solution is x = -2, -3

x^2 + 3x +2 = 0  Use quadratic equation to solve this problem.

solution:

formula for quadratic equation x = ( -b ± √ (b^2 - 4ac)) / 2a

Given details are  a = 1

b = 3

c = 2

x = (-3 ± √ (9 - 4(1)(2))) / 2(1)

x = (-3 ± √ ( 9 - 8 )) / 2

x = (-3 ± √ 1) /2

Therefore, the solution is (-3 + √ 1) / 2, (-3 - √ 1) /2

Solve Algebra Practice Exams

Solve Algebra Practice-Exams:

Algebra is one of the main branches of arithmetic. It explains the interaction and properties of quantity by means of letters and other signs. The basic algebra has the following subtopics are

Variables,
Expressions,
Terms,
Polynomials,
Equations
There is some solve algebra problems listed below for the exam practice:


Solve Algebra Practice for Exams- Example 1:


Solve for x: 5 x - 6 = 3 x – 10

Solution:

Subtract 3x from both sides of the equation

5x – 3x – 6 = 3x – 3x – 10

2x – 6 = - 10

Add 6 to both sides of the equation

2x – 6 + 6 = - 10 + 6

2x = - 4

Divided by 2 both side of the equation

x = - 2

The answer x is – 2

Solve Algebra Practice for Exams- Example 2:

Solve for x: 4x - 6 = 12x – 40

Solution:

4x - 6 = 12x – 40

Subtract 4x from both sides of the equation:

4x – 4x – 6 = 12x – 4x – 40

– 6 = 8x – 40

Add 40 to both sides of the equation

– 6 + 40 = 8x – 40 + 40

36 = 8x

Divided by 8 both side of the equation

x = 4.5

The answer x is 4.5.

Solve Algebra Practice - Example 3:

Solve the equation: 5(-3x - 2) - (x - 3) = -4(4x + 5) + 13

Solution:

Given the equation

5(-3x - 2) - (x - 3) = -4(4x + 5) + 13

Multiply factors

-15x - 10 - x + 3 = -16x - 20 +13

Group like terms

-16x - 7 = -16x - 7

Add 7 on both sides

−16x = − 16x

Here both sides are equal so x will have infinite number of solutions.

Solve Algebra Practice - Example 4:

Simplify the expression

2(a -3) + 4b - 2(a -b -3) + 5

Solution:

Given the algebraic expression

2(a -3) + 4b - 2(a -b -3) + 5

Multiply factors

= 2a - 6 + 4b -2a + 2b + 6 + 5

Group like terms

= 6b + 5

Solve Algebra Practice - Example 5:

Solve for x: 5x - 6 = 12x – 50

Solution:

4x - 6 = 12x – 50

Subtract 5x from both sides of the equation:

5x – 5x – 6 = 12x – 5x – 50

– 6 = 8x – 50

Add 50 to both sides of the equation

– 6 + 50 = 8x – 50 + 50

44 = 8x

Divided by 8 both side of the equation

x = 5.5

The answer x is 5.5.

Practice Digit

Introduction:

Let us we will discuss about practice digit. A digit should be symbol that is used in numerals to denote numbers in positional numeral systems. The name "digit" comes from the practice fact that the 10 digits of the hand communicate to the 10 symbols of the universal support 10 number system. In given number system, if the base is an integer, number of digits necessary is always equivalent to the fixed value of sustain. Having problem with Numerical Differentiation keep reading my upcoming posts, i will try to help you.


Computation Of Place Values


Every digit in a number method represents an integer.
In decimal the digit "1" characterizes the integer one. But in practice hexadecimal scheme, the letter "A" denotes the number ten.
A positional number practice system should have a digit denoting the integers from zero up to, but not include the radix of the number system.
The 0 is instantaneously to the left of the partition, so it is in the one's place.
The 1 to the left of the zero has a set value of one, and is in the ten's place.
The 3 is to the right of the one's place, so it is in the tenths place.
The 4 to the right of the tenths situate is in the hundredths place.
Therefore, the total value of number should be 1 ten, 0 ones, 3 tenths, and 4 hundredths.
Note that the zero that gives no value to number. They are denoting that the 1 is in the tens place rather than the one's place.
The computation practice engages the multiplication of the given digit by the base lift up by exponent n-1. Where 'n' denotes the position of digit from the separator.
From the right side, the digit should be multiplied by the base lift up by negative (-) n.

Please express your views of this topic rational expressions adding by commenting on blog.

Example:
Number 11.74 (written in base ten).

Here, 1 could be second to left of separator. So the calculation will be,

n - 1 = 2 - 1 = 1

1 × 101 = 10

Number 4 is second to right of the separator. Therefore, the calculation will be,

n = -2

4 × 10-2 = 4/100

Value of n should be positive (+). But this is only when digit is to left of separator.

Practice 8th grade Pre Algebra

Introduction of practicing 8th grade pre algebra:

The word algebra is derived from the Arabic word al–jabr. In Arabic language, ‘al’ means ‘the’ and ‘jabr’ means ‘reunion of broken parts’. The usage of the word can be understood by a simple example. In the equation x + 5 = 9, the left hand side is the addition (sum) of two parts x and 5. If we add (unite) (–5) to each side of the equation in pre algebra.

We get  (x + 5) + (–5) = 9 + (–5) or x + [5 + (–5)] = 9 – 5 or x + 0 = 4 or x = 4.  Here 9 and -5 are reunited to get 4. This type of mathematics is called pre algebra.

Let us practice some 8th grade pre algebra problems.

I like to share this help in algebra 2 with you all through my article.

Practicing example problems for 8th grade pre algebra:

Example 1:

In pre algebraic expression  of given 5(-3x - 2) - (x - 3) = -4(4x + 5) + 13

Solution:

5(-3x - 2) - (x - 3) = -4(4x + 5) + 13

-15x - 10 - x + 3 = -16x - 20 +13

-16x - 7 = -16x – 7

0 = 0

Example 2:

Reduce the expression in pre algebra: 2(a -3) + 4b - 2(a -b -3) + 5

Solution:

= 2(a -3) + 4b - 2(a -b -3) + 5

= 2a - 6 + 4b -2a + 2b + 6 + 5

= 6b + 5

= -x -22

Example 3:

Estimate in pre algebra: f (2) - f (1), f(x) = 6x + 1

Solution:

f (2) - f (1) = (6*2 + 1) - (6*1 + 1)

= 6

Example 4:

Solve |-2x + 2| -3 = -3

Solution:

|-2x + 2| -3 = -3

|-2x + 2| = 0

x = 1

Problem 5:

If Jim and Jerry work together they  finish a job in 4 hours. If work alone takes, Jim takes 10 hours to finish the job, how many hours would it take Jerry to do the job alone.

a. 16

b. 5.6

c. 6.7

d. 6.0

Solution:

Take the hours is x and it will take to Jerry  the job alone. In 1 hour Jim can do `(1)/(10)` of the work, and Jerry can do `(1)/(x)` of the work. As an equation this looks like `(1)/(10)`+`(1)/(x)`=`(1)/(4)`

`(1)/(4)`represents of the job they can complete in one hour together.Lcm of 10 and x will be 10x and multiply with = 40x.

4x + 40 = 10x.

Subtract 4x from both sides of the equation.

4x- 4x + 40 = 10x - 4x.

This simplifies to 40 = 6x.

Divide each side of the equation by 6.

X = 40 / 6.

Therefore, x = 0.6666 and it would take Jerry about 6.7 hours to complete the job alone.

The above are some practice problems for 8th grade pre algebra with answers.

Understanding Formula Volume of a Cylinder is always challenging for me but thanks to all math help websites to help me out.

Practice problem for 8th grade pre algebra:


Problem 1:

The square of a positive number is 64. What is the number?

Answer: 8

Problem 2:

If Jim and Jerry work together, they  finish a job in 4 hours. If work alone takes, Jim takes 10 hours to finish the job, how many hours would it take Jerry to do the job alone.

Answer: Jerry about 6.7 hours to complete the job alone.

Multipying Radicals

Multiplication is one of the basic operations in math. In arithmetic we multiply numbers whereas in algebra we multiply variables and expressions. Radicals are one type of expressions and they are also called roots with indices.The index of a square root is 2 but it is generally not indicated in the symbol.Let us discuss the concept of multiplying radicals. I like to share this Rules of Radicals with you all through my article.

The multiplication of expressions which are not in radical form is always defined. But it is not the case in radicals multiplication. There are certain rules for multiplying radicals for the multiplication to be defined. The most fundamental and the most important rule is that multiplication radicals are defined only if the indices of the radicals are same. A square root can be multiplied only with another square root and not with a cube root. Secondly, the radicands of radicals with even number indices cannot be negative.
With the above restrictions we can proceed to see how multiplication of roots is done.When the radical indices are same, the radical symbol can be ‘factored out’. That is, the radicands can be multiplied under one radical symbol. This is a great advantage and it makesthe multiplication simpler. In many cases the result may turn out to be an integer. I have recently faced lot of problem while learning Product Rule for Radicals, But thank to online resources of math which helped me to learn myself easily on net.

For example, consider the multiplication of √(8) by√(2). Both of them are irrational numbers. But as per the concept we explained, √(8)*√(2) = √(8*2) = √(16) = 4, which is an integer. Even in cases where the final answer may not be integers, we are still supposed to simplify the final radicand by factoring.
For example, √(6)*√(2) = √(6*2) = √(12). Though √(12) is irrational, 12 can be factored as 4*3 and 4 being a perfect square, it can be taken out. Thus the correct way to work it out is,   √(6)*√(2) = √(6*2) = √(12) = √(4*3) = √(4)*√(3) = 2√(3).
The same concept is used in case of multiplying radicals with exponents, especially when variables are involved. Let us illustrate as to how it works.
√(x3)*√(x)=  √(x3*x) = √(x4) = x2.

Even in cases where the radical symbol cannot be avoided, we still should try to keep the minimum exponent inside the symbol. For example,
√(x5)*√(x3) =  √(x5*x3) = √(x15) = √(x14*x) = √(x14)*√(x) =  x7*√(x)
As mentioned earlier, radicals of even number indices having negative radicands are not real numbers. Hence the multiplications in such cases have to be done by special techniques using the concept of imaginary numbers.
All imaginary numbers can be factored with √(-1) to remove the imaginary part and the letter ‘i’ is used as a symbol for √(-1).

Multiplying Polynomials

Polynomials are expressions containing finite number of terms. None of the terms can have a division by a variable and also the exponents of any term must only be a non-negative integer. These types of expressions can be added, subtracted, multiplied or can be divided.
In this lesson let discuss about multiplying polynomials. The method of how to multiply polynomials is based on the concept of distributive property. Please express your views of this topic Operations with Polynomials by commenting on blog.

A polynomial with only terms is called as binomials. We all know that two binomials are multiplied by the technique FOIL. The same concept is slightly modified and extended in case of polynomial multiplications. As one of the multiplying polynomials examples, let us consider the following with just one variable. (a0xn + a1xn-1 + a2xn-2 + ….. + an-1x + an)* (b0xm + b1xm-1 + b2xm-2 + ….. + bm-1x + bm) = ?
Pick up the first term a0xn of the first expression and multiply that with all the terms of the second expression and add the products as per the distributive property of multiplication over addition. This is first set of expression of the entire product. Now take the second term a1xn-1 of the first expression and repeat the same process.

The result will be the second set of the expression for the entire product. The method is repeated till the last term of the first expression is multiplied with all the terms of the second expression. This is the final set of expression of the entire product. Now add all sets of expressions and simplify the sum by algebraically adding like terms. You may notice that the degree of the final product is sum of the degrees of the given expressions. Is this topic What is an Algebraic Expression? hard for you? Watch out for my coming posts.

For better clarity let us take an actual case in multiplying polynomials problems.                                                       
(3x^2 – 2x + 4)*( x^3 + 2x^2 – x + 5) = ?
Step 1: (3x^2)*( x^3 + 2x^2 – x + 5) = 3x5 + 6x^4 – 3x^3 + 15x^2
Step 2: (-2x)*( x^3 + 2x^2 – x + 5) = -2x^4 – 4x^3 + 2x^2 – 10x
Step 3: (4)*( x^3 + 2x^2 – x + 5) = 4x^3 + 8x^2 – 4x + 20
Adding the expressions obtained in all the steps, we can say that
(3x^2 – 2x + 4)*( x^3 + 2x^2 – x + 5) = 3x5 + 6x^4 – 3x^3 + 15x^2- 2x^4 – 4x^3 + 2x^2 – 10x + 4x^3 + 8x^2– 4x + 20
Simplifying by adding the like terms, the final product can be written as,
3x5 + 4x^4 – 3x^3 + 25x^2 - 14x + 20
The given expressions had the degrees as 2 and 3 respectively and it may be seen the degree of the the product is 5 which is 2 + 3.

Arithmetic Test

Introduction to Arithmetic Test:

Arithmetic or arithmetic is the oldest and most elementary branch of mathematics, used by almost everyone, for tasks ranging from simple day-to-day counting to advanced science and business calculations. It involves the study of quantity and especially as the result of combining numbers. In common usage and it can be refers to the simpler properties when using the traditional operations of addition, subtraction, multiplication and division with smaller values of numbers. Now let us see about the practice arithmetic test.  (Source in Wikipedia).

Please express your views of this topic Sum of Infinite Arithmetic Series by commenting on blog.

simple arithmetic Test:


In simple arithmetic operations.We perform the addition,subtraction,division and multiplication


Practice problem  in Addition:

689 + 750

Solution:

When we add 689 + 750 we get  1439 as the final answer

Practice problem in Subtraction:

885 – 695

Solution:  When we subtract  885 – 695  we get 190 as the final answer

Practice problem in Division:

72 ÷ 6

Solution: When we divide 72/ 6 we get  12 as the final answer

Practice problem in Multiplication:

85 x 76

Solution: When we divided 85 x 76 we get  6460 as the final answer.

Practice Arithmetic Test in word Problems


Practice Arithmetic Test Problem 1:

The first term of an arithmetic sequence is equal to 18 and the common difference is equal to 3. Find a formula for the n th term and the value of the 70 th term

Answer: 213

Practice Arithmetic Test Problem 2:

The first term of an arithmetic chain is 200 and the common difference is equal to -10. Find the value of the 20 th term

Answer: 10

Practice Arithmetic Test Problem 3:

The first term of an arithmetic sequence is equal to 20 and the common difference is equal to 4. Find a formula for the n th term and the value of the 80 th term

Answer: = 1280

Practice Arithmetic Test Problem 4:

The first term of an arithmetic sequence is equal to 25 and the common difference is equal to 5. Find a formula for the n th term and the value of the 95 th term

Answer:480

Algebra Practice Questions

INTRODUCTION:

In Algebra a letter such as a, b, c...x, y, z stands for an unknown number which is called variable. Algebra is used to create a mathematical model of real-world situations. Algebra is a method of determining and solving the puzzles in our daily life. It has an algebraic expressions and properties, variables with patterns. Algebra can be applied on real numbers, complex numbers, matrices, vectors etc .algebra has a set of operation with an identity elements. Let us see online algebra test answers in this article. I like to share this System Equations with you all through my article.


Algebra Online test and its answers:


Consider some problems on online algebra test and its answers.

When Ravi attempts a free throw with succeeds 59.4% of a time and prabhu attempts 31 free throws. Then how many times prabhu will succeed.
a) 15 b) 18 c) 28 d) 52

Ans: B

Solve: 2x - 27 = 15
a) -6 b) 12 c) 21 d) 60

Ans: C

What is the value of the expression 2(x +12) when x = -4 ?
a) -27 b) 1 c) 12 d) 16

Ans: D

What is the reciprocal of -115?
a) -55 b) -1/115 c) 1/115 d) 115

Ans: B

Find the correct sets of numbers are ordered from least to greatest?
a) -4/2, -4, 0, 3/2 b) -4, -4/2, 0, 3/2 c) 0, 3/2, -4/3, -4 d) 0, -4/3, -4, 3/4

Ans: B

What is the opposite of 7/2?
a) -2/7 b) -7/2 c) 2/7 d) -1

Ans: B

Some more Online Problems in algebra test:


A flower grows 40 centimeters per month. It was 20 centimeters tall on April On what date will it most likely be 80 centimeters tall?
a) April 15 b) June 1 c) may 15 d) June 1

Ans: B

In a lunch period, biscuits sold at a rate of 5 bags every 3 minutes. The lunch period lasted for 45 minutes. How many bags of biscuits were sold?
a) 27 b) 53 c) 75 d) 225

Ans: C

The area of a triangle is given by the equation h 2 − 6h = 27
where h is the height of the triangle. What is the value of h?

a) 4 b) 9 c) 12 d) 28

Ans: B

The area of an isosceles right triangle is described by the equation x2 = 729
where x is the height in centimeters of the triangle. What is the height of the triangle?

a) 12 b) 27 c) 96 d) 192

Ans: B

These are the online algebra test and its answers.