practice Constant

The values can’t be change. Constants it’s also called variable. In math, constant is a number, But sometimes we can also take the variable as a constant. For example,  In this equation x2+5x+3 = 0, 3 is a constant.  In this equation x+5=-20, 5 and -20 are constants. Now we are doing to practice some constant problems.

Understanding Chain Rule Practice is always challenging for me but thanks to all math help websites to help me out.

Practice constant Problems:

Practice problem 1:

X2+4x+k= 24. If x= 2, solve for k .

Solution:

Step 1: Substitute x value in the given equation.

Step 2: So we get (2)2+4(2)+k=24.

Step 3: Here we need to simplify this.

Step 4: 4+8+k=24.

Step 5: When we add we get 12+k=24.

Step 6: Subtract 12 on both the sides so, 12-12+k= 24-12.

Step 7: Therefore answer is k=12.


Practice problem 2:

X+4y-3z+r= 51...if x=9,y=3 and z=2 ,solve r.

Solution:

Step 1: Substitute x, y and z value in the given equation.

Step 2: So we get, 9+4(3)-3(2)+r= 51.

Step 3: Now we need to simplify this equation.

Step 4: 9+12-6+r=51.

Step 5: When we simplify we get 15+r= 51.

Step 6: Divide using 15 on both the sides.

Step 7: Therefore, the answer is 51/15.


Practice problem 3:

M3+m2+6m+S= 72. If m= 2, solve for S.

Solution:

Step 1: Substitute m value in the given equation.

Step 2: So we get (2) ^3+ (2) ^2+6(2) +S=72.

Step 3: Now we need to simplify this equation.

Step 4: 8+4+12+S=72.

Step 5: When we add we get 24+S=72.

Step 6: Subtract  24 on both the sides so, 24-24=72-24 .

Step 7: Therefore the value of s is 3.

My forthcoming post is on Conditional Probability Venn Diagram and Associative Property of Multiplication Example will give you more understanding about Algebra.


Practice problem 4:

C2+15c+g=195. If c= 3 then find g.

Solution:

Step 1: Substitute m value in the given equation.

Step 2: So we get 32+15(3)+g=195.

Step 3: Here we need to simplify this equation.

Step 4: 9+45+g=195.

Step 5: After addition we get 54+g=195.

Step 6: Divide using 54 on both the sides.

Step 7: So the value of g =195/54.

work out problems:

X2+19x+t=121 if x=12, find the value of t?
A3+4a+z= 147 if a =5, find the z value?
H2+h+m= 12 if h=6, find the m value?
F4+8f+n=546 if f=4, find the n value?

Math Operations

Operator in math is a function which acts on two values ( operands). These operators are known as binar operators as they need two operands. There are four binary operators namely +  ,  -   ,  x   , /. Operator are mainly consists of number of terms. For example, the assignment operator are used for assigning the values to a particular variable.

The four fundamental operations on numbers are Addition, Subtraction, Multiplication and Division. The operator of Addition is ‘+’, Operator of Subtraction is ‘-’, Operator for multiplication is ‘x’ and Operator for division is ‘÷’. Let us see about arithmetic operations with appropriate operator.
Explanation for various types of operator statement:

1. '+' is an  operator statement:

For this operator, the addition operation and the string concatenation operation are included.
Addition operation is used for adding the two values that are given.
String concatenation are used for joining the two statements.

2. ++ is an increment operator:

For this, the values are incremented according to the given values.

3. -- is an decrement operator:

For this, the values are decrement according to the given values.

4. x  is a multiplication operator:

I like to share this Mode Math Definition with you all through my article.

For this, the multiplication function and the pointer function are included.
Multiplication function is used to multiply the two given values.
Pointer are used for denoting the reference variable.

5. / is a division operator:

For this, the values are divided for getting the remainder and the quotient factor.

6.  ^  exponential operator:

Exponential operator is used for raising the power to a function.

7. -  mathematical operator:

For this, the subtraction and the negation process are included in this operator statement.
Subtraction operation is used for finding the difference of the given two numbers.


Examples using math operations:

Addition Operator ‘+’: For adding two or more numbers we use the ‘+’ operator. This operator is known as plus.Normally additions have two operands and one operator. For example x + y here ‘x’ and ‘y’ is called as operator and the symbol’+’ is called as operand.

Ex: Add 5 and 6.

Sol:

5 + 6 = 11

Where, ‘+’ is the operator of addition.

5 and 6 are operands.

11 is the resultant of the operation of addition.

Subtraction Operator ‘-’: For subtracting two or more numbers, the operator used is ‘-’.  This operator is known as minus.

Ex : Subtract 11 and 5.

Sol:   11 – 5 = 6.

Where, ‘-’ is the operator of subtraction.

11 and 5 are operands.

6 is the resultant of the operation of subtraction.

Multiplication Operator ‘x’: For Multiplying two or more numbers, the operator used is ‘x’.  This operator is known as into.

Ex : Multiply 12 and 5.

Sol:             12 x 5 = 60.

Where, ‘x’ is the operator of multiplication.

12 and 5 are operands.

60 is the resultant of the operation of multiplication.

Division Operator ‘÷’: For dividing two or more numbers, the operator used is ‘÷’.  There is no special name for division operator.

Ex : Divide 10 by 5.

Sol:   10 ÷5 = 2.

Where, ‘÷’ is the operator of Division.

10 and 5 are operands.

2 is the resultant of the operation of division.

Additional Operators:

‘√’:

This operator is known as the square root.

Ex: Find Square root of 4?

Sol: √4 = 2.

Where, ‘√’ is the operator of square root.

4 is the operand.

2 is the resultant of the operation of square root.

Order of operations in math:

The order of operation is square root, division, multiplication, Subtraction and addition.

Ex: Calculate 5 + 6 – [(√4 ÷2) x 3].

Sol:  5 + 6 – [(2 ÷ 2) x 3]

5 + 6 – [1 x 3]

5 + 6 – 3

5 + 3

7

Steps and examples for addition operation in math:

Simple addition uses the following steps to adding the numbers.

Step 1: Add the first place number from the right (first number) first after that take over the carry to next step.

Step 2:   Add the ten’s place number (means second digit) as well as also add the carry from the first step. Now note down the answer and over the carry.

Step 3  :Carry on this process on until reach the end place of the given number.

The followings are the some of the types of addition

Add single digit number with single digit
Add single digit number with double digit
Add three one digit numbers
Add double digit number with three digit number
Add double digit number with double digit number
Add treble digit number with treble digit number etc….

Example for Simple Addition :

Add single digit number with single digit:

Ex: Add 5 with 9

Sol:

5
(+) 9
-----------     14

Ex 2: Adding the following three one digit numbers.

9, 6 and 3

Sol:

Given numbers 9, 6 and 3

Step 1:

Add the first two numbers

9

(+)      6

----------------------

1 5

Step 2:

Add the next number with the previous step sum or total.

Previous step total =15

Third number =3

1 5

(+)        3

----------------------

1 8

Therefore 9 + 6 +3 =18

Algebra is widely used in day to day activities watch out for my forthcoming posts on solve this math problem for me and cbse guide for class 10. I am sure they will be helpful.

Ex 3: Add the following two numbers: 54 with 16

Sol:

Given numbers 54 and 16

Step 1:

Add the first digit (ones place) numbers (4+ 6 = 10) then over the carry (here 1 is the carry)

5 4
(+)   1 6
-----------
0

Step 2:

Add the second digit (tens place) numbers (5 +1 = 6) the add with previous step carry (6 +1 =7)

5 4
(+)   1 6
-----------
7 0

Finally we get 54 +16 = 70

Ex 4: Add the following two numbers: 15948 and 69741

Sol:

Given 15948 and 69741

1 5 9 4 8
(+)     6 9 7 4 1
------------------

Step 1:

Add the one place number then over the carry (8+1=9 carry =0)

1 5 9 4 8
(+)    6 9 7 4 1
------------------
9

Step 2:

Add the tens place number then over the carry (4+4 = 8 carry 0)

1 5 9 4 8
(+)    6 9 7 4 1
------------------
8 9

Step 3:

Add the hundreds place number then over the carry (9+7 =16 carry 1)

1 5 9 4 8
+)    6 9 7 4 1
------------------
6 8 9

Step 4:

Add the thousands place number (5+9 = 14) then add with carry from the above step (14 +1 = 15 carry 1)

1 5 9 4 8
(+)    6 9 7 4 1
------------------
5 6 8 9

Step 5:

Add the ten thousands place number (1+ 6=7) then add with previous step carry (7 + 1 =8)

1 5 9 4 8
(+)    6 9 7 4 1
------------------
8 5 6 8 9

Therefore 15948 + 69741 = 85689

Ex 5: Add 72 with 4.12

Sol:

Given numbers 72 and 4.12

Step 1:

Change the whole number into decimal form

Whole number =72

Decimal form = 72.00

Step 2:

7 2 .0 0

(+)       4. 1 2

--------------------



Start the addition process from the right

Step 3:

Add the first place number from the right and over the carry to the next step (0 +2 =2 carry =0).

7 2 .0 0

(+)       4. 1 2

--------------------

2



Step 4:

Add the next place number from the right and then add the carry with is step total. Finally over the carry to the next step (here 0 +1 =1 carry =0)

7 2 .0 0

(+)       4. 1 2

--------------------

. 1 2

Step 5:

Add the next place number from the right and then add the carry with is step total. Finally over the carry to the next step (here 2 +4 =6 carry =0)

7 2 .0 0

(+)       4. 1 2

--------------------

6. 1 2

Step 6:

Add the next place number from the right and then add the carry with is step total. Finally over the carry to the next step (here 7 +0 =7, carry =0)

7 2 .0 0

(+)       4. 1 2

--------------------

7 6 .1 2

72 + 4.12 =76. 12

Examples on multiplaction operation in math:

Ex 1: Multiply 10 with 5.

Sol:     10 x 5 = 15.

Where, ‘x’ is the operator of multiplication.

10 and 5 are the operands of multiplication.

15 is the resultant of the multiplication operation.



Ex 2: Simplify 10(5) + 20(2) + 9(3) + 4(1).

Sol:     15 + 40+ 27+ 4

86

Ex 3: A train has 9 carriages. There are 42 seats in each carriage. How many seats are there on the train?

Sol:   Number of carriage = 9

Number of seats in each carriage = 42

Multiply number of carriages with number of seats in each carriage.

Number of seats in the train = 9 x 42

= 378.

Ex 4: There are five cupboards in a room. Each cupboard has 7 racks. Find how many racks totally the room has.

Sol:      Number of cupboards = 5.

Number of racks in each cupboard = 7.

To find total number of rack, multiply the number of cupboard with number of racks in each cupboard.

Total number of rack = 5 x 7

= 35.

Ex 5: There are six baskets full of apple in a lorry. 1st basket contains 35 apples, 2nd and 3rd basket contains 40 apples, 4th, 5thth basket contains 50 apples. Calculate the total number of apples in the lorry. and 6

Sol:      35 apples in the 1 basket

40 apples in the 2 basket

50 apples in the 3 basket

Total Number of Apples = 35(1) + 40(2) + 50(3)

= 35 + 80 +150

= 265

Number Percentage Calculator

Percent means ‘for every 100 ’. So, when we say P% .it means P  out of 100 . Thus, P%=P/100 . It is often denoted by symbol “% ”. Any percentage can be expressed as a fraction. For example, 40%=40/100=2/5 .

Percentages are used to find whether one quantity is large or small compared with another quantity. The first term usually represents a part of, or a change in the second term, which should be greater than zero.

Percentages are usually used to express numbers between zero and one, any dimensionless proportionality can be expressed as a percentage.

Please express your views of this topic Percentage of Difference by commenting on blog.

Let us now express some percentages as fractions:

a.       5%=5/100=1/20 .

b.      10%=10/100=1/10 .

c.      25%=25/100=1/4 .

d.     75%=75/100=3/4 .

e.      125%=125/100=5/4 .

f.        175%=175/100=7/4 .

g.       (3 1/8)%=25/800=1/32 .

h.     (6 1/4)%=25/400=1/16 .

i.        (8 1/3)%=25/300=1/12 .

j.        (16 2/3)%=50/300=1/6 .

k.       (66 2/3)%=200/300=2/3 .

l.        (87 1/2)%=175/200=7/8 .



Calculation of Percentage:

Calculation of percentage:

The percent symbol can be treated as being equivalent to the pure number constant 1/100=0.01,  while performing calculations with percentage.

If a number is first changed byP%  and then changed by Q% , then the net change in the number =[P+Q+((PQ)/100)] . Remember that any decreasing value in the formula should be taken as ‘negative’ and increasing value should be taken as ‘positive’.

Similarly, if A’s salary is P%  less than B’s salary, then the percentage by which B’s salary is more than A’s salary is(100P)/(100-P) .

If expenditure also, then percentage change in expenditure or revenue=[P+Q+((PQ)/100)] . Where ‘P’ is the percentage change in price and ‘Q’ is the percentage change in consumption.

Problems on number percentages:

Ex1 : What percentage of 1600 is 40?

Sol:  Let 40  be "P% of 1600.

So, 40=P%   of 1600 =(P/100)(1600)=16P .

Thus, P=40/16=2.5% .

Ex2 : Calculate 40%  of 625 .

Sol: 40%  of a number =2/5  of the number =2/5  of 625=(2/5)(625)=250 .

Ex :3 A number is first increased by 30%  and then decreased by 20% . Find the net change in the number.

Sol:  Let the original number be 100 .

Increasing by 30%, " it becomes " 130 .

Now, if 130  is decreased by 20% , it becomes 104 .

Thus, the net change =(104-100)=4%  increase.

Algebra is widely used in day to day activities watch out for my forthcoming posts on Divide Fractions by Whole Numbers and sample paper of class 9 cbse sa2. I am sure they will be helpful.

Practice problems on number percentages:

Q:1  A’s salary is 25%  more than B’s salary. By what percent is B’s salary less than A’s?

Sol:  Let B’s salary be .

Since A’s salary is 25%  more than that of B, his salary will be Rs. 125 .

Thus, B’s salary is Rs. 25 less than the A’s salary.

So, in percentage: (25125)(100)=20% . Hence, B’s salary is 20%  less than A’s salary.

Multiplication Properties

In this page we are going to discuss about multiplication properties concept. Integer used to measuring and counting in general things. Positive integer and negative integer is the two types of integers. Integer is nothing but using notational symbol represents numbers. Multiplying integer’s mean scaling a number by another.

3 x 4 = 3 + 3 + 3 + 3 = 12

3 x 4 = 4 + 4 + 4 = 12

Properties of multiplication

To study multiplying integers, we have to know five basic properties such as associative, commutative, distributive, and multiplicative identity. These multiplication properties are described as below.

Commutative property
Associative property
Multiplicative identity property
Distributive property
Zero property



Commutative property:

This is the product the two number is same even we change the order. That is, a * b = b * a.

Example: 5 * 4 = 4 * 5

= 20

Associative Property:

This is the product the three more number is same even we change the order. That is, (a * b) * c = a * (b * c).

Example: 5 *( 4 * 3 ) = (5 * 4) * 3

= 60

Identity Property:

This is the multiplication of any number by 1 is that number. That is, a * 1 = a.

Example: 9 * 1 = 9.

Distributive property:

Multiplication of a number with the addition of two numbers is same as addition of two products. That is, a* (b + c)=ab + ac.

Example: 5 * (4 + 3) = 5*4 + 5*3

= 35

Zero property:

Any thing multiply with zero is zero. That is, a * 0 = 0.

Example: 7 * 0 = 0

Multiplication rules

Positive * Positive = Positive

Negative * Negative =Positive

Positive * Negative = Negative

Negative * Positive = Negative

The following examples are to study multiplying integers with multiplication rule

5 * 4 = 20

-5 * -4 = 20

5 * -4 = -20

-5 * 4 = -20

Multiplication grid

Multiplication grid

Methods to multiplying integers

Below are the methods to multiplying integers-

Method 1:

Multiplying integers of 345 * 6

Solution:

Here, 5 is at unit place, so we do    5 * 1       =    5  * 6 =    30

4 is at 10s place, so we do   4 * 10     =  40  * 6 =   240

3 is at 100s place, so we do 3   * 100 = 300 * 6 = 1800
--------
The product result  =  2070
--------

Method 2:

Multiplying integers of 345 * 6

Solution:

345

x 6
--------------
30
240
1800
--------------
2070
--------------

Sample Space

In statistics, learning sample space basically represents the presentation and interpretation of the events of the possible outcomes that occur in a planned learning or scientific investigation. The learning sample space helps to refer all recording of information's are numerical or categorical, as an observation. The learning sample space assists in the following three cases, designed experiments, observational studies, and retrospective studies, the end result was a set of data that of course is subject to uncertainty.


Definition for Sample Space

The set of all favorable combinations of outcomes of a statistical experiment is labeled the sample space and is denoted by the mathematical symbol S.

Each possible outcome of a sample space is labeled a member of the sample space, in other words it is labeled the sample point. The trials of a sample space has a finite number of elements are separated by commas (,) and enclosed in braces ({}).

Thus the: sample space of possible outcomes when a coin is tossed, may be written S = {H, T),

Where, H means that “heads" and T means that “tails,”.


Examples for Sample Space:

Example 1:

Determine the sample space for the event of rolling a die, using the learning sample space.

Solution:

In rolling a die the number that shows on the top face.

The required sample space S1 = {1, 2, 3, 4, 5, 6}.

Example 2:

Determine the sample space for the number is even or odd.

Solution:

The required sample space S2 = {even, odd}.


More than one sample space:

If we know which element in S1 occurs, we can tell which outcome in S2 occurs; however, knowledge of what happens in S2 is of little help in determining which element in S1 occurs. Provides more information than S

Ex:  A coin is tossed twice. What is the Sample space?

Sol: The sample space; for this experiment is

S= {HH, HT, TH, TT}.

My previous blog post was on Multiplication Rule please express your views on the post by commenting.

Binary Number Representation

The binary numeral system, or base-2 number system, represents numeric values using two symbols, 0 and 1. More specifically, the usual base-2 system is a positional notation with a radix of 2. Owing to its straightforward implementation in digital electronic circuitry using logic gates, the binary system is used internally by all modern computers.

Understanding Binary Logistic Regression is always challenging for me but thanks to all math help websites to help me out.

The representaion of binary numbers are uses in mathematics are defined as follow,

Binary numbers representation in math:

In the binary number representation consists of octal, decimal, and hexa decimal numbers in the column. We can be represent the binary numbers by use of its operation. In the binary number representation, the decimal is easiest method of understanding binary numbers.

For example we can represent: 4567,

4 is represent the 1000’s

5 is represent the 100’s

6 is represent the 10’s

7 is represent the 1’s

Which means the representaion of 4567 is given as follow,
4567 = 1x1000 + 2x100 + 3x10 + 4x1

Given binary number representation,

1000


= 103 = 10x10x10

100


= 102 = 10x10

10


= 101 = 10

1


= 100 (any number to the exponent zero is 1)

The table above can be represented as the binary numbers,

Such that,

4567 = 4x1000 + 5x100    + 6x10     + 7x1

= 4x103 + 5x102 + 6x101 + 7x100

Examples for binary number representation in math:

The examples of binary number representation in math is given as follow:

Example:1

To determine the decimal number in 10102?

Solution:

Step 1: 1 => 1×2×2×2 = 8

Step 2:  0 => 0×2×2 = 0

Step 3: 1 => 1×2 (=2)

Step 4: 0 => 0

Answer is: 1010 = 8+0+2+0 = 10.

Example 2:

To determine the decimal number in 10112?

Solution:

Step 1:  1=> 1×2×2×2 (=8)

Step 2: 0 => 0×2×2 (=0)

Step 3: 1 => 1×2 (=2)

Step 4: 1 => 1

Answer is: 1001 = 8+0+2+1 = 11.

My forthcoming post is on how to find the prime factorization of a number and neet entrance exam syllabus will give you more understanding about Algebra.

Example 3:

To determine the decimal number in 1.112?

Solution:

Step 1:  1 => 1

Step 2:  1 => 1×(1/2)

Step 3: 1 => 1×(1/4)

Answer is :1.75.

Example 4:

To determine the decimal number in 11.112?

Solution:

Step 1: 1 => 1×2 (=2)

Step 2: 1 => 1

Step 3:  1 => 1×(1/2)

Step 4: 1 =>  1×(1/4)

So, 11.11 is 2+1+1/2+1/4 = 3.75 in Decimal

Answer is: 3.75.

Examples of Poisson Distribution

Definition: A random variable X is  a Poisson distribution if the probability mass function of X is P(X = x) =e−λ  λx / x!,                                       x = 0,1,2, …for some λ > 0

The mean of  Poisson Distribution denoted by λ, and the variance is denoted by λ.
The parameter of Poisson distribution is λ.

The Poisson distribution is a restrictive case of Binomial distribution under the following conditions.
(i)   Number of trials(n) is indefinitely huge(large), that is n → ∞.
(ii)  The constant probability(p) of success in each trial is very less.
ie., p → 0.
(iii) np = λ is finite where λ is a positive real number. When an event occurs rarely, the distribution of such event may be assumed to follow a Poisson distribution. The example problems of poisson distribution is   given below

Examples of Poisson distribution:

Some Examples of poisson distributions are given below

(1) The number of gamma particles emitted by a radio active source in a given time interval.
(2) The number of phone calls received at a telephone exchange in a given time interval.
(3) The number of defective articles in a packet of 250, produced by a good industries limited.
(4) The number of printing errors at each page of a book by a good publication centre.
(5) The number of road accidents reported in a city at a particular time.

Example Problems for Poission distribution:

Example problem 1: If a publisher of  technical books takes a great pain to ensure that his books are free of typological errors, so that the probability of any given page containing atleast one such error is 0.005 and errors are independent from page to page

(i) what is the probability of its 400 page novels will contain exactly one page with error.

(ii) atmost three pages with errors.
[e−2 = 0.1353 ; e−0.2. = 0.819].


Solution :

n = 400 , p = 0.005
np = 2 = λ
(i)  P(one page with error) = P(X = 1) = e−λ λ1/1! = e-2 21/1!
= 0.1363 × 2 = 0.2726
(ii)  P(atmost 3 pages with error) = P(X ≤ 3)
= Σ e−λ λx / x!    [limits 0 to 3]
= `sum` e−2 (2)x  /  x!
= e2 [1 +2/1! + 22/2! + 23/3!]
= e−2 (19/3 )= 0.8569