Introduction:
Two different quantities having different percentage of needed substances can be mixed together to get the desired amount. This is called mix word problem in math. For example: Let X and Y be two different solutions. Each are having different percentage of Salt in it. Those two X and Y can be mixed together to get the required percentage of the final solutions.
Now let us see few problems of this kind.
Example Problems on Mix Word Problems Math:
Ex 1: Two salt solutions having 12% and 36% of salts respectively in them. They are added together to form 40% of salt solution. If 280 liters of 12% salt solution is added, how many liters of 36% has to be added?
Soln: Let x be the salt solution.
Given: `36/100` x + `12/100` `xx` 280 = `40/100` (X + 80)
`=>` 36x + 3360 = 40x +3200
`=>` 40x – 36x = 3360 – 3200
`=>` 4x = 160 `=>` x = 40 liters.
Ex 2: Two oils have to be mixed together. One of the oil has 15% of the required content and the other has 30% of the required content. How many liters of 30% of oil have to be added to 290 liters of 15% of the oil, so that we can have a mix of 42% of the oil content?
Soln: Let X be the amount of oil with 30% content in it.
Given: `30/100` (x) + `15/100` (290) = `42/100` (x + 90)
`=>` 30x + 15`xx` 290 = 42x + 42`xx` 90
`=>` 12x = 4350 – 3780
`=>` x = `570/12` = 47.50 liters
More Example Problem on Mix Word Problems Math:
Ex 3: Two herbals have to be mixed together. One of them has 25% of the required vitamins and the other has 35% of the required vitamins. How many liters of 25% of herbal have to be added to 120 liters of 45% of herbal, so that we can have a mix of 40% of vitamins?
Soln: Let x be the amount of herbal with 25% vitamin in it.
Given: `25/100` (x) + `45/100` (120) = 40/100 (x+120)
`=>` 25x + 5400 = 40x + 4800
`=>` 15x = 5400 – 4800 = 600 `=>`x = `600/15`
x = 40liters
Thursday, September 5, 2013
Friday, June 7, 2013
Horizontal Lines in Math
Introduction to Horizontal Lines in Math:
Horizontal lines in math are used for arithmetic operations which denotes the next step in this operations. Generally it does not have any meaning or symbols or notation or formula but it is expressed for easy understanding of the problems. In every math problems horizontal lines are present especially with the dividing symbol. In fraction, horizontal lines divides numerator and denominator in math. A horizontal line in math separates the equations when substituting the equation steps.
Applications of Horizontal lines in math:
In all math problems we denote horizontal lines.Some of the applications are,
+14
26
Horizontal lines in math are used for arithmetic operations which denotes the next step in this operations. Generally it does not have any meaning or symbols or notation or formula but it is expressed for easy understanding of the problems. In every math problems horizontal lines are present especially with the dividing symbol. In fraction, horizontal lines divides numerator and denominator in math. A horizontal line in math separates the equations when substituting the equation steps.
Applications of Horizontal lines in math:
In all math problems we denote horizontal lines.Some of the applications are,
- In addition we are using horizontal lines when we adding numbers we expressed the horizontal lines for sum answers
+14
26
- In subtraction we apply as same as addition for the subtracted answers
Example 20
-12
8
-12
8
- In multiplication horizontal lines express for the multiply value with the multiplicand
Examples 12
` xx ` 14
48
12
168
` xx ` 14
48
12
168
- In division horizontal lines are shows the divide symbols which denotes quotient and steps involved in the division.
Example 12 by 3
`12/3` =4
`12/3` =4
- In fraction we denotes the horizontal lines for separations of numerator and denominator by use of horizontal lines.
Example` 5/3, 4/3` here 5 is the numerator and 4 is denominator.
- In algebraic equations when we equating two equations horizontal lines express the next step for equations in both elimination and substitution method we have horizontal lines for equations.
Example 2x+y =5 .........(1)
x-y =4........(2)
3x =9
x=`9/3 `
x=3
substitute x=3 in (2)
3-y =4
y=`4/3`
Example Problems for Horizontal lines in math:
Example 1:
Add 24 and 10.
Solution:
24
+ 10
34
Example 2:
Subtract -24 and 10
Solution:
24
- 10
14
Example 3:
Multiply 10 ` xx`
Solution:
10
`xx` 4
40
Example 4:
Divide 25 ÷ 5
Solution:
`25/5` = 5
Example 5:
Simplify `4/3+ 2/3`
Solution:
`4/3+2/3= 6/3`
Example 6:
solve x+y =3
2x-y =0
Solution:
x+y =3 .........(1)
2x-y =0........(2)
3x =3
x=`3/3 `
x=1
substitute x=1 in (2)
1-y =4
y=`4/1`
y=4
x-y =4........(2)
3x =9
x=`9/3 `
x=3
substitute x=3 in (2)
3-y =4
y=`4/3`
Example Problems for Horizontal lines in math:
Example 1:
Add 24 and 10.
Solution:
24
+ 10
34
Example 2:
Subtract -24 and 10
Solution:
24
- 10
14
Example 3:
Multiply 10 ` xx`
Solution:
10
`xx` 4
40
Example 4:
Divide 25 ÷ 5
Solution:
`25/5` = 5
Example 5:
Simplify `4/3+ 2/3`
Solution:
`4/3+2/3= 6/3`
Example 6:
solve x+y =3
2x-y =0
Solution:
x+y =3 .........(1)
2x-y =0........(2)
3x =3
x=`3/3 `
x=1
substitute x=1 in (2)
1-y =4
y=`4/1`
y=4
Thursday, June 6, 2013
Step Functions In Math
Introduction to step function:
A step function in math is one of the concepts of functions in math. The line is the collection of x and y points. At the same time x and y value can be increased values. We can plot this points on the graph we can get the equation of the line. That forms the step figure. So these functions are known as step functions.
Definition of Step functions in math
Definition of Step function:
A function in the real analysis R can be defined as the step function it can be written as linear combination of semi open intervals. Open interval can be defined as [a, b) in R. Then the step function f(x) can be described as,
f(x) =a1 f1(x) + a2 f2(x) + …………………… + an fn(x)
Brief Description of the Heaviside Step functions in math
This involves the following steps.
They are,
Step 1: First we can form the equation into y = m x + b form.
Step 2: Substitute any value for x we can get the y values.
Step 3: Plot the (x, y) points on the graph.
Step 4: This line plot increases sequentially.
Step 5: It look like as a step format.
These are the main steps used in math step functions.
Consider the equation of the line y = x -1
Substitute the values for x we can get the y values.
Substitute x=1 in the given equation y = x -1.
That is y= 1 – 1 = 0
Therefore the (x, y) co-ordinate is (1, 0)
Substitute x=2 in the given equation y = x -1.
That is y= 2 – 1 = 1
Therefore the (x, y) co-ordinate is (2, 1)
Substitute x=0 in the given equation y = x -1.
That is y= 0 – 1 = -1
Therefore the (x, y) co-ordinate is (0, -1)
Substitute x=-1 in the given equation y = x -1.
That is y= =-1 – 1 = -2
Therefore the (x, y) co-ordinate is (-2, -2)
Substitute x=-2 in the given equation y = x -1.
That is y= -2 – 1 = -3
Therefore the (x, y) co-ordinate is (-1, -3)
We can tabulate these points
X: -2 -1 0 1 2
Y: -3 -2 -1 0 1
Here the x and y values are correspondingly varies. So this function is namely known as step function.
The graph of the step function in math described as
This is also known as Heaviside step functions in math.
A step function in math is one of the concepts of functions in math. The line is the collection of x and y points. At the same time x and y value can be increased values. We can plot this points on the graph we can get the equation of the line. That forms the step figure. So these functions are known as step functions.
Definition of Step functions in math
Definition of Step function:
A function in the real analysis R can be defined as the step function it can be written as linear combination of semi open intervals. Open interval can be defined as [a, b) in R. Then the step function f(x) can be described as,
f(x) =a1 f1(x) + a2 f2(x) + …………………… + an fn(x)
Brief Description of the Heaviside Step functions in math
This involves the following steps.
They are,
Step 1: First we can form the equation into y = m x + b form.
Step 2: Substitute any value for x we can get the y values.
Step 3: Plot the (x, y) points on the graph.
Step 4: This line plot increases sequentially.
Step 5: It look like as a step format.
These are the main steps used in math step functions.
Consider the equation of the line y = x -1
Substitute the values for x we can get the y values.
Substitute x=1 in the given equation y = x -1.
That is y= 1 – 1 = 0
Therefore the (x, y) co-ordinate is (1, 0)
Substitute x=2 in the given equation y = x -1.
That is y= 2 – 1 = 1
Therefore the (x, y) co-ordinate is (2, 1)
Substitute x=0 in the given equation y = x -1.
That is y= 0 – 1 = -1
Therefore the (x, y) co-ordinate is (0, -1)
Substitute x=-1 in the given equation y = x -1.
That is y= =-1 – 1 = -2
Therefore the (x, y) co-ordinate is (-2, -2)
Substitute x=-2 in the given equation y = x -1.
That is y= -2 – 1 = -3
Therefore the (x, y) co-ordinate is (-1, -3)
We can tabulate these points
X: -2 -1 0 1 2
Y: -3 -2 -1 0 1
Here the x and y values are correspondingly varies. So this function is namely known as step function.
The graph of the step function in math described as
This is also known as Heaviside step functions in math.
Monday, June 3, 2013
Live Online Chat About Math
Introduction to live online chat about math:
Live online chat about math is nothing but the help to students by tutors of live tutoring website through online. Many students find difficult in doing the math homework problems. Parents also don't find time to help their students with homework problems. So they were in need of a website that provide live online chat about math. Tutor vista is a website to get help in math anytime. Many students find maths help vista website with the highly qualified tutors live at any time. Below are some of the model problems solved by tutors for live online chat about math.
Live online chat about math:
1. Add 322 + 343 + 346
Solution
The given is an addition problem with three 3 digit numbers. Addition is the basic arithmetic operations
3 2 2
3 4 3
3 4 6 +
---------
10 1 1
---------
2. Subtract the numbers given 444 – 240
Solution
Similar to add start the subtract with the ones digit first
4 4 4
2 4 0 –
----------
2 0 4
---------
3. Solve and find the value of b 7b+3 = 24
Solution
This is an algebra problem. Here we have to find the value of b.
7b + 3 = 24
Subtract 3 on both sides
7b+3 -3 = 24 – 3
7b = 21
Divide by 7 on both sides to find the value of b
`(7b)/7` =`21/7`
b= 3
4. Solve 10 + 4 x 3 + 22-(1x4)
Solution
Here more than one arithmetic operation is given in the problem. So we perform this problem by using arithmetic operation.
10 + 4x3+22 - 4-------Parenthesis
10 + 4x3 + 4 – 4 -----Exponents
10 + 12 + 4 - 4 ------Multiplication
22+ 4 - 4 ------------No division so we do addition
22 ------------Subtraction
So the answer is 22
Live online chat about math:
Add 868 + 325 + 110
Subtract 875-340
Find the value of s in the equation shown below 4s – 6 = 26
By using PEMDAS rule perform 14 + 2 – 3 +(4x5) – 20
5. Find the value of u in the equation 3p = 42
Answers:
1303
535
s = 8
13
p = 14
Live online chat about math is nothing but the help to students by tutors of live tutoring website through online. Many students find difficult in doing the math homework problems. Parents also don't find time to help their students with homework problems. So they were in need of a website that provide live online chat about math. Tutor vista is a website to get help in math anytime. Many students find maths help vista website with the highly qualified tutors live at any time. Below are some of the model problems solved by tutors for live online chat about math.
Live online chat about math:
1. Add 322 + 343 + 346
Solution
The given is an addition problem with three 3 digit numbers. Addition is the basic arithmetic operations
3 2 2
3 4 3
3 4 6 +
---------
10 1 1
---------
2. Subtract the numbers given 444 – 240
Solution
Similar to add start the subtract with the ones digit first
4 4 4
2 4 0 –
----------
2 0 4
---------
3. Solve and find the value of b 7b+3 = 24
Solution
This is an algebra problem. Here we have to find the value of b.
7b + 3 = 24
Subtract 3 on both sides
7b+3 -3 = 24 – 3
7b = 21
Divide by 7 on both sides to find the value of b
`(7b)/7` =`21/7`
b= 3
4. Solve 10 + 4 x 3 + 22-(1x4)
Solution
Here more than one arithmetic operation is given in the problem. So we perform this problem by using arithmetic operation.
10 + 4x3+22 - 4-------Parenthesis
10 + 4x3 + 4 – 4 -----Exponents
10 + 12 + 4 - 4 ------Multiplication
22+ 4 - 4 ------------No division so we do addition
22 ------------Subtraction
So the answer is 22
Live online chat about math:
Add 868 + 325 + 110
Subtract 875-340
Find the value of s in the equation shown below 4s – 6 = 26
By using PEMDAS rule perform 14 + 2 – 3 +(4x5) – 20
5. Find the value of u in the equation 3p = 42
Answers:
1303
535
s = 8
13
p = 14
Friday, May 31, 2013
Define Hypothesis Math Term
Define hypothesis math term:
The term hypothesis refers the condition statement just after the word if. In the given conditional statement.
For example Suppose the condition is “in the given quadrilateral figure - four sides are equal in measure then the shape is said to be square” the hypothesis of the given statement is “all four sides are equal in measure. In the given example,” all four sides are equal in measure. “Always, never or sometimes true.
Options:
1)Always
2)Never
3)Sometimes.
Answer is (3)
Solution:
Answer 3) sometimes it is correct because, it’s true if only the given shape is square, In case of other shapes like rectangle, rhombus then condition is not satisfied.
Example 1)
Find the math term hypothesis of the given equation.
If x = 1 then x2 +3x +5 = 9
Option:
1)True
2)false
Answer : 1)True.
Solution:
Substitute x value in the given equation,
X+3x+5 =9
1+3(1) +5 = 9
1+3+5 =9
9=9
Therefore the given condition is true.
Example 2:
Find the math term hypothesis in t he given statement is “Every integer greater than 1 is product of prime number “
Solution:
Let us consider the if x is true statement .
Therefore ,
X`=>` y is a true statement and y is a true conditional statement.
Now,we need to find the series of the statement x1,x2 …..etc.,and then verifies the given conditional statement” Every integer greater than 1 is product of prime number”.
X=x1
X1=x2,
X2=x3,
….
….
Xn-1 =xn
Xn =Y are ture.
Because assume that the given statement X is true then all repeated statement of the simple rule Y is true..
Example 3:
The conditional statement is “if n is an integer then n2 is even and then n is odd.”
Solution:
N is odd number .
so n-1 is even number .
Therefore , we have,
n-1 =2k
where ,k is the constant .
n-1 = 2k ,k `in` Z
n=2k+1
Similarly we have by hypothesis,
n2 =x; x `in` Z
And then,
n2 = 4k2 +4k+1
=2(2k2 +2k)+1
In the above equation, 2(2k2+2k) is even number.
So 2(2k2+2k) +1 is the odd number.Therefore our hypothesis is violated.
So the integer number “n “ must be even number.
So,
X`=>` S =” n2 is even ”
Y `=>` ” n is even ”
Not –Y `=>` ” n is odd ”
Practice problems related to the term hypothesis in math :
1) Prove that if m and n are even integers, then n +m is even.
2) Prove that if n is an odd integer, then n 2 is an odd integer.
The term hypothesis refers the condition statement just after the word if. In the given conditional statement.
For example Suppose the condition is “in the given quadrilateral figure - four sides are equal in measure then the shape is said to be square” the hypothesis of the given statement is “all four sides are equal in measure. In the given example,” all four sides are equal in measure. “Always, never or sometimes true.
Options:
1)Always
2)Never
3)Sometimes.
Answer is (3)
Solution:
Answer 3) sometimes it is correct because, it’s true if only the given shape is square, In case of other shapes like rectangle, rhombus then condition is not satisfied.
Example 1)
Find the math term hypothesis of the given equation.
If x = 1 then x2 +3x +5 = 9
Option:
1)True
2)false
Answer : 1)True.
Solution:
Substitute x value in the given equation,
X+3x+5 =9
1+3(1) +5 = 9
1+3+5 =9
9=9
Therefore the given condition is true.
Example 2:
Find the math term hypothesis in t he given statement is “Every integer greater than 1 is product of prime number “
Solution:
Let us consider the if x is true statement .
Therefore ,
X`=>` y is a true statement and y is a true conditional statement.
Now,we need to find the series of the statement x1,x2 …..etc.,and then verifies the given conditional statement” Every integer greater than 1 is product of prime number”.
X=x1
X1=x2,
X2=x3,
….
….
Xn-1 =xn
Xn =Y are ture.
Because assume that the given statement X is true then all repeated statement of the simple rule Y is true..
Example 3:
The conditional statement is “if n is an integer then n2 is even and then n is odd.”
Solution:
N is odd number .
so n-1 is even number .
Therefore , we have,
n-1 =2k
where ,k is the constant .
n-1 = 2k ,k `in` Z
n=2k+1
Similarly we have by hypothesis,
n2 =x; x `in` Z
And then,
n2 = 4k2 +4k+1
=2(2k2 +2k)+1
In the above equation, 2(2k2+2k) is even number.
So 2(2k2+2k) +1 is the odd number.Therefore our hypothesis is violated.
So the integer number “n “ must be even number.
So,
X`=>` S =” n2 is even ”
Y `=>` ” n is even ”
Not –Y `=>` ” n is odd ”
Practice problems related to the term hypothesis in math :
1) Prove that if m and n are even integers, then n +m is even.
2) Prove that if n is an odd integer, then n 2 is an odd integer.
Thursday, May 30, 2013
Find Domain off
Introduction of Domain function
The domain of definition or simply the domain of a function is the set of "input" or argument values for which the function is defined. That is, the function provides an "output" or "value" for each member of the domain. The domain of cosine is the set of all real numbers, while the domain of the square root consists only of numbers greater than or equal to 0 when the Function is represented in an xy
Examples
Problem 1: Find the domain of the real valued linear function f given below.
f(x) = x + 1
Substitute x=1, 2,3,4,5
x=1 f(x) =1+1
By solving it we get
f(x) =2
The ordered pair is (1, 2)
x=2 f(x) =2+1
By solving it we get
f(x) =3
The ordered pair is = (2, 3)
x=3 f(x) =3+1
By solving it we get
f(x) =4
The ordered pair is = (3, 4)
x=4 f(x) =4+1
By solving it we get
f(x) =5
The ordered pair is = (4, 5)
x=5 f(x) =5+1
By solving it we get
f(x) =6
The ordered pair is = (5, 6)
The domain function is 1, 2,3,4,5
Problem 2: Find the domain of the real valued linear function f given below.
f(x) = 2x + 1
Substitute x=1, 2,3,4,5
x=1 f(x) =2(1)+1
By solving it we get
f(x) =3
The ordered pair is (1, 3)
x=2 f(x) =(2*2)+1
By solving it we get
f(x) =5
The ordered pair is = (2, 5)
x=3 f(x) = (3*2)+1
By solving it we get
f(x) =7
The ordered pair is = (3, 7)
x=4 f(x) =(4*2)+1
By solving it we get
f(x) =9
The ordered pair is = (4, 9)
x=5, f(x) =(5*2)+1
By solving it we get
f(x) =11
The ordered pair is = (5, 11)
The domain function is 1, 2,3,4,5
Practice problem
Problem : Find the domain of the real valued linear function f given below.
f(x) = 2x + 5
The domain of definition or simply the domain of a function is the set of "input" or argument values for which the function is defined. That is, the function provides an "output" or "value" for each member of the domain. The domain of cosine is the set of all real numbers, while the domain of the square root consists only of numbers greater than or equal to 0 when the Function is represented in an xy
Examples
Problem 1: Find the domain of the real valued linear function f given below.
f(x) = x + 1
Substitute x=1, 2,3,4,5
x=1 f(x) =1+1
By solving it we get
f(x) =2
The ordered pair is (1, 2)
x=2 f(x) =2+1
By solving it we get
f(x) =3
The ordered pair is = (2, 3)
x=3 f(x) =3+1
By solving it we get
f(x) =4
The ordered pair is = (3, 4)
x=4 f(x) =4+1
By solving it we get
f(x) =5
The ordered pair is = (4, 5)
x=5 f(x) =5+1
By solving it we get
f(x) =6
The ordered pair is = (5, 6)
The domain function is 1, 2,3,4,5
Problem 2: Find the domain of the real valued linear function f given below.
f(x) = 2x + 1
Substitute x=1, 2,3,4,5
x=1 f(x) =2(1)+1
By solving it we get
f(x) =3
The ordered pair is (1, 3)
x=2 f(x) =(2*2)+1
By solving it we get
f(x) =5
The ordered pair is = (2, 5)
x=3 f(x) = (3*2)+1
By solving it we get
f(x) =7
The ordered pair is = (3, 7)
x=4 f(x) =(4*2)+1
By solving it we get
f(x) =9
The ordered pair is = (4, 9)
x=5, f(x) =(5*2)+1
By solving it we get
f(x) =11
The ordered pair is = (5, 11)
The domain function is 1, 2,3,4,5
Practice problem
Problem : Find the domain of the real valued linear function f given below.
f(x) = 2x + 5
Tuesday, May 28, 2013
Vista Math Help
Introduction to vista math help:
Tutorvista is one of the online tutoring services providers; Tutorvista provides helps in various subjects like English, math, and science. Tutor vista helps at any time around the clock regardless of place where they are located. Student can learn and do homework from their home at any time. Here tutor to explain step by step so that the students can easily understand much easier. In this article we shall discuss tutorvista help in math problems.
Tutorvista math help example problem
Example 1:
Sum the polynomial 5x4 – 6x2 + 9x + 5 and 5x + 6x3 – 3x2 – 1.
Solution:
Use the associative property and distributive properties add real numbers
(5x4 – 6x2 + 9x + 5) + (6x3 – 3x2 + 5x – 1) = 5x4 + 6x3 – 6x2 – 3x2 + 9x + 5x + 5 – 1
= 5x4 + 6x3 – (6+3)x2 + (9+5)x + 4
= 5x4 + 6x3 – 9x2 + 14x + 4.
An another method for adding two polynomials
5x4 +0x3– 6x2 + 9x + 5
0x4 + 6x3 – 3x2 + 5x – 1
____________________
8x4 + 6x3– 9x2 + 14x + 4
____________________
Example 2:
Consider e1 and e2 are two events with a random trial such that P(E2)=0.25, P(E1 or E2)= 0.95 and P(E1 and E2) = 0.35, find the value of P(E1).
Solution:
Let P (E1) = x then,
P (E1 or E2) = P (E) + P (E2) – P (E1 and E2)
0.95 = x + 0.25 – 0.35
Adding the trial value
x = (0.95-0.25 + 0.35) = 0.65
Hence the value of P (E1) = 1.05
Example 3:
Find the increases percentage value from 35 to 50.5?
Solution:
Changes in value = new value – original value
50.5 – 35 = 15.5
Increased percentage = `15.5 / 35` x100%
Answer:
The increased percentage value is = 44.2%
Tutorvista math help practice problem
Problem:
Adding the polynomial 8x4 – 5x2 + 9x + 4 and 4x + 6x3 – 3x2 – 1.
Answer: 8x4 + 6x3– 8x2 + 13x + 3
Problem:
Find how many percentage increases from 30 to 55.5?
Answer: 85%
Problem:
Consider e1 and e2 are two events with a random trial such that P(E2)=0.35, P(E1 or E2) = 0.75 and P(E1 and E2) = 0.45, find the value of P(E1).
Answer:
P (E1) = 0.85
Tutorvista is one of the online tutoring services providers; Tutorvista provides helps in various subjects like English, math, and science. Tutor vista helps at any time around the clock regardless of place where they are located. Student can learn and do homework from their home at any time. Here tutor to explain step by step so that the students can easily understand much easier. In this article we shall discuss tutorvista help in math problems.
Tutorvista math help example problem
Example 1:
Sum the polynomial 5x4 – 6x2 + 9x + 5 and 5x + 6x3 – 3x2 – 1.
Solution:
Use the associative property and distributive properties add real numbers
(5x4 – 6x2 + 9x + 5) + (6x3 – 3x2 + 5x – 1) = 5x4 + 6x3 – 6x2 – 3x2 + 9x + 5x + 5 – 1
= 5x4 + 6x3 – (6+3)x2 + (9+5)x + 4
= 5x4 + 6x3 – 9x2 + 14x + 4.
An another method for adding two polynomials
5x4 +0x3– 6x2 + 9x + 5
0x4 + 6x3 – 3x2 + 5x – 1
____________________
8x4 + 6x3– 9x2 + 14x + 4
____________________
Example 2:
Consider e1 and e2 are two events with a random trial such that P(E2)=0.25, P(E1 or E2)= 0.95 and P(E1 and E2) = 0.35, find the value of P(E1).
Solution:
Let P (E1) = x then,
P (E1 or E2) = P (E) + P (E2) – P (E1 and E2)
0.95 = x + 0.25 – 0.35
Adding the trial value
x = (0.95-0.25 + 0.35) = 0.65
Hence the value of P (E1) = 1.05
Example 3:
Find the increases percentage value from 35 to 50.5?
Solution:
Changes in value = new value – original value
50.5 – 35 = 15.5
Increased percentage = `15.5 / 35` x100%
Answer:
The increased percentage value is = 44.2%
Tutorvista math help practice problem
Problem:
Adding the polynomial 8x4 – 5x2 + 9x + 4 and 4x + 6x3 – 3x2 – 1.
Answer: 8x4 + 6x3– 8x2 + 13x + 3
Problem:
Find how many percentage increases from 30 to 55.5?
Answer: 85%
Problem:
Consider e1 and e2 are two events with a random trial such that P(E2)=0.35, P(E1 or E2) = 0.75 and P(E1 and E2) = 0.45, find the value of P(E1).
Answer:
P (E1) = 0.85
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