Write a Mixed Decimal

Mixed number: Mixed number is formed by associating a whole number with fraction numbers.

Decimal number: A number along decimal a point in it is called decimal number.

Mixed decimal number:  A number having a decimal point in its mixed format is mixed decimal number.

Example: 7.5 `3/6`.  Let us see how to write mixed decimal number.


Writing a mixed decimal:

Write a mixed decimal:

In the whole number part of mixed number the decimal point appears show that the number is mixed decimal.

It should be written as,

133.63 `1/5`

Conversion of mixed decimal into fraction:

Step 1: First see how many numbers present after the decimal point.

Step 2: Calculate number of 10’s equal to the counted digits after the decimal point.

Step 3: Put number of 10’s as denominator and the number is mixed decimal number.

Step 4: Convert the numerator mixed decimal number into fraction.

Step 5: Now simplify the terms by a common term.

Example problems for writing a mixed decimal:

Example: 1

Write the following mixed decimal number into fraction number.

12.5 2/3

Solution:

Given, mixed decimal number =12.5 2/3.

We have to add the 10 as the denominator and the mixed decimal number as numerator.

That is, `(125 2/3)/10`

= `(377/3)/10`

= `377/30`

=12`17/30`

Example: 2

Write the following mixed decimal number into fraction number.

5.22 1/2

Solution:

Given, mixed decimal number =5.22 1/2.

We have to add the 100 as the denominator and the mixed decimal number as numerator.

That is, `(522 1/2)/100`

= `(1045/2)/100`

= `1045/200`

=`209/20`

=1`9/20` .

Example: 3

Write the following mixed decimal number into fraction number.

15.123 1/5

Solution:

Given, mixed decimal number =15.123 1/5.

We have to add the 1000 as the denominator and the mixed decimal number as numerator.

That is, `(15123 1/5)/1000`

= `(75616/5)/1000`

= `75616/5000`

= 25 `616/5000`

= 25 `77/625` .

Answer: 25 `77/625` .

Practice problems for write a mixed decimal number:

Problem: 1

Write the following mixed decimal into mixed number.

5.9 3/4

Answer: 5`39/40`.

Problem: 2

Write the following mixed decimal into mixed number.

5.633 1/2 .

Answer: 5`1267/2000`.

Hypothesis Testing Variance

Hypothesis testing is the use of statistics  found in the probability that a specified hypothesis is correct. Hypothesis is specified as declaration which may or may not be accurate. In statistics two hypothesis testing are used. They are null hypothesis and alternative hypothesis. These two hypothesis tested are opposed to each other. In statistics the significance level is symbolized through alpha. Let us see about the probability of hypothesis testing variance.

Hypothesis testing variance

There are five constituent to either statistical test:

Null Hypothesis

Alternate Hypothesis

Test Statistic

Level of significance

Conclusion

Hypothesis testing variance

Consider the population is standard, we are able to test the variance of the method using the chi-square distribution through (n – 1) degrees of freedom.

To test a variance or standard deviation of a population to be exact normally distributed, we can utilize the χ2–test.

The χ2- test for a variance or standard deviation is not as robust as the samples for the population mean otherwise the population proportion.

Therefore it is necessary to while performing the χ2–test used for a variance that the population is usually circulated. The results can be deceptive if the population is not standard.

Examples for hypothesis testing variance

χ2- test for population variance

In a sample of size 16 drawn from a normal population standard deviation is 4 can you say that population standard deviation is 5.

Solution

Null hypothesis:

H0: The population standard deviation is 5

Test statistic:

χ2   =`(ns^2)/sigma^2` ~ χ2n-1

Level of significance:

α= 0.05 at 5% level of χ2  table values for 16 degrees of freedom is 24.996

Calculation:

n = 16, s = 4, σ = 5

χ2   =`(16xx16)/25 = 256/25`

χ2   = 10.24

Calculated value = 10.24

Table value = 24.996

Calculated value < Table value

Therefore the null hypothesis is accepted.

The population standard deviation is 5

Result

The population standard deviation is 5

Whole Set Fraction

This article is about whole set fraction. Whole set fraction is nothing but it is about the set of fractions. A fraction is a number that can be represented by an ordered pair of whole numbers `a/b` where `b!= 0`. Here a is represented as numerator and b as denominator. The tutors in tutor vista are always ready to help the students in any topics like whole set fraction. The online tutors help students in online. Tutor vista is the best tutoring website where many students follow this. Below we can see about the whole set of fraction.

whole set fractions

In set notation, the set of fractions is

F ={ `a/b` where a and b are whole numbers,  b` !=` 0 }

Two fractions that represent the same relative amount are termed to be equivalent fractions.

Proper Fraction:

When the numerator is less than the denominator then those fractions are called as Proper fraction.

Example: `2/3 `

Improper fraction:

When the numerator is greater than the denominator then this fraction is called as Improper fraction

Example: `7/5`

All the integers are simply a improper fraction

Example

3 is nothing but `3 / 1` which is an improper fraction

Mixed fraction:

Mixed fraction is a whole number with proper fraction

Example:  2 `1/3`

whole set fractions

Example 1:

Add `5/3`  + ` 8/3`

Here both the denominator is same

Add numerator alone.

`(5+8)/3`

`13/3`

Adding improper fraction with different denominator

Example 2:

Add `5/2`  + `4/3`

Here find LCM and solve to make the denominator equal

LCM of 2 and 3 is 6

`(5xx3)/ (2xx3) = 15/6`

`(4xx2)/ (3xx2) = 8/6`

So `(15+8)/6 = 23/6`

whole set fractions

Example 3: Add `3/11` +`6/11`

Solution

Here the denominators are same. So just add the numerator alone

` (3+6)/11`

`9/11` is the answer.

Example 4: Subtract `8/9` – `4/9`

Solution

Here the denominators are same. So just subtract the numerator alone.

`(8 - 4)/9`

`4/9`

Example 5: Multiply` 3/5` x `3/ 4 `

Solution

`3 / 5 ` x `3 / 4`

`(3xx3)/(5xx4) `

= `9 / 20 `

Addition Angle Formulas

Addition angle formula is based on trigonometric functions. We are having  the addition angle formulas to find the value of  the trigonometric equations and values for the trigonometric angles.
                         Cos (A+B) = Cos A Cos B - Sin A Sin B
                         Cos (A-B) = Cos A Cos B + Sin A Sin B
                         Sin (A+B) = Sin A Cos B + Cos A Sin B
                         Sin (A-B) = Sin A Cos B - Cos A Sin B
                         Tan (A+B) = `(Tan A + Tan B)/(1 - Tan A Tan B)`
                         Tan (A-B) =  `(Tan A - Tan B)/(1 + Tan A Tan B)`
                         Here we will some problems based on addition angle formulas.

Addition Angle Problems:

Problem 1:
         Solve the following trigonometric function using Addition angle formula Sin 75^o
Solution:
            Sin 75^o
                 We can write sin 75^o as Sin (45^o + 30^o)
                              We have the formula for Sin (A+B) = Sin A Cos B + Cos A Sin B
                              Where A = 45^o and  B = 30^o
                              Sin 45^o = Cos 45 = `(1)/(sqrt(2))`
                              Sin 30 = `(1)/(2)`    Cos 30 = `(sqrt(3))/(2)`
                              Sin (45^o + 30^o) = Sin 45^o Cos 30^o + Cos 45^o Sin 30^o
                                                          =  `(1)/(sqrt(2))` `(sqrt(3))/(2)` `(1)/(sqrt(2))` `(1)/(2)`
                                                          = `(sqrt(3))/(2sqrt(2))` + `(1)/(2sqrt(2))`
                                                          = `(sqrt(3)+1)/(2sqrt(2))`
Problem 2:
       Solve the following trigonometric function using Addition angle formula Cos 135^o
Solution:
               Cos 135^o
                We can write Cos 135^o as Sin (90^o + 45^o)
                              We have the formula for Cos (A+B) = Cos A Cos B - Sin A Sin B
                              Where A = 90^o and  B = 45^o
                              Sin 45^o = Cos 45 = `(1)/(sqrt(2))`
                              Sin 90^o = 1  Cos 90^o = 0
                              Cos (90^o + 45^o) = Cos 90^o Cos 45^o - Sin 90^o Sin 45^o
                                                      = 0 . `(1)/(sqrt(2))` - 1 . `(1)/(sqrt(2))`
                                                      = 0 - `(1)/(sqrt(2))`
                                                      = - `(1)/(sqrt(2))`

Problem 3:

       Solve the following trigonometric function using Addition angle formula Tan 135^o
Solution:
               Tan 135^o
               We can write Tan 135^o as Tan (180^o - 45^o)
                              We have the formula Tan (A - B) = `(Tan A - Tan B)/(1 + Tan A Tan B)`
                              Where A = 180^o and  B = 45^o
                              Tan 45^o = 1 Tan 180^o = 0
                              Tan 135^o = `(Tan 180^o - Tan 45^o)/(1 + Tan 180^o Tan 45^o)`
                              Tan 135^o = `"(0 - 1)/(1 + (0) (1)) `
                              Tan 135^o = `(- 1)/(1)`
                              Tan 135^o = -1

Compute Percentages

In mathematics, a percentage is a way of expressing a number as a fraction of 100 (per cent meaning "per hundred" in French). It is often denoted using the percent sign, "%", or the abbreviation "pct". For example, 45% (read as "forty-five percent") is equal to 45 / 100, or 0.45. Let us see how to do percentages.                                                                                                                                           - Source from Wikipedia

How to compute percentages:

STEP 1: While begin the percentage` x / 100 ` = `(is) / (of)` . Out of hundred x may be the percentage, "is" denotes as fraction, and "of" denotes as whole.
STEP 2: If we are having the questions 80 : 40 percentage then we will write it as X = 40, is = 40 ("80 is"), and then of = as unknown value. Now the value is possible to write like this `40 / 100 ` = `80 / x` .
STEP 3: Let us do the cross multiplication. Now we can get a Constant value on 1 side and then multiply it with another side. Now we will get a result likes this 40x = 8,000.
STEP 4: Now you have to find out the x value. Where, x = `8000 / 40` = 200, now the x value will be 200.

How do you solve percentages some examples here:

Problem 1:
           In a question paper there is 80 questions. Laura took that test. If she gets 75% correct, how many questions did Laura missed?
Solution:
            Therefore total correct answers are 75% of 80 or else `75 / 100` × 80
            ` 75 / 100` × 80 = 60%
            So the question paper contains 80 questions and Laura got 60 exact answers, the number of questions Laura left is 80 − 60 = 20.
            Therefore Laura missed 20 questions.
Problem 2:
            Compute this, what is 85% of 15?
Solution:
Step 1: Compute the percent.
           The percent value is 85.
            P = 85
Step 2: Find out the base.
           The base is the number next the word OF, 15
            b = 15
Step 3: Identify the quantity.
           The calculation is the unknown.
            a =?
Step 4: Enter the value in the percent proportion formula.
         `a/ 15` = `85 / 100`
Step 5: Exercises the equation for the unknown.
           The Least Common Divisible of 15 and 100 is 100
     `100 / 1` * `a / 15` = `100 / 1` * `85 / 100`
            6.6a = 85
          ` (6.6a) / (6.6)` = `85 / (6.6)`
            a = 12.8
            12.8 is 85% of 15.


Practice problem for compute percentages:
Problem 1:
            What is the percentage of 67%?
Solution:
             = 0. 67.
Problem 2:
            What is percentage of 87% of 18?
Solution:
             = 15.66
            Therefore 15.66 is 87% of 18.

Taylor Polynomial Series

In mathematics, the Taylor series is a representation of a function as an infinite sum of terms calculated from the values of its derivatives at a single point. It is named after the English mathematician Brook Taylor. If the series is centered at zero, the series is also called a Maclaurin series, named after the Scottish mathematician Colin Maclaurin. It is common practice to use a finite number of terms of the series to approximate a function. The Taylor series may be regarded as the limit of the Taylor polynomials.  In the article we shall discuss about Taylor polynomial series.(Source: Wikipedia)

Taylor polynomial series:

Find the Taylor series for sin x.

    Sin x = `sum_(0)^oo` `((-1)^k)/((2k + 1)!)` x^2k+1

The remainder term is not expressible in any simple way but can be estimated by using the Lagrange's form of the remainder. The coefficients

   `((-1)^k)/((2k + 1)!)`

are easily verified by calculating successive derivatives of f(x) = sin x and using the formulas

 a_k=`(f^(k)(0))/(k!)`

To check convergence of the series, apply Lagrange's form for R_a(x); For each x`in` R. there exists Z such that

R_n(x) = `(f^(n + 1)(z))/((n + 1)!)` xⁿ⁺¹

Now |fⁿ⁺¹(z)| equals either |cos z| or |sin z| So, in either case,|fⁿ⁺¹ (z)|`<=`1 ,and

       |R_n(x) |`<=` | x |ⁿ⁺¹ /(n+1)!

Since | x |ⁿ⁺¹ /(n+1)!`->`0 as n `|->` `oo`  for all x`in` R, we can see that the remainder term |R_n(x)|`|->` 0 as n `|->` `oo`

 for all x`in` R. Thus the series representation is completely justified for all real x.

  Observe that our estimate for |R_n (x)|,

                                                    |R_n (x)|`<=`|x|^_n+1/(n+1)!

gives also a sense of the rate of convergence of the series for fixed x, for example, for | x|`<=` 1, we find

                                                    |R_n (x)|`<=`1/(n+1)!

Thus, if we want to calculate sin x on (-1, 1) to within .01, we need take only the first five terms of the series (n = 4) to achieve that degree of accuracy.

 Had we used the integral form for R_n (x) we would have obtained a similar estimate.

Sample problem for Taylor polynomial series:

Pro:  Evaluate the definite integral `int_0^1` Sin (x) dx

Sol:  The integrand has no anti derivative expressible in the terms of familiar functions. Howebver, we know how to find its Taylor series. we know that

      Sin t = t -`(t^3)/(3!)` + `(t^5)/(5!)` - `(t^7)/(7!)` + ----

Now if we substitute t = x, we have

     Sin (x) = X - `(x^3)/(3!)` + `(x^5)/(5!)` - `(x^7)/(7!)` + ----

In spite of the fact that we cannot antidifferentiate the function, we can antidifferentiate the Taylor series:

    `int_0^1` Sin (x) dx = `int_0^1` (X - `(x^3)/(3!)` + `(x^5)/(5!)` - `(x^7)/(7!)` + ----) dx

                            =(`(x^2)/(2)` -`(x^)/(4*3!)` + `(x^6)/(6*5!)`-`(x ^ 7)/(8*7!)`+ -----) |₀¹

                            = (`(x^2)/(2)` -`(x^)/(4*3!)` + `(x^6)/(6*5!)`+`(x^7)/(8*7)` + ---)

Notice that this is an alternating series so we know that it converges. if we add up the first four terms, the pattern becomes ckear: the series converges to 0.2871

Negative Integer Exponents

The exponents are which integer is placed in the power, of base numbers. It can be easily represent as, “a small number to the right side and above of base number”. It is called as exponents. These exponents have some of important rules and laws. Power with negative integer exponents is also one of the rules. Here we are going to explain about this negative integer exponent rule.

If we are having variables, which is containing the exponents and it have equal bases means, we can do some mathematical operations with the exponents. Those operations are called as the “laws of exponents” or “rules of exponents”. In this rule based negative integer rule of exponent is defined as following ways,

Definition for negative integer exponents:

It is otherwise called as power with negative exponent rules. This negative exponent rule is defined as, if m is a positive integer and x is a non-zero rational number, then it can be denoted as,

X-m = (1/x)^m (or)

= (1/x)^m

Which is (x)^-m is the reciprocal of (x)^m

And we adopt the same rule for rational exponents also. If p/q is a positive rational number means, and x>0 is a rational number, then

 X^ - (p/q) = (1/x)^ (p/q) = (1/x)^ (p/q) .

Which is, (x)^-(p/q) is the reciprocal of (x) ^(p/q) or the number obtained by raising the reciprocal of x to the exponent p/q.

For example: 1). 3^-2

= (1/3)^2

= (1/3)^-2

= -6 .

2). (4)^-(2/3)

= (1/4)^-(2/3)

= (1/4)^ (2/3)

This kind of exponentiation used for discovers the negative integer exponents and simplify the problems.


Example problems for negative integer exponents:

1) Solve: (8)^-(2/3)

Solution:

Given: (8)^-(2/3)

= (1/8)^ (2/3)

= [(1/8)^ (1/3)]^ 2

= (1/2)^2, since (1/2)^3 = 1/8

= 1/4 .

2) Solve: (32/243)^-(4/5)

Solution:

Given: (32/243)^-(4/5)

= (243/32)^(4/5)

= [(243/32)^(1/5)]^4

= [(3^5/2^5)^(1/5)]^4

= [((3/2)^5)^(1/5)]^4

= (3/2)^4

= 81/16.

3) Evaluate, and find the following negative integer exponent value:

Evaluate: (27/125)^-(2/3)  xx    (27/125)^-(4/3)

Solution:

 (27/125)^- (2/3) xx (27/125)^-(4/3)

= (125/25)^(2/3) xx (125/27)^(4/3)

= [(5^3/3^3)^(1/3)]^ 2 xx [(5^3/3^3)^(1/3)]^4

= [((5/3)^3)^(1/3)]^2 xx [((5/3)^3)^(1/3)]^4

= (5/3)^2 xx (5/3)^4

= (5/3)^6

= 15625/729.

These all are the explanations and example problems may clear about the negative integer exponents.