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3 years ago

What is 1.83 repeating as a fraction?

Mathematics

2 answers:

Alina [70]3 years ago

8 0

<h3>Further explanation</h3>

Firstly , let us learn about types of sequence in mathematics.

Arithmetic Progression is a sequence of numbers in which each of adjacent numbers have a constant difference.

$T_n = a + (n-1)d$

$S_n = \frac{1}{2}n ( 2a + (n-1)d )$

<em>Tn = n-th term of the sequence</em>

<em>Sn = sum of the first n numbers of the sequence</em>

<em>a = the initial term of the sequence</em>

<em>d = common difference between adjacent numbers</em>

<em />

Geometric Progression is a sequence of numbers in which each of adjacent numbers have a constant ration.

$T_n = a ~ r^{n-1}$

$S_n = \frac{a( 1 - r^n ) }{1 - r}$

<em>Tn = n-th term of the sequence</em>

<em>Sn = sum of the first n numbers of the sequence</em>

<em>a = the initial term of the sequence</em>

<em>r = common ratio between adjacent numbers</em>

Let us now tackle the problem!

Since the question is not clearly stated, then let me assume the problem is to convert 1.83838383... as a fraction.

Let :

X = 1.8383838383... = 1.<u>83</u>

100X = 183.83838383... = 183.<u>83</u>

100X - X = 183.<u>83</u> - 1.<u>83</u>

99X = 182

X = 182 ÷ 99

$\large {\boxed {X = \frac{182}{99} } }$

If the problem is to convert 1.833333... as a fraction , then :

Let :

X = 1.833333... = 1.8<u>3</u>

10X = 18.33333... = 18.<u>3</u>

100X = 183.3333... = 183.<u>3</u>

100X - 10X = 183.<u>3</u> - 18.<u>3</u>

90X = 165

X = 165 ÷ 90

$\large {\boxed {X = \frac{11}{6} } }$

If the problem is to convert 1.83183183183... as a fraction , then :

Let :

X = 1.83183183183... = 1.83<u>183</u>

100X = 183.183183... = 183.<u>183</u>

100000X = 183183.183... = 183183.<u>183</u>

100000X - 100X = 183183.<u>183</u> - 183.<u>183</u>

99900X = 183000

X = 183000 ÷ 99900

$\large {\boxed {X = \frac{610}{333} } }$

<h3>Learn more</h3>

Geometric Series : brainly.com/question/4520950
Arithmetic Progression : brainly.com/question/2966265
Geometric Sequence : brainly.com/question/2166405

<h3>Answer details</h3>

Grade: Middle School

Subject: Mathematics

Chapter: Arithmetic and Geometric Series

Keywords: Arithmetic , Geometric , Series , Sequence , Difference , Term

maks197457 [2]3 years ago

4 0

The answer to this question would be: 1 5/6

To change a decimal number into fraction, you need to divide the number on the right of decimal point with 1. In this case, the number is 0.83.
This number is hard since .83 doesn't have many factors. To find the answer you can try to multiply the decimal with some number until it close to 1(no decimal left)
0.83* 2= 1.66
0.83* 3= 2.49 ---> close to half, if you find this number, you can try to double it
0.83* 4= 3.32
0.83* 5= 4.15
0.83* 6= 4.98---> close to 1, that means there is high probability that the number can be divided by 6
0.83 would be 4.98/6, but if we assume that the number is 0.8333...... then 0.83 would be 5/6. So, 1.83 would be 1 5/6

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Step-by-step explanation:

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2 years ago

In a binomial distribution, n = 8 and π=0.36. Find the probabilities of the following events. (Round your answers to 4 decimal p

skelet666 [1.2K]

Answer:

$\mathbf{P(X=5) =0.0888}$

P(x ≤ 5 ) = 0.9707

P ( x ≥ 6) = 0.0293

Step-by-step explanation:

The probability of a binomial mass distribution can be expressed with the formula:

$\mathtt{P(X=x) =(^{n}_{x} ) \ \pi^x \ (1-\pi)^{n-x}}$

$\mathtt{P(X=x) =(\dfrac{n!}{x!(n-x)!} ) \ \pi^x \ (1-\pi)^{n-x}}$

where;

n = 8 and π = 0.36

For x = 5

The probability $\mathtt{P(X=5) =(\dfrac{8!}{5!(8-5)!} ) \ 0.36^5 \ (1-0.36)^{8-5}}$

$\mathtt{P(X=5) =(\dfrac{8!}{5!(3)!} ) \ 0.36^5 \ (0.64)^{3}}$

$\mathtt{P(X=5) =(\dfrac{8 \times 7 \times 6 \times 5!}{5!(3)!} ) \times \ 0.0060466 \ \times 0.262144}$

$\mathtt{P(X=5) =(\dfrac{8 \times 7 \times 6 }{3 \times 2 \times 1} ) \times \ 0.0060466 \ \times 0.262144}$

$\mathtt{P(X=5) =({8 \times 7 } ) \times \ 0.0060466 \ \times 0.262144}$

$\mathtt{P(X=5) =0.0887645}$

$\mathbf{P(X=5) =0.0888}$ to 4 decimal places

b. x ≤ 5

The probability of P ( x ≤ 5) $\mathtt{P(x \leq 5) = P(x = 0)+ P(x = 1)+ P(x = 2)+ P(x = 3)+ P(x = 4)+ P(x = 5})$

${P(x \leq 5) = ( \dfrac{8!}{0!(8!)} \times (0.36)^0 \times (1-0.36)^8 \ ) + \dfrac{8!}{1!(7!)} \times (0.36)^1 \times (1-0.36)^7 \ +$ $\dfrac{8!}{2!(6!)} \times (0.36)^2 \times (1-0.36)^6 \ + \dfrac{8!}{3!(5!)} \times (0.36)^3 \times (1-0.36)^5 + \dfrac{8!}{4!(4!)} \times (0.36)^4 \times (1-0.36)^4 \ + \dfrac{8!}{5!(3!)} \times (0.36)^5 \times (1-0.36)^3 \ )$

P(x ≤ 5 ) = 0.0281+0.1267+0.2494+0.2805+0.1972+0.0888

P(x ≤ 5 ) = 0.9707

c. x ≥ 6

The probability of P ( x ≥ 6) = 1 - P( x ≤ 5 )

P ( x ≥ 6) = 1 - 0.9707

P ( x ≥ 6) = 0.0293

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2 years ago

A) How many ways can 2 integers from 1,2,...,100 be selected

Anna007 [38]

Answer with explanation:

→Number of Integers from 1 to 100

=100(50 Odd +50 Even)

→50 Even =2,4,6,8,10,12,14,16,...............................100

→50 Odd=1,3,,5,7,9,..................................99.

→Sum of Two even integers is even.

→Sum of two odd Integers is odd.

→Sum of an Odd and even Integer is Odd.

(a)

Number of ways of Selecting 2 integers from 50 Integers ,so that their sum is even,

=Selecting 2 Even integers from 50 Even Integers , and Selecting 2 Odd integers from 50 Odd integers ,as Order of arrangement is not Important, ,

$=_{2}^{50}\textrm{C}+_{2}^{50}\textrm{C}\\\\=\frac{50!}{(50-2)!(2!)}+\frac{50!}{(50-2)!(2!)}\\\\=\frac{50!}{48!\times 2!}+\frac{50!}{48!\times 2!}\\\\=\frac{50 \times 49}{2}+\frac{50 \times 49}{2}\\\\=1225+1225\\\\=2450$

=4900 ways

(b)

Number of ways of Selecting 2 integers from 100 Integers ,so that their sum is Odd,

=Selecting 1 even integer from 50 Integers, and 1 Odd integer from 50 Odd integers, as Order of arrangement is not Important,

$=_{1}^{50}\textrm{C}\times _{1}^{50}\textrm{C}\\\\=\frac{50!}{(50-1)!(1!)} \times \frac{50!}{(50-1)!(1!)}\\\\=\frac{50!}{49!\times 1!}\times \frac{50!}{49!\times 1!}\\\\=50\times 50\\\\=2500$

=2500 ways

7 0

3 years ago