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**Hint:**Writing decimals in expanded form simply means writing each number according to its place value. This is done by multiplying each digit by its place value and adding them together.

The use of decimal numbers is limited to situations where accuracy and precision are required. The concept of expanded form is useful in understanding the numerical value of a quantity as well as the value of a number.

**Complete step by step solution:**

We are given the decimal is \[67.83\]

In words, we would say it as Sixty-Seven and Eighty-three hundredths.

The whole number \[67\] is said as sixty-seven, the decimal is said as ‘and’, and the decimal part, which is \[83\], is said as a regular number: eighty-three. But since it is a decimal, we would say the last number as its place value.

The last number \[3\] is in the hundredths place, so we must say eighty-three hundredths.

i.e. \[67.83\]

\[

= 6tens + 7ones + 8tenths + 3hundreths \\

= (6 \times 10) + (7 \times 1) + (8 \times \dfrac{1}{{10}}) + (3 \times \dfrac{1}{{100}}) \\

\]

Now, simplify it as follows,

\[

= 60 + 7 + \dfrac{8}{{10}} + \dfrac{3}{{100}} \\

\therefore 67.83 = 60 + 7 + \dfrac{8}{{10}} + \dfrac{3}{{100}} \\

\]

**Hence, the following decimal in expanded form: \[67.83 = 60 + 7 + \dfrac{8}{{10}} + \dfrac{3}{{100}}\].**

**Additional Information:**

Weights in the ratios of \[120,{\text{ }}110,{\text{ }}15,{\text{ }}12\] and multiples of ten were used for trade in the Indus Valley civilization.

During 305 BC, rod calculus in ancient China used bamboo strips with the decimal system for math operations such as multiplication.

Archimedes, a mathematician, devised a decimal positional system for Sand Reckoner to represent large numbers that were multiples of ten.

**Note:**

Let's look at an example: \[2.435\]

In words, we would say this as two and four hundred thirty-five thousandths.

The whole number \[2\] is said as two, the decimal is said as 'and,' and the decimal part, which is \[435\], is said as a regular number: four hundred thirty-five. But since it is a decimal, we would say the last number as its place value.

The last number \[5\] is in the thousandths place, so we must say four hundred thirty-five thousandths.

i.e. \[2.435\]

\[

= 2ones + 4tenths + 3hundredths + 5thousandths \\

= (2 \times 1) + (4 \times \dfrac{1}{{10}}) + (3 \times \dfrac{1}{{100}}) + (5 \times \dfrac{1}{{1000}}) \\

= 2 + \dfrac{4}{{10}} + \dfrac{3}{{100}} + \dfrac{5}{{1000}} \\

2.435 = 2 + \dfrac{4}{{10}} + \dfrac{3}{{100}} + \dfrac{5}{{1000}} \\

\]

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