How do you multiply \[\dfrac{{(9.52 \times {{10}^4}) \times (2.77 \times {{10}^{ - 5}})}}{{(1.39 \times {{10}^7}) \times (5.83 \times {{10}^2})}}\] ?

Hint: In order to solve the above question we need to multiply the fractional number with another fractional number. We will use simple calculations to solve this problem after rewriting the equation and removing parenthesis. Here firstly we will multiply the numerators and then later on denominator will be multiplied. Lastly to get the desired results we will further simplify the fraction into its simplest form.

Formula used:
To multiply the fraction given in the above question we will use the formula as mentioned.
Product of fraction \[ = \]Product of numerator \[ \div \] Product of denominator
\[\dfrac{a}{b} \times \dfrac{c}{d} = \dfrac{{a \times c}}{{b \times d}}\]

Complete step by step answer:
As we need to multiply \[\dfrac{{(9.52 \times {{10}^4}) \times (2.77 \times {{10}^{ - 5}})}}{{(1.39 \times {{10}^7}) \times (5.83 \times {{10}^2})}}\]
So we will remove the parentheses firstly,
\[\dfrac{{9.52 \times {{10}^4} \times 2.77 \times {{10}^{ - 5}}}}{{1.39 \times {{10}^7} \times 5.83 \times {{10}^2}}}\]
Now after moving the numbers around we will get,
\[\dfrac{{9.52 \times 2.77 \times {{10}^4} \times {{10}^{ - 5}}}}{{1.39 \times 5.83 \times {{10}^7} \times {{10}^2}}}\]
Later after multiplication we will get,
\[\dfrac{{26.3704 \times {{10}^{ - 1}}}}{{8.1037 \times {{10}^{ - 9}}}}\]
After separation and further simplification we get,
\[\dfrac{{26.3704}}{{8.1037}} \times \dfrac{{{{10}^{ - 1}}}}{{{{10}^{ - 9}}}} \\
\therefore 3.25411 \times {10^{ - 10}} \]

Note:Here in the above problem remember that simplification or reduction of the obtained result into its simpler form by cancelling out the common factors from the numerator and denominator is required or else the desired results will not be obtained. Remember that multiplication of fractions results in fractions. There can be the possibility of committing calculation mistakes so we should simplify the obtained results carefully.