Answer
\[8\times {{10}^{-5}}\].
Work Step by Step
If a number is expressed as \[a\times {{10}^{n}}\], where a is a number greater than or equal to 1 and less than 10, and n is an integer, then the number is said to be expressed in scientific notation.
Hence all the four numbers, \[1.6\times {{10}^{4}}\], \[7.2\times {{10}^{-3}}\],\[3.6\times {{10}^{8}}\] and \[4\times {{10}^{-3}}\], involved in the given expression are expressed in scientific notation.
Now, to compute the given expression, break the computation into three parts. First, compute the multiplications\[\left( 1.6\times {{10}^{4}} \right)\left( 7.2\times {{10}^{-3}} \right)\] and \[\left( 3.6\times {{10}^{8}} \right)\left( 4\times {{10}^{-3}} \right)\] separately, and then divide the result of the first multiplication\[\left( 1.6\times {{10}^{4}} \right)\left( 7.2\times {{10}^{-3}} \right)\] by the result of the second multiplication \[\left( 3.6\times {{10}^{8}} \right)\left( 4\times {{10}^{-3}} \right)\].
Then, perform the multiplication \[\left( a\times b \right)\] as the usual multiplication of any two numbers is performed and the multiplication \[\left( {{10}^{x}}\times {{10}^{y}} \right)\] using the product rule for exponents, explained below. And then write the two results obtained after performing the above two multiplications, together, with the symbol \[\times \] between them.
Now, apply the procedure mentioned above and the product rule for exponents to compute the required multiplications:
\[\begin{align}
& \left( 1.6\times {{10}^{4}} \right)\left( 7.2\times {{10}^{-3}} \right)=\left( 1.6\times 7.2 \right)\left( {{10}^{4}}\times {{10}^{-3}} \right) \\
& =\left( 11.52 \right)\left( {{10}^{4-3}} \right) \\
& =11.52\times {{10}^{1}}
\end{align}\]
Hence, \[\left( 1.6\times {{10}^{4}} \right)\left( 7.2\times {{10}^{-3}} \right)=11.52\times {{10}^{1}}\].
Similarly,
\[\begin{align}
& \left( 3.6\times {{10}^{8}} \right)\left( 4\times {{10}^{-3}} \right)=\left( 3.6\times 4 \right)\left( {{10}^{8}}\times {{10}^{-3}} \right) \\
& =\left( 14.4 \right)\left( {{10}^{8-3}} \right) \\
& =14.4\times {{10}^{5}}
\end{align}\]
Hence, \[\left( 3.6\times {{10}^{8}} \right)\left( 4\times {{10}^{-3}} \right)=14.4\times {{10}^{5}}\].
Now, divide the result of the first multiplication\[\left( 1.6\times {{10}^{4}} \right)\left( 7.2\times {{10}^{-3}} \right)\] by the result of the second multiplication\[\left( 3.6\times {{10}^{8}} \right)\left( 4\times {{10}^{-3}} \right)\]:
\[\frac{\left( 1.6\times {{10}^{4}} \right)\left( 7.2\times {{10}^{-3}} \right)}{\left( 3.6\times {{10}^{8}} \right)\left( 4\times {{10}^{-3}} \right)}=\frac{11.52\times {{10}^{1}}}{14.4\times {{10}^{5}}}\]
Then, perform the division \[\left( \frac{a}{b} \right)\] as the usual division of any two numbers is performed and the division \[\left( \frac{{{10}^{x}}}{{{10}^{y}}} \right)\] using the quotient rule for exponents, explained below. And then write the two results obtained after performing the above two divisions, together, with the symbol \[\times \] between them.
According to the quotient rule for exponents:
\[\frac{{{a}^{x}}}{{{a}^{y}}}={{a}^{x-y}}\]
Now, apply the procedure mentioned above and the quotient rule for exponents to compute the required division:
\[\begin{align}
& \frac{11.52\times {{10}^{1}}}{14.4\times {{10}^{5}}}=\left( \frac{11.52}{14.4} \right)\times \left( \frac{{{10}^{1}}}{{{10}^{5}}} \right) \\
& =\left( \frac{\left( \right)\left( 0.8 \right)}{} \right)\times \left( {{10}^{1-5}} \right) \\
& =0.8\times {{10}^{-4}}
\end{align}\]
The result obtained is still not expressed in scientific notation because \[0.8\] is less than 1.
Note that \[0.8\] can be rewritten as \[8\times {{10}^{-1}}\], where \[{{10}^{-1}}=\frac{1}{10}\].
Hence,
\[\begin{align}
& 0.8\times {{10}^{-4}}=\left( 8\times {{10}^{-1}} \right)\times {{10}^{-4}} \\
& =8\times \left( {{10}^{-1}}\times {{10}^{-4}} \right) \\
& =8\times \left( {{10}^{-1-4}} \right) \\
& =8\times {{10}^{-5}}
\end{align}\]
Hence,
\[\begin{align}
& \frac{\left( 1.6\times {{10}^{4}} \right)\left( 7.2\times {{10}^{-3}} \right)}{\left( 3.6\times {{10}^{8}} \right)\left( 4\times {{10}^{-3}} \right)}=\frac{11.52\times {{10}^{1}}}{14.4\times {{10}^{5}}} \\
& =0.8\times {{10}^{-4}} \\
& =8\times {{10}^{-5}}
\end{align}\]
