First, we ignore the decimal point and multiply as if both numbers were whole numbers: \begin{align*} \require{cancel} %%%%%%%%%% %%% Step A %%% %%%%%%%%%% & \begin{array}{ccccc} & & & \!\!\!\!\! \color{lightgray} \substack{ \fbox{$\color{blue}1$} \\[2pt] \fbox{$\color{blue}\phantom{0}$} } \!\!\!\! & \\ & & & \!\!\!\! 1 \!\!\!\! & \!\!\!\!\!\!\! . 2 \!\!\!\! \\ \!\!\!\!\times\!\!\!\! & \!\!\!\!\phantom{0}\!\!\!\! & \!\!\!\!\phantom{0}\!\!\!\! & \!\!\!\! 5 \!\!\!\! & \!\!\!\! 1 \!\!\!\! \\ \hline & & \!\!\!\! \!\!\!\! & \!\!\!\! 1 \!\!\!\! & \!\!\!\! 2 \!\!\!\! \\ \!\!\!\!+\!\!\!\! & \!\!\!\! \!\!\!\! & \!\!\!\! 6 \!\!\!\! & \!\!\!\! 0 \!\!\!\! & \!\!\!\! 0 \!\!\!\! \\ \hline & \!\!\!\! \!\!\!\! & \!\!\!\! 6 \!\!\!\! & \!\!\!\! 1 \!\!\!\! & \!\!\!\! 2 \!\!\!\! \\ \end{array} \end{align*}

We now count the total number of decimal places in the two factors.

There is $\color{blue}1$ decimal place in $1.2,$ so the product will also have ${\color{blue}{1}}$ decimal place.

So, we take our value of $612$ and insert a decimal point to make a number with $1$ decimal place: \[ 61\,\overset{\color{red}\downarrow}{\color{red}\bbox[2px, lightgray]{.}}\!\!\!\underbrace{2}_{\large\text{$\color{blue}1$ digit}} \]

Therefore, $1.2 \times 51 = 61.2\,.$

First, we ignore the decimal point and multiply as if both numbers were whole numbers:

\begin{align*} \require{cancel} %%%%%%%%%% %%% Step A %%% %%%%%%%%%% & \begin{array}{ccccc} & & & \!\!\!\!\! \color{lightgray} \substack{ \fbox{$\color{blue}2$} \\[2pt] \fbox{$\color{blue}2$} } \!\!\!\! & \\ & & & \!\!\!\! 3 \!\!\!\! & \!\!\!\! 4 \!\!\!\! \\ \!\!\!\!\times\!\!\!\! & \!\!\!\!\phantom{0}\!\!\!\! & \!\!\!\!\phantom{0}\!\!\!\! & \!\!\!\! 7 \!\!\!\! & \!\!\!\!\!\!\! . 5 \!\!\!\! \\ \hline & & \!\!\!\! 1 \!\!\!\! & \!\!\!\! 7 \!\!\!\! & \!\!\!\! 0 \!\!\!\! \\ \!\!\!\!+\!\!\!\! & \!\!\!\! 2 \!\!\!\! & \!\!\!\! 3 \!\!\!\! & \!\!\!\! 8 \!\!\!\! & \!\!\!\! 0 \!\!\!\! \\ \hline & \!\!\!\! 2 \!\!\!\! & \!\!\!\! 5 \!\!\!\! & \!\!\!\! 5 \!\!\!\! & \!\!\!\! 0 \!\!\!\! \\ \end{array} \end{align*}

We now count the total number of decimal places in the two factors.

There is $\color{blue}1$ decimal place in $7.5,$ so the product will also have ${\color{blue}{1}}$ decimal place.

So, we take our value of $2550$ and insert a decimal point to make a number with $1$ decimal place:

\[ 255\,\overset{\color{red}\downarrow}{\color{red}\bbox[2px, lightgray]{.}}\!\!\!\underbrace{0}_{\large\text{$\color{blue}1$ digit}} \]

Therefore, $34 \times 7.5 = 255.0 = 255\,.$

First, we ignore the decimal point and multiply as if both numbers were whole numbers:

\begin{align*} & \begin{array}{ccccc} & & & \!\!\!\!\! \color{lightgray} \substack{ \fbox{$\color{blue}3$} \\[2pt] \fbox{$\color{blue}3$} } \!\!\!\! & \!\!\!\!\! \color{lightgray} \substack{ \fbox{$\color{blue}4$} \\[2pt] \fbox{$\color{blue}3$} } \!\!\!\! & \\ & & & \!\!\!\! 1 \!\!\!\! & \!\!\!\! 7 \!\!\!\! & \!\!\!\! 8 \!\!\!\! \\ \!\!\!\!\times\!\!\!\! & \!\!\!\!\phantom{0}\!\!\!\! & \!\!\!\!\phantom{0}\!\!\!\! & \!\!\!\!\phantom{0}\!\!\!\! & \!\!\!\! 5 \!\!\!\! & \!\!\!\!\!\!\! . 4 \!\!\!\! \\ \hline & & \!\!\!\!\!\!\!\! & \!\!\!\!7\!\!\!\! & \!\!\!\!1\!\!\!\! & \!\!\!\!2\!\!\!\! \\ \!\!\!\!+\!\!\!\! & \!\!\!\!\!\!\!\! & \!\!\!\!8\!\!\!\! & \!\!\!\!9\!\!\!\! & \!\!\!\!0\!\!\!\! & \!\!\!\!0\!\!\!\! \\ \hline & \!\!\!\!\!\!\!\! & \!\!\!\!9\!\!\!\! & \!\!\!\!6\!\!\!\! & \!\!\!\!1\!\!\!\! & \!\!\!\!2\!\!\!\! \\ \end{array} \end{align*}

We now count the total number of decimal places in the two factors.

There is $\color{blue}1$ decimal place in $5.4,$ so the product will also have ${\color{blue}{1}}$ decimal place.

So, we take our value of $9612$ and insert a decimal point to make a number with $1$ decimal place:

\[ 961\,\overset{\color{red}\downarrow}{\color{red}\bbox[2px, lightgray]{.}}\!\!\!\underbrace{2}_{\large\text{$\color{blue}1$ digit}} \]

Therefore, $178 \times 5.4 = 961.2\,.$

First, we ignore the decimal point and multiply as if both numbers were whole numbers: \begin{align*} & \begin{array}{ccccc} & & & \!\!\!\!\! \color{lightgray} \substack{ \fbox{$\color{blue}1$} \\[2pt] \fbox{$\color{blue}1$} } \!\!\!\! & \!\!\!\!\! \color{lightgray} \substack{ \fbox{$\color{blue}2$} \\[2pt] \fbox{$\color{blue}2$} } \!\!\!\! & \\ & & & \!\!\!\! 9 \!\!\!\! & \!\!\!\! 2 \!\!\!\! & \!\!\!\! 5 \!\!\!\! \\ \!\!\!\!\times\!\!\!\! & \!\!\!\!\phantom{0}\!\!\!\! & \!\!\!\!\phantom{0}\!\!\!\! & \!\!\!\!\phantom{0}\!\!\!\! & \!\!\!\! 4 \!\!\!\! & \!\!\!\!\!\!\! . 5 \!\!\!\! \\ \hline \!\!\!\!\!\!\!\! & \!\!\!\!\!\!\!\! & \!\!\!\!4\!\!\!\! & \!\!\!\!6\!\!\!\! & \!\!\!\!2\!\!\!\! & \!\!\!\!5\!\!\!\! \\ \!\!\!\!+\!\!\!\! &\!\!\!\!3\!\!\!\! & \!\!\!\!7\!\!\!\! & \!\!\!\!0\!\!\!\! & \!\!\!\!0\!\!\!\! & \!\!\!\!0\!\!\!\! \\ \hline & \!\!\!\!4\!\!\!\! & \!\!\!\!1\!\!\!\! & \!\!\!\!6\!\!\!\! & \!\!\!\!2\!\!\!\! & \!\!\!\!5\!\!\!\! \\ \end{array} \end{align*}

We now count the total number of decimal places in the two factors.

There is $\color{blue}1$ decimal place in $4.5,$ so the product will also have ${\color{blue}{1}}$ decimal place.

So, we take our value of $41625$ and add a decimal point to make a number with $\color{blue}1$ decimal place.

\[ 4162\,\overset{\color{red}\downarrow}{\color{red}\bbox[2px, lightgray]{.}}\!\!\!\underbrace{5}_{\large\text{$\color{blue}1$ digit}} \]

Therefore, $925 \times 4.5 = 4,162.5.$

First, we ignore the decimal point and multiply as if both numbers were whole numbers:

\begin{align*} & \begin{array}{ccccc} & & & & \!\!\!\!\! \color{lightgray} \substack{ \fbox{$\color{blue}1$} \\ \fbox{$\color{blue}2$} \\[2pt] \fbox{$\color{blue}1$} } \!\!\!\!\! & \!\!\!\! \color{lightgray} \substack{ \fbox{$\color{blue}2$} \\ \fbox{$\color{blue}6$} \\[2pt] \fbox{$\color{blue}3$} } \!\!\!\! & \\ & & & & \!\!\!\! 2 \!\!\!\! & \!\!\!\! 3 \!\!\!\! & \!\!\!\! 9 \!\!\!\! \\ \!\!\!\!\times\!\!\!\! & \!\!\!\!\phantom{0}\!\!\!\! & \!\!\!\!\phantom{0}\!\!\!\! & \!\!\!\!\phantom{0}\!\!\!\! & \!\!\!\! 3 \!\!\!\! & \!\!\!\!\!\!\! . 7 \!\!\!\! & \!\!\!\! 4 \!\!\!\! \\ \hline & & & \!\!\!\!\!\!\!\! & \!\!\!\!9\!\!\!\! & \!\!\!\!5\!\!\!\! & \!\!\!\!6\!\!\!\! \\ \!\!\!\!+\!\!\!\! & & \!\!\!\!1\!\!\!\! & \!\!\!\!6\!\!\!\! & \!\!\!\!7\!\!\!\! & \!\!\!\!3\!\!\!\! & \!\!\!\!0\!\!\!\! \\ \!\!\!\!+\!\!\!\! & \!\!\!\!\!\!\!\! & \!\!\!\!7\!\!\!\! & \!\!\!\!1\!\!\!\! & \!\!\!\!7\!\!\!\! & \!\!\!\!0\!\!\!\! & \!\!\!\!0\!\!\!\! \\ \hline & \!\!\!\! \!\!\!\! & \!\!\!\! 8 \!\!\!\! & \!\!\!\! 9 \!\!\!\! & \!\!\!\! 3 \!\!\!\! & \!\!\!\! 8 \!\!\!\! & \!\!\!\! 6 \!\!\!\! \\ \end{array} \end{align*}

We now count the total number of decimal places in the two factors.

There are $\color{blue}2$ decimal places in $3.74,$ so the product will also have ${\color{blue}{2}}$ decimal places.

So, we take our value of $89386$ and insert a decimal point to make a number with $\color{blue}2$ decimal places:

\[ 893\,\overset{\color{red}\downarrow}{\color{red}\bbox[2px, lightgray]{.}}\!\!\!\underbrace{86}_{\large\text{$\color{blue}2$ digits}} \]

Therefore, $239 \times 3.74 = 893.86\,.$

First, we ignore the decimal point and multiply as if both numbers were whole numbers:

\begin{align*} & \begin{array}{ccccc} & & & & \!\!\!\!\! \color{lightgray} \substack{ \fbox{$\color{blue}\phantom{0}$} \\ \fbox{$\color{blue}\phantom{0}$} \\[2pt] \fbox{$\color{blue}\phantom{0}$} } \!\!\!\!\! & \!\!\!\! \color{lightgray} \substack{ \fbox{$\color{blue}\phantom{0}$} \\ \fbox{$\color{blue}\phantom{0}$} \\[2pt] \fbox{$\color{blue}1$} } \!\!\!\! & \\ & & & & \!\!\!\! 7 \!\!\!\! & \!\!\!\!\!\!\! . 0 \!\!\!\! & \!\!\!\! 3 \!\!\!\! \\ \!\!\!\!\times\!\!\!\! & \!\!\!\!\phantom{0}\!\!\!\! & \!\!\!\!\phantom{0}\!\!\!\! & \!\!\!\!\phantom{0}\!\!\!\! & \!\!\!\! 2 \!\!\!\! & \!\!\!\! 1 \!\!\!\! & \!\!\!\! 6 \!\!\!\! \\ \hline & & & \!\!\!\!4\!\!\!\! & \!\!\!\!2\!\!\!\! & \!\!\!\!1\!\!\!\! & \!\!\!\!8\!\!\!\! \\ \!\!\!\!+\!\!\!\! & & \!\!\!\!\!\!\!\! & \!\!\!\!7\!\!\!\! & \!\!\!\!0\!\!\!\! & \!\!\!\!3\!\!\!\! & \!\!\!\!0\!\!\!\! \\ \!\!\!\!+\!\!\!\! & \!\!\!\!1\!\!\!\! & \!\!\!\!4\!\!\!\! & \!\!\!\!0\!\!\!\! & \!\!\!\!6\!\!\!\! & \!\!\!\!0\!\!\!\! & \!\!\!\!0\!\!\!\! \\ \hline & \!\!\!\! 1 \!\!\!\! & \!\!\!\! 5 \!\!\!\! & \!\!\!\! 1 \!\!\!\! & \!\!\!\! 8 \!\!\!\! & \!\!\!\! 4 \!\!\!\! & \!\!\!\! 8 \!\!\!\! \\ \end{array} \end{align*}

We now count the total number of decimal places in the two factors.

There are $\color{blue}2$ decimal places in $7.03,$ so the product will also have ${\color{blue}{2}}$ decimal places.

So, we take our value of $151848$ and insert a decimal point to make a number with $\color{blue}2$ decimal places:

\[ 1518\,\overset{\color{red}\downarrow}{\color{red}\bbox[2px, lightgray]{.}}\!\!\!\underbrace{48}_{\large\text{$\color{blue}2$ digits}} \]

Therefore, $7.03 \times 216 = 1,518.48\,.$

To determine the distance Mary covers over $18$ days, we need to multiply $18$ by $3.6.$

First, we ignore the decimal point and multiply as if both numbers were whole numbers:

\begin{align*} \require{cancel} %%%%%%%%%% %%% Step A %%% %%%%%%%%%% & \begin{array}{ccccc} & & & \!\!\!\!\! \color{lightgray} \substack{ \fbox{$\color{blue}2$} \\[2pt] \fbox{$\color{blue}4$} } \!\!\!\! & \\ & & & \!\!\!\! 1 \!\!\!\! & \!\!\!\! 8 \!\!\!\! \\ \!\!\!\!\times\!\!\!\! & \!\!\!\!\phantom{0}\!\!\!\! & \!\!\!\!\phantom{0}\!\!\!\! & \!\!\!\! 3 \!\!\!\! & \!\!\!\!\!\!\! . 6 \!\!\!\! \\ \hline & & \!\!\!\! 1 \!\!\!\! & \!\!\!\! 0 \!\!\!\! & \!\!\!\! 8 \!\!\!\! \\ \!\!\!\!+\!\!\!\! & \!\!\!\! \!\!\!\! & \!\!\!\! 5 \!\!\!\! & \!\!\!\! 4 \!\!\!\! & \!\!\!\! 0 \!\!\!\! \\ \hline & \!\!\!\! \!\!\!\! & \!\!\!\! 6 \!\!\!\! & \!\!\!\! 4 \!\!\!\! & \!\!\!\! 8 \!\!\!\! \\ \end{array} \end{align*}

We now count the total number of decimal places in the two factors.

There is $\color{blue}1$ decimal place in $3.6,$ so the product will also have ${\color{blue}{1}}$ decimal place.

So, we take our value of $648$ and insert a decimal point to make a number with $1$ decimal place: \[ 64\,\overset{\color{red}\downarrow}{\color{red}\bbox[2px, lightgray]{.}}\!\!\!\underbrace{8}_{\large\text{$\color{blue}1$ digit}} \]

Therefore, $18 \times 3.6 = 64.8.$

So, Mary will run a total of $64.8$ kilometers in $18$ days.

To find the distance between Davenport and Chicago, we need to multiply $153$ by $1.6.$

First, we ignore the decimal point and multiply as if both numbers were whole numbers:

\begin{align*} & \begin{array}{ccccc} & & & \!\!\!\!\! \color{lightgray} \substack{ \fbox{$\color{blue}\phantom{0}$} \\[2pt] \fbox{$\color{blue}3$} } \!\!\!\! & \!\!\!\!\! \color{lightgray} \substack{ \fbox{$\color{blue}\phantom{0}$} \\[2pt] \fbox{$\color{blue}1$} } \!\!\!\! & \\ & & & \!\!\!\! 1 \!\!\!\! & \!\!\!\! 5 \!\!\!\! & \!\!\!\! 3 \!\!\!\! \\ \!\!\!\!\times\!\!\!\! & \!\!\!\!\phantom{0}\!\!\!\! & \!\!\!\!\phantom{0}\!\!\!\! & \!\!\!\!\phantom{0}\!\!\!\! & \!\!\!\! 1 \!\!\!\! & \!\!\!\!\!\!\! . 6 \!\!\!\! \\ \hline & & \!\!\!\!\!\!\!\! & \!\!\!\!9\!\!\!\! & \!\!\!\!1\!\!\!\! & \!\!\!\!8\!\!\!\! \\ \!\!\!\!+\!\!\!\! & \!\!\!\!\!\!\!\! & \!\!\!\!1\!\!\!\! & \!\!\!\!5\!\!\!\! & \!\!\!\!3\!\!\!\! & \!\!\!\!0\!\!\!\! \\ \hline & \!\!\!\!\!\!\!\! & \!\!\!\!2\!\!\!\! & \!\!\!\!4\!\!\!\! & \!\!\!\!4\!\!\!\! & \!\!\!\!8\!\!\!\! \\ \end{array} \end{align*}

We now count the total number of decimal places in the two factors.

There is $\color{blue}1$ decimal place in $1.6,$ so the product will also have ${\color{blue}{1}}$ decimal place.

So, we take our value of $2448$ and insert a decimal point to make a number with $1$ decimal place: \[ 244\,\overset{\color{red}\downarrow}{\color{red}\bbox[2px, lightgray]{.}}\!\!\!\underbrace{8}_{\large\text{$\color{blue}1$ digit}} \]

So, $153 \times 1.6 = 244.8.$

We conclude that the distance between the two cities is $244.8$ kilometers.