To determine how much strawberry Mary needs, we need to calculate $1\,\dfrac{1}{3} \times 5.$

First, we write $1\,\dfrac{1}{3} $ as an improper fraction: \[ 1\,\dfrac{1}{3}=\dfrac{(1 \times 3)+1}{3}= \dfrac{4}{3} \]

We also write $5$ as an improper fraction: \[ 5=\dfrac{5}{1} \]

Now, we can multiply the numbers. We multiply the numerators, and we multiply the denominators: \[ \dfrac{4}{3} \times \dfrac{5}{1} = \dfrac {4 \times 5} {3 \times 1} = \dfrac{20}{3} \]

Finally, we write the resulting improper fraction as a mixed number: \[ \dfrac{20}{3} = 6\,\textrm{R}\, 2 = 6\,\dfrac{2}{3} \]

So, Mary needs $6\,\dfrac{2}{3}\,\textrm{lb}$ of strawberries.

To determine the amount of flour that Nancy needs, we need to calculate $2\,\dfrac{1}{4} \times 6.$

First, we write $2\,\dfrac{1}{4} $ as an improper fraction: \[ 2\,\dfrac{1}{4}=\dfrac{(2 \times 4)+1}{4}= \dfrac{9}{4} \]

We also write $6$ as an improper fraction: \[ 6=\dfrac{6}{1} \]

So now, we need to solve the following multiplication problem:

\[ \dfrac{9}{\color{red}4} \times \dfrac{\color{blue}6}{1} \]

Notice that the first fraction's denominator ($\color{red}4$) and the second fraction's numerator ($\color{blue}6$) have a common factor of $2.$

Therefore, we can simplify our problem by swapping the denominators first and then multiplying the resulting fractions:

\begin{align*} \dfrac{9}{\color{red}4} \times \dfrac{\color{blue}6}{1} &= \dfrac{9}{1} \times \dfrac{\color{blue}6}{\color{red}4} \\[5pt] &= \dfrac{9}{1} \times \dfrac{3}{2} \\[5pt] &= \dfrac{9 \times 3}{1 \times 2} \\[5pt] &=\dfrac{27}{2} \end{align*}

Finally, we write the resulting improper fraction as a mixed number: \[ \dfrac{27}{2} = 13\,\textrm{R}\, 1 = 13\,\dfrac{1}{2} \]

So, Nancy needs $13\,\dfrac{1}{2}$ cups of flour.

To find out the weight of the raisins left in the box, we need to calculate $3 \, \dfrac{1}{3} \times \dfrac{2}{3}.$

First, we write $3 \,\dfrac{1}{3}$ as an improper fraction:

\[ 3 \,\dfrac{1}{3} = \dfrac{(3 \times 3) + 1}{3} = \dfrac{10}{3} \]

Now, we can multiply the two numbers. We multiply the numerators, and we multiply the denominators:

\[ \dfrac{10}{3} \times \dfrac{2}{3} = \dfrac{10 \times 2}{3 \times 3} = \dfrac{20}{9} \]

Finally, converting this to a mixed number, we get

\[ 20\div 9 = 2\,\textrm{R} \, 2 = 2\,\dfrac{2}{9}. \]

So, the box now weighs $2\,\dfrac{2}{9}\,\textrm{oz}.$

To find out Julie's height, we need to calculate $4 \, \dfrac{1}{2} \times \dfrac{4}{5}.$

First, we write $4 \,\dfrac{1}{2}$ as an improper fraction:

\[ 4 \,\dfrac{1}{2} = \dfrac{(4 \times 2) + 1}{2} = \dfrac{9}{2} \]

So now, we need to solve the following multiplication problem:

\[ \dfrac{9}{\color{red}2} \times \dfrac{\color{blue}4}{5} \]

Notice that the first fraction's denominator ($\color{red}2$) and the second fraction's numerator ($\color{blue}4$) have a common factor of $2.$

Therefore, we can simplify our problem by swapping the denominators first and then multiplying the resulting fractions:

\begin{align*} \dfrac{9}{\color{red}2} \times \dfrac{\color{blue}4}{5} &= \dfrac{9}{5} \times \dfrac{\color{blue}4}{\color{red}2} \\[5pt] &= \dfrac{9}{5} \times \dfrac{2}{1} \\[5pt] &= \dfrac{9 \times 2}{5 \times 1} \\[5pt] &=\dfrac{18}{5} \end{align*}

Finally, converting this to a mixed number, we get

\[ 18 \div 5 = 3\,\textrm{R} \, 3 = 3\,\dfrac{3}{5}. \]

So, Julie is $3\,\dfrac{3}{5}\,\textrm{ft}$ tall.

Notice that Debbie had $2+ \dfrac{1}{4} = 2 \, \dfrac{1}{4}$ boxes of cornflakes. So, to find out the total weight, we need to calculate $1 \, \dfrac{1}{2} \times 2 \, \dfrac{1}{4}.$

First, we write $1\,\dfrac{1}{2}$ as an improper fraction: \[ 1 \, \dfrac{1}{2} = \dfrac{(1 \times 2) + 1}{2}= \dfrac{3}{2} \]

We also write $2 \, \dfrac{1}{4} $ as an improper fraction: \[ 2 \, \dfrac{1}{4}=\dfrac{(2 \times 4)+1}{4}= \dfrac{9}{4} \]

Now, we can multiply the two numbers. We multiply the numerators, and we multiply the denominators: \[ \dfrac{3}{2} \times \dfrac{9}{4} = \dfrac {3 \times 9}{2 \times 8} = \dfrac{27}{8} \]

Finally, we write the resulting improper fraction as a mixed number: \[ \dfrac{27}{8} =3\,\textrm{R}\, 3 = 3 \, \dfrac{3}{8} \]

So, Debbie ate a total of $3 \, \dfrac{3}{8}\,\textrm{oz}$ of cornflakes cereal.

To find out how long the lump of play-doh would be, we need to calculate $1 \, \dfrac{1}{3} \times 1 \, \dfrac{1}{8}.$

First, we write $1\,\dfrac{1}{3}$ as an improper fraction: \[ 1 \, \dfrac{1}{3} = \dfrac{(1 \times 3) + 1}{3}= \dfrac{4}{3} \]

We also write $1 \, \dfrac{1}{8} $ as an improper fraction: \[ 1 \, \dfrac{1}{8}=\dfrac{(1 \times 8)+1}{8}= \dfrac{9}{8} \]

So now, we need to solve the following multiplication problem:

\[ \dfrac{\color{red}4}{3} \times \dfrac{9}{\color{blue}8} \]

Notice that the first fraction's numerator ($\color{red}4$) and the second fraction's denominator ($\color{blue}8$) have a common factor of $4.$

Therefore, we can simplify our problem by swapping the denominators first and then multiplying the resulting fractions:

\begin{align*} \dfrac{\color{red}4}{3} \times \dfrac{9}{\color{blue}8} &= \dfrac{\color{red}4}{\color{blue}8} \times \dfrac{9}{3} \\[5pt] &= \dfrac{1}{2} \times \dfrac{3}{1} \\[5pt] &= \dfrac{1 \times 3}{2 \times 1} \\[5pt] &=\dfrac{3}{2} \end{align*}

Finally, we write the resulting improper fraction as a mixed number: \[ \dfrac{3}{2} =1\,\textrm{R}\, 1 = 1 \, \dfrac{1}{2} \]

So, the play-doh would be $1 \, \dfrac{1}{2}\,\textrm{in}$ long.