I chow you how to solve for (d) and (i) and you could do the rest by yourself:
The best way to solve this operations is convert the numbers to a one form and then compare.
For (d)
[tex]\begin{gathered} \frac{7}{3}=\frac{6+1}{3}=\frac{6}{3}+\frac{1}{3}=2+\frac{1}{3}=2\frac{1}{3}=2.333 \\ \frac{13}{5}=\frac{10+3}{5}=\frac{10}{5}+\frac{3}{5}=2+\frac{3}{5}=2\frac{3}{5}=2.6 \\ So, \\ \frac{7}{3}<\frac{13}{5} \end{gathered}[/tex]
Now for (i), take into account that this numbers are negative:
[tex]\begin{gathered} -11.5=-11.5\cdot\frac{4}{4}=-\frac{11.5\cdot4}{4}=-\frac{46}{4} \\ So,\text{ } \\ -\frac{46}{4}<-\frac{31}{4} \end{gathered}[/tex]
Note that 46/4 is greater than 31/4, but -46/4 is lower than -31/4.
Also note that in this example I find to equalize the denominator of the numbers adn then you can compare the numerators.
