So,
#1: \(n \geq 3 \frac{11}{16} + 4 \frac{1}{2}\)
Convert to like improper fractions.
\(n \geq \frac{59}{16} + \frac{72}{16}\)
Add.
\(n \geq \frac{131}{16}\ or\ 8 \frac{3}{16}\)
So, one solution could be \(8 \frac{3}{16}\).
Another solution could by 9. There is also 10, 11, 12, etc., and all numbers in between.
#2: \(k \ \textless \ 6 \frac{2}{5} * 15\)
Convert into improper fraction form.
\(k \ \textless \ \frac{32}{5} * 15\)
Multiply.
\(\frac{(2^5)(3)(5)}{5}\)
Cross-cancel, and we have our final result.
\((2^5)(3) = 96\)
k < 96
96 is not a solution.
95 is a solution.
So is 94, 93, 92, etc, and all numbers in between.
