Solve the following problems: 21. -3 - 4 = ? 22. Joshua had 3 1/2 kg of sugar. He used 1 1/4 kg and bought 2 1/2 kg more. How much sugar does he have now? 23. Convert 0.1888... to a fraction.

The final position on the number line is $-7$.
The integer is $\boxed{-7}$.

22.

Initial sugar: $3\frac{1}{2} = \frac{3 \times 2 + 1}{2} = \frac{7}{2}$ kg
Sugar used: $1\frac{1}{4} = \frac{1 \times 4 + 1}{4} = \frac{5}{4}$ kg
Additional sugar bought: $2\frac{1}{2} = \frac{2 \times 2 + 1}{2} = \frac{5}{2}$ kg

Remaining sugar = Initial sugar - Sugar used + Additional sugar bought

Joshua finally remained with $\boxed{4\frac{3}{4} \text{ kg}}$ of sugar.

23.

The number $0.1\dot{8}$ as a fraction is $\boxed{\frac{17}{90}}$.

24.
The problem asks to calculate the distance covered, but the image of the athletic field does not provide any dimensions (e.g., length of straight sections, radius of semicircles). Without these dimensions, it is impossible to calculate the perimeter of the track or the total distance covered.

25.

Height ($h$) = 30 cm
Base length ($l$) = 20 cm (since it's a square base)
Base width ($w$) = 20 cm

The formula for the volume of a rectangular prism is $V = l \times w \times h$.

We know that $1 \text{ m} = 100 \text{ cm}$.
Therefore, $1 \text{ m}^3 = (100 \text{ cm})^3 = 100 \times 100 \times 100 \text{ cm}^3 = 1,000,000 \text{ cm}^3$.

The volume of the paint that can fit in the container is $\boxed{0.012 \text{ m}^3}$.

26.

Distance ($D$) = 432 Km
Time ($T$) = 6 hours

$1 \text{ Km} = 1000 \text{ m}$

$1 \text{ hour} = 60 \text{ minutes}$
$1 \text{ minute} = 60 \text{ seconds}$

Wangina's average speed is $\boxed{20 \text{ m/s}}$.