current attempt in progress the figure below shows the rate of change of the quantity of water in a water…

Explanation:

Step1: Approximate area under curve

The change in the quantity of water is given by the area under the rate - of - change curve. We can approximate the area under the curve using rectangles. The interval from $t = 0$ to $t=30$ is divided into sub - intervals of width $\Delta t=6$ days.

Step2: Calculate area of rectangles

For $0\leq t < 6$: The rate is approximately $r_1=- 50$ liters/day, and the area of the rectangle $A_1=r_1\times\Delta t=-50\times6=-300$. For $6\leq t < 12$: The rate is approximately $r_2 = 0$ liters/day, and the area of the rectangle $A_2=r_2\times\Delta t=0\times6 = 0$. For $12\leq t < 18$: The rate is approximately $r_3 = 100$ liters/day, and the area of the rectangle $A_3=r_3\times\Delta t=100\times6 = 600$. For $18\leq t < 24$: The rate is approximately $r_4 = 150$ liters/day, and the area of the rectangle $A_4=r_4\times\Delta t=150\times6=900$. For $24\leq t < 30$: The rate is approximately $r_5 = 100$ liters/day, and the area of the rectangle $A_5=r_5\times\Delta t=100\times6 = 600$.

Step3: Sum of areas

The total change in the quantity of water $\Delta Q=A_1 + A_2+A_3+A_4+A_5=-300 + 0+600+900+600=1800$ liters.

Step4: Find final quantity

The initial quantity of water $Q_0 = 16000$ liters. The final quantity of water $Q=Q_0+\Delta Q$. So $Q=16000 + 1800=17800$ liters. Rounding to the nearest hundred liters, $Q\approx17800$ liters.

Answer:

17800