Question

water pressure
the water - pressure at different depths in a swimming pool is measured and recorded in the table. use the data provided in the table to answer the questions.

depth (feet)	water pressure (psi)
5	17.1
10	19.5
15	21.9
20	24.3
25	26.7

part a
what is the water pressure at a depth of 12 feet? (round to the nearest hundredth)
part b
at what depth does the water pressure reach 25 psi? (round to the nearest hundredth)

Explanation:

Step1: Find the linear relationship

Assume the relationship between depth $d$ (in feet) and water - pressure $P$ (in psi) is linear, of the form $P = md + b$. When $d = 0$, $P=14.7$, so $b = 14.7$. To find $m$, we use two points, say $(d_1,P_1)=(0,14.7)$ and $(d_2,P_2)=(5,17.1)$. The slope $m=\frac{P_2 - P_1}{d_2 - d_1}=\frac{17.1 - 14.7}{5-0}=\frac{2.4}{5}=0.48$. So the equation is $P = 0.48d+14.7$.

Step2: Solve Part A

For a depth $d = 12$ feet, substitute $d = 12$ into the equation $P = 0.48d+14.7$. Then $P=0.48\times12 + 14.7=5.76+14.7 = 20.46\approx20.46$ psi.

Step3: Solve Part B

We want to find $d$ when $P = 25$ psi. Set $P = 25$ in the equation $P = 0.48d+14.7$. Then $25=0.48d+14.7$. Subtract 14.7 from both sides: $25 - 14.7=0.48d$, so $10.3 = 0.48d$. Solve for $d$: $d=\frac{10.3}{0.48}\approx21.46$ feet.

Answer:

Part A: 20.46 psi
Part B: 21.46 feet