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.

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

Explanation:

Step1: Find the pattern

We observe that for every increase of 5 feet in depth, the water pressure increases by \(17.1 - 14.7=2.4\) psi. So the rate of change (slope) \(m=\frac{2.4}{5}=0.48\) psi per foot. Using the point - slope form \(y - y_1=m(x - x_1)\), with the point \((0,14.7)\) (where \(x = 0\) depth and \(y = 14.7\) psi), the equation is \(y=0.48x + 14.7\)

Step2: Solve Part A

For \(x = 12\) feet (depth), substitute \(x = 12\) into the equation \(y=0.48x+14.7\).
\(y=0.48\times12 + 14.7\)
\(y = 5.76+14.7\)
\(y=20.46\) psi

Step3: Solve Part B

We want to find \(x\) when \(y = 25\) psi. Substitute \(y = 25\) into \(y=0.48x + 14.7\)
\(25=0.48x+14.7\)
Subtract 14.7 from both sides: \(25 - 14.7=0.48x\)
\(10.3 = 0.48x\)
Divide both sides by 0.48: \(x=\frac{10.3}{0.48}\approx21.46\) feet

Answer:

Part A: \(20.46\) psi
Part B: \(21.46\) ft