ExplanationFirst, solve for the constant k using the price information of the 2-bedroom, 2-bath unit (m = 800, r = t = 2, and f = 1):

\(800 = k(\frac{5^2+10\times 2}{1+5})\)

\(800 = k(\frac{20 + 20}{6})\)

\(800 = k(\frac{40}{6})\)

\(120 = k\)

Next, solve for the rent on the 3-bedroom, 1-bath unit on the 3rd floor (r = 3, t = 1, and f = 3):

\(m = 120(\frac{5(3)^2+10\times 1}{3+5})\)

\(m = 120(\frac{45+10}{3+5})\)

m = 825

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Sandy

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