Ask Question
16 November, 09:48

A business owner opens one store in town A. The equation p (x) = 10,000 (1.075) represents the anticipated profit after t years. The business owner opens a store in town B six months later and predicts the profit from that store to increase at the same rate. Assume that the initial profit from the store in town B is the same as the initial profit from the store in town A. At any time after both stores have opened, how does the profit from the store in town B compare with the profit from the store in town A?

+4
Answers (1)
  1. 16 November, 13:01
    0
    The rigth equation to anticipate the profit after t years is p (t) = 10,000 (1.075) ^t

    So, given that both store A and store B follow the same equations but t is different for them, you can right:

    Store A: pA (t) 10,000 (1.075) ^t

    Store B: pB (t') : 10,000 (1.075) ^t'

    => pA (t) / pB (t') = 1.075^t / 1.075^t'

    => pA (t) / pB (t') = 1.075 ^ (t - t')

    And t - t' = 0.5 years

    => pA (t) / pB (t') = 1.075 ^ (0.5) = 1.0368

    or pB (t') / pA (t) = 1.075^ (-0.5) = 0.964

    => pB (t') ≈ 0.96 * pA (t)

    Which means that the profit of the store B is about 96% the profit of store A at any time after both stores have opened.
Know the Answer?
Not Sure About the Answer?
Find an answer to your question ✅ “A business owner opens one store in town A. The equation p (x) = 10,000 (1.075) represents the anticipated profit after t years. The ...” in 📘 Mathematics if you're in doubt about the correctness of the answers or there's no answer, then try to use the smart search and find answers to the similar questions.
Search for Other Answers