Anonymous
Anonymous asked in Science & MathematicsMathematics · 1 year ago

the cost c of a gallon of milk was $0.54 in 1941. in 1985, it was $2.31. assuming the exponential growth model applies, find the cost of a?

gallon of milk in 2000 ; in 2010

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  • david
    Lv 7
    1 year ago
    Favorite Answer

    2.31 = 0.54 x (k^44)

    k = ~1.0336

    C(2000) = 0.54 x (1.0336^56) = $3.44

    C(2010) = 0.54 x (1.0336^66) = $4.78

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  • 1 year ago

    The cost c of a gallon of milk was $0.54 in 1941: → c₁₉₄₁ = a * k^(1941) = 0.54

    The cost c of a gallon of milk was $2.31 in 1985: → c₁₉₈₅ = a * k^(1985) = 2.31

    c₁₉₈₅/c₁₉₄₁ = 2.31/.54

    [a * k^(1985)] / [a * k^(1941)] = 2.31/0.54

    k^(1985) / k^(1941) = 2.31/0.54

    k^(1985) * k^(- 1941) = 2.31/0.54

    k^(1985 - 1941) = 2.31/0.54

    k^(44) = 2.31/0.54

    k = (2.31/0.54)^(1/44)

    k = (0.77/0.18)^(1/44)

    k = (77/18)^(1/44)

    Recall: c₁₉₄₁ = a * k^(1941) = 0.54

    a * k^(1941) = 0.54

    a = 0.54/k^(1941)

    a = 0.54 * k^(- 1941)

    The cost c of a gallon of milk in 2000:

    c₂₀₀₀ = a * k^(2000) → recall: a

    c₂₀₀₀ = 0.54 * k^(- 1941) * k^(2000)

    c₂₀₀₀ = 0.54 * k^(- 1941 + 2000)

    c₂₀₀₀ = 0.54 * k^(59) → recall: k

    c₂₀₀₀ = 0.54 * [(77/18)^(1/44)]^(59)

    c₂₀₀₀ = 0.54 * (77/18)^(59/44)

    c₂₀₀₀ = $3.7914

    The cost c of a gallon of milk in 2010:

    c₂₀₁₀ = a * k^(2010) → recall: a

    c₂₀₁₀ = 0.54 * k^(- 1941) * k^(2010)

    c₂₀₁₀ = 0.54 * k^(- 1941 + 2010)

    c₂₀₁₀ = 0.54 * k^(69) → recall: k

    c₂₀₁₀ = 0.54 * [(77/18)^(1/44)]^(69)

    c₂₀₁₀ = 0.54 * (77/18)^(69/44)

    c₂₀₁₀ = $5.2754

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