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
2 Answers
- ?Lv 72 years agoFavorite 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
- la consoleLv 72 years 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