Anonymous
Anonymous asked in Science & MathematicsMathematics · 2 years 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

Relevance

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

• 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