Question 823913
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y(t) = a * e^(kt)
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where y(t) = value at time t
a = value at the start
k = rate of growth (when k > 0) or decay (when k < 0)
t = time (years in this case)
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number of households with dogs 1-year from now:
43000000 * 1.15 = 49450000
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y(1) = 49450000 = 43000000 * e^(k(1))
y(1) = 49450000/43000000 = e^(k(1))
ln( 49450000/43000000 ) = ln( e^k )
k = ln( 49450000/43000000 )
k = 0.13976194237515862651
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y(10) = 43000000 * e^(10k)
y(10) = 173958983
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answer:
in 10 years there should be about 173.958983 million households with dogs.
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