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p(t)=80(1.05)^(t)
The function models 
p, the population, in thousands, of City 
Xt years after 2000 . Based on the model, what is the approximate population, in thousands, of City 
X in 2010 ?
Choose 1 answer:
(A) 88
(B) 
130
(c) 
212
(D) 840

p(t)=80(1.05)t p(t)=80(1.05)^{t} \newlineThe function models p p , the population, in thousands, of City XX tt years after 20002000. Based on the model, what is the approximate population, in thousands, of City X X in 20102010 ?\newlineChoose 11 answer:\newline(A) 8888\newline(B) 130130 \newline(C) 212212 \newline(D) 840840

Full solution

Q. p(t)=80(1.05)t p(t)=80(1.05)^{t} \newlineThe function models p p , the population, in thousands, of City XX tt years after 20002000. Based on the model, what is the approximate population, in thousands, of City X X in 20102010 ?\newlineChoose 11 answer:\newline(A) 8888\newline(B) 130130 \newline(C) 212212 \newline(D) 840840
  1. Identify the given function: Identify the given function and what it represents.\newlineThe function p(t)=80(1.05)tp(t) = 80(1.05)^t models the population of City X, in thousands, tt years after 20002000.
  2. Determine the value of t: Determine the value of t for the year 20102010.\newlineSince t represents the number of years after 20002000, for the year 20102010, t = 20102010 - 20002000 = 1010 years.
  3. Substitute the value of t: Substitute the value of t into the function to find the population in 20102010.\newlinep(1010) = 8080(11.0505)^{1010}
  4. Calculate the population in 20102010: Calculate the population in 20102010 using the function. p(10)=80(1.05)1080×1.62889130.3112p(10) = 80(1.05)^{10} \approx 80 \times 1.62889 \approx 130.3112
  5. Round the population to the nearest thousand: Round the population to the nearest thousand as the answer choices are in whole numbers.\newlineThe approximate population in thousands is 130130.

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