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P(t)=20(0.95)^(t)
The function models 
P, the population of Leetown in thousands, 
t years after 2007 . What was the population of Leetown in 2007 ?
Choose 1 answer:
(A) 5 thousand
(B) 19 thousand
(C) 20 thousand
(D) 95 thousand

P(t)=20(0.95)tP(t)=20(0.95)^t\newlineThe function models PP, the population of Leetown in thousands, tt years after 20072007. What was the population of Leetown in 20072007?\newlineChoose 11 answer:\newline(A)\text{(A)} 55 thousand\newline(B)\text{(B)} 1919 thousand\newline(C)\text{(C)} 2020 thousand\newline(D)\text{(D)} 9595 thousand

Full solution

Q. P(t)=20(0.95)tP(t)=20(0.95)^t\newlineThe function models PP, the population of Leetown in thousands, tt years after 20072007. What was the population of Leetown in 20072007?\newlineChoose 11 answer:\newline(A)\text{(A)} 55 thousand\newline(B)\text{(B)} 1919 thousand\newline(C)\text{(C)} 2020 thousand\newline(D)\text{(D)} 9595 thousand
  1. Understand function, determine population: Understand the function and determine the initial population.\newlineThe function P(t)=20(0.95)tP(t) = 20(0.95)^t models the population of Leetown in thousands, tt years after 20072007. To find the population in 20072007, we need to evaluate the function at t=0t = 0, because 00 years after 20072007 is the year 20072007 itself.
  2. Substitute t=0t = 0: Substitute t=0t = 0 into the function to find the initial population.P(0)=20(0.95)0P(0) = 20(0.95)^0Since any number raised to the power of 00 is 11, we have:P(0)=20×1P(0) = 20 \times 1P(0)=20P(0) = 20
  3. Interpret result: Interpret the result.\newlineThe population of Leetown in 20072007 was 2020 thousand.

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