Use a graphing calculator and the following scenario. The population P of a fish farm in t years is modeled by the equation P(t)=(1700)/(1+9e^(-0.6 t)). To the nearest tenth, how long will it take for the population to reach 900 ? ◻ yr

Identify Equation: Identify the equation that models the population of the fish farm over time. The given equation is P(t)=(1700)/(1+9e^(-0.6 t)), which is a logistic growth model.Set Population to 900: Set the population P(t) to 900 to solve for the time t when the population will reach that number. So, we have 900 = (1700)/(1+9e^(-0.6 t)).Isolate Exponential Part: Isolate the exponential part of the equation by multiplying both sides by the denominator and then dividing by 900. (900)(1+9e^(-0.6 t)) = 1700 1+9e^(-0.6 t) = 1700/900 1+9e^(-0.6 t) = 1.8889 (rounded to four decimal places for simplicity).Subtract to Isolate: Subtract 1 from both sides to isolate the term with the exponential. 9e^(-0.6 t) = 0.8889Divide to Solve: Divide both sides by 9 to solve for the exponential term. e^(-0.6 t) = 0.8889/9 e^(-0.6 t) = 0.0988 (rounded to four decimal places for simplicity).Take Natural Logarithm: Take the natural logarithm of both sides to solve for the exponent. -0.6 t = ln(0.0988)Divide to Solve for t: Divide both sides by -0.6 to solve for t. t = ln(0.0988)/(-0.6)Calculate Value of t: Use a calculator to find the value of t. t ≈ ln(0.0988)/(-0.6) t ≈ 2.3148 (rounded to four decimal places for simplicity).Round to Nearest Tenth: Round the answer to the nearest tenth as the question prompt asks for. t ≈ 2.3 years

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Use a graphing calculator and the following scenario.
The population P of a fish farm in t years is modeled by the equation P(t)=(1700)/(1+9e^(-0.6 t)).
To the nearest tenth, how long will it take for the population to reach 900 ?
◻ yr

Use a graphing calculator and the following scenario. $\newline$ The population $P$ of a fish farm in $t$ years is modeled by the equation $P(t)=\frac{1700}{1+9 e^{-0.6 t}}$ . $\newline$ To the nearest tenth, how long will it take for the population to reach $900$ ? $\newline$ $\square$ yr

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Algebra 1

Write linear and exponential functions: word problems

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Q. Use a graphing calculator and the following scenario.

\newline

The population

P

of a fish farm in

t

years is modeled by the equation

P(t)=\frac{1700}{1+9 e^{-0.6 t}}

\newline

To the nearest tenth, how long will it take for the population to reach

900

\newline

\square

Identify Equation: Identify the equation that models the population of the fish farm over time. $\newline$ The given equation is $P(t)=\frac{1700}{1+9e^{-0.6 t}}$ , which is a logistic growth model.
Set Population to $900$ : Set the population $P(t)$ to $900$ to solve for the time $t$ when the population will reach that number. $\newline$ So, we have $900 = \frac{1700}{1+9e^{-0.6 t}}$ .
Isolate Exponential Part: Isolate the exponential part of the equation by multiplying both sides by the denominator and then dividing by $900$ . $\newline$ ($900) $(1+9e^{-0.6 t}) = 1700$ $ $\newline$ $1+9e^{-0.6 t} = \frac{1700}{900}$ $\newline$ $1+9e^{-0.6 t} = 1.8889$ (rounded to four decimal places for simplicity).
Subtract to Isolate: Subtract $1$ from both sides to isolate the term with the exponential. $\newline$ $9e^{-0.6 t} = 0.8889$
Divide to Solve: Divide both sides by $9$ to solve for the exponential term. $\newline$ $e^{(-0.6 t)} = \frac{0.8889}{9}$ $\newline$ $e^{(-0.6 t)} = 0.0988$ (rounded to four decimal places for simplicity).
Take Natural Logarithm: Take the natural logarithm of both sides to solve for the exponent. $\newline$ $-0.6 t = \ln(0.0988)$
Divide to Solve for t: Divide both sides by $-0.6$ to solve for t. $\newline$ $t = \frac{\ln(0.0988)}{-0.6}$
Calculate Value of t: Use a calculator to find the value of $t$ . $\newline$ $t \approx \ln(0.0988)/(-0.6)$ $\newline$ $t \approx 2.3148$ (rounded to four decimal places for simplicity).
Round to Nearest Tenth: Round the answer to the nearest tenth as the question prompt asks for. $t \approx 2.3$ years