Questions: A population can be modeled by the exponential equation y=220,000 * e^(-0.0756 * t), where t= years since 1990 and y= population. Complete parts (a) through (d). a. What is the continuous decay rate per year? The population is decreasing at a continuous rate of % per year. (Round to two decimal places as needed.) b. What is the annual decay rate (not continuous)? The population is decreasing at an annual rate of % per year. (Round to two decimal places as needed.) c. Rewrite the equation in the form P=P0 *(1+r)^t. P= (Simplify your answer. Use integers or decimals for any numbers in the expression. Round to four decimal places as needed.) d. How many people will there be after 8 years? people (Round to the nearest whole number as needed.)

A population can be modeled by the exponential equation y=220,000 * e^(-0.0756 * t), where t= years since 1990 and y= population. Complete parts (a) through (d).
a. What is the continuous decay rate per year?

The population is decreasing at a continuous rate of % per year.
(Round to two decimal places as needed.)
b. What is the annual decay rate (not continuous)?

The population is decreasing at an annual rate of % per year.
(Round to two decimal places as needed.)
c. Rewrite the equation in the form P=P0 *(1+r)^t.

P=
(Simplify your answer. Use integers or decimals for any numbers in the expression. Round to four decimal places as needed.)
d. How many people will there be after 8 years?
people
(Round to the nearest whole number as needed.)

Transcript text: A population can be modeled by the exponential equation $\mathrm{y}=220,000 \cdot e^{-0.0756 \cdot t}$, where $\mathrm{t}=$ years since 1990 and $\mathrm{y}=$ population. Complete parts (a) through (d). a. What is the continuous decay rate per year? The population is decreasing at a continuous rate of $\square$ \% per year. (Round to two decimal places as needed.) b. What is the annual decay rate (not continuous)? The population is decreasing at an annual rate of $\square$ \% per year. (Round to two decimal places as needed.) c. Rewrite the equation in the form $P=P_{0} \cdot(1+r)^{t}$. \[ \mathrm{P}=\square \] (Simplify your answer. Use integers or decimals for any numbers in the expression. Round to four decimal places as needed.) d. How many people will there be after 8 years? $\square$ people (Round to the nearest whole number as needed.)

Solution

Solution Steps

Solution Approach

a. The continuous decay rate is given directly by the exponent in the equation, which is -0.0756. The decay rate as a percentage is the positive value of this number multiplied by 100.

b. To find the annual decay rate, convert the continuous decay rate to an annual rate using the formula $ r = e^k - 1 $, where $ k $ is the continuous decay rate.

c. Rewrite the exponential equation in the form $ P = P_0 \cdot (1 + r)^t $ by finding the equivalent annual decay rate and using it in the equation.

Step 1: Continuous Decay Rate

The continuous decay rate is given by the exponent in the equation $ y = 220,000 \cdot e^{-0.0756 \cdot t} $. Thus, the continuous decay rate is: \[ \text{Continuous Decay Rate} = -k = 0.0756 \] Expressed as a percentage: \[ \text{Continuous Decay Rate (\%)} = 0.0756 \times 100 = 7.56\% \]

Step 2: Annual Decay Rate

To find the annual decay rate, we use the formula: \[ r = e^k - 1 \] Substituting $ k = -0.0756 $: \[ r = e^{-0.0756} - 1 \approx -0.0728129928028105 \] Expressed as a percentage: \[ \text{Annual Decay Rate (\%)} = |r| \times 100 \approx 7.2813\% \]

Step 3: Rewrite the Equation

The original equation can be rewritten in the form $ P = P_0 \cdot (1 + r)^t $. Here, $ P_0 = 220,000 $ and $ r \approx -0.0728129928028105 $: \[ P = 220000 \cdot (1 - 0.0728129928028105)^t \] This simplifies to: \[ P \approx 220000 \cdot (0.9271870071971895)^t \]

Step 4: Population After 8 Years

To find the population after 8 years, we substitute $ t = 8 $ into the equation: \[ P = 220000 \cdot e^{-0.0756 \cdot 8} \approx 120160.40353839587 \] Rounding to the nearest whole number gives: \[ P \approx 120160 \]