Question

a city has a population of 133,000. the population y increases by 4% each year.
a. write an exponential function that represents the population after t years.
b. what will the population be after 11 years? round your answer to the nearest thousand.
a. y = \square
b. about \square

Explanation:

Part (a)

Step1: Recall exponential growth formula

The general form of an exponential growth function is \( y = a(1 + r)^t \), where \( a \) is the initial amount, \( r \) is the growth rate (as a decimal), and \( t \) is time.
Here, the initial population \( a = 133000 \), and the growth rate \( r = 4\% = 0.04 \).

Step2: Substitute values into the formula

Substitute \( a = 133000 \) and \( r = 0.04 \) into the formula: \( y = 133000(1 + 0.04)^t \), which simplifies to \( y = 133000(1.04)^t \).

Step1: Use the function from part (a)

We have \( y = 133000(1.04)^t \), and we need to find \( y \) when \( t = 11 \).

Step2: Calculate the value

First, calculate \( (1.04)^{11} \). Using a calculator, \( (1.04)^{11}\approx1.53945 \).
Then, multiply by 133000: \( y = 133000\times1.53945\approx204746.85 \).

Step3: Round to the nearest thousand

Rounding 204746.85 to the nearest thousand gives 205000 (since the hundreds digit is 7, which is more than 5, we round up the thousands place: 204000 + 1000 = 205000).

Answer:

\( y = 133000(1.04)^t \)

Part (b)