Sovi.AI - AI Math Tutor

Scan to solve math questions

QUESTION IMAGE

after years of maintaining a steady population of 32,000, the populatio…

Question

after years of maintaining a steady population of 32,000, the population of a town begins to grow exponentially. after 1 year and an increase of 8% per year, the population is 34,560. which equation can be used to predict, y, the number of people living in the town after x years? (round population values to the nearest whole number.)
\\( \circ \\ y = 32,000(1.08)^x \\)
\\( \circ \\ y = 32,000(0.08)^x \\)
\\( \circ \\ y = 34,560(1.08)^x \\)
\\( \circ \\ y = 34,560(0.08)^x \\)

Explanation:

Step1: Recall Exponential Growth Formula

The general formula for exponential growth is $y = a(1 + r)^x$, where $a$ is the initial amount, $r$ is the growth rate (as a decimal), and $x$ is the time.

Step2: Identify Values

Here, the initial population $a = 32,000$. The growth rate $r = 8\%= 0.08$, so $1 + r=1 + 0.08 = 1.08$. The time is $x$ years, and $y$ is the population after $x$ years.

Step3: Form the Equation

Substituting the values into the exponential growth formula, we get $y=32000(1.08)^x$. We can check by plugging $x = 1$: $y=32000(1.08)^1=32000\times1.08 = 34560$, which matches the given information. The other options use incorrect initial values or growth factors (using $0.08$ instead of $1.08$ for growth, or wrong initial population like $34560$ which is the population after 1 year, not the initial population).

Answer:

A. $y = 32,000(1.08)^x$