after years of maintaining a steady population of 32,000, the population of a town begins to grow…

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.)\no y = 32,000(1.08)^x\no y = 32,000(0.08)^x\no y = 34,560(1.08)^x\no y = 34,560(0.08)^x

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.)\no y = 32,000(1.08)^x\no y = 32,000(0.08)^x\no y = 34,560(1.08)^x\no y = 34,560(0.08)^x

Answer

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 number of time - periods.

Step2: Identify the initial population and growth rate

The initial population of the town is $a = 32000$ (since it was steady at 32000 before the exponential growth started), and the growth rate $r=0.08$ (because 8% = 0.08).

Step3: Substitute values into the formula

Substituting $a = 32000$ and $r = 0.08$ into the formula $y=a(1 + r)^x$, we get $y = 32000(1 + 0.08)^x=32000(1.08)^x$.

Answer:

$y = 32000(1.08)^x$