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.) y = 32,000(1.08)x y = 32,000(0.08)x y = 34,560(1.08)x y = 34,560(0.08)x

Answer 1

Answer:

The value of population after x years = y =  32,000 × $(1.08)^{x}$  .

Step-by-step explanation:

Given as :

The initial population = p = 32,000

The final population after 1 year = P = 34,560

The rate of increase of population = r = 8%

The number of people after x years

According to question

The population after n years = initial population × $(1 + \dfrac{\tetxrm rate}{100})^{\textrm time}$

Or, P = p × $(1 + \dfrac{\tetxrm r}{100})^{\textrm t}$

Or, 34,560 = 32,000 × $(1 + \dfrac{\tetxrm 8}{100})^{\textrm x}$

Or, 34,560 = 32,000 × $(1.08)^{x}$

Or, y =  32,000 × $(1.08)^{x}$

So, The value of population after x years = y =  32,000 × $(1.08)^{x}$

Hence, The value of population after x years = y =  32,000 × $(1.08)^{x}$  . Answer

Answer 2

Answer:

y=32,000(1.08)^x

Step-by-step explanation:

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