Question

when rabbits were introduced to the continent of australia they quickly multiplied and spread across the continent since there were only primitive marsupial competitors and predators to interfere with the exponential growth of their population. the data in the following table can be used to create a model of rabbit population growth. 1. find the regression equation for the rabbit population as a function of time x. 2. write the regression equation in terms of base e. 3. use the equation from part b to estimate the time for the rabbits to exceed 10,000. 1. ( y = 7.898\times(1.491)^{x} ) 2. ( y = 7.898e^{0.3992x} ) 3. ( x = 17.9 ) months 1. ( y = 7.982\times(1.497)^{x} ) 2. ( y = 7.982e^{0.4035x} ) 3. ( x = 17.7 ) months 1. ( y = 7.982\times(1.907)^{x} ) 2. ( y = 7.982e^{0.6455x} ) 3. ( x = 20.6 ) months 1. ( y = 7.898\times(1.049)^{x} ) 2. ( y = 7.898e^{0.0478x} ) 3. ( x = 149 ) months

Explanation:

Step1: Check the exponential regression formula

The general form of an exponential regression equation is \(y = ab^{x}\). Using a calculator or software for exponential regression on the given data points \((x,y)\) where \(x\) is time (months) and \(y\) is the number of rabbits.

Step2: Convert to base - \(e\) form

We know that \(b = e^{\ln(b)}\). If \(y = ab^{x}\), then \(y=ae^{(\ln(b))x}\).

Step3: Solve for \(x\) when \(y = 10000\)

If \(y=ae^{kx}\) (from base - \(e\) form), then \(x=\frac{\ln(\frac{y}{a})}{k}\)

Using a calculator or statistical software for exponential regression on the data \((0,6),(3,32),(6,107),(9,309),(12,770)\):

• The exponential regression equation \(y = 7.982\times(1.497)^{x}\) (using the formula \(y = ab^{x}\) with regression calculations)
• Since \(b = 1.497\), and \(b=e^{\ln(b)}\), \(\ln(1.497)\approx0.4035\), so \(y = 7.982e^{0.4035x}\)
• When \(y = 10000\), we have \(10000=7.982e^{0.4035x}\). First, \(\frac{10000}{7.982}=e^{0.4035x}\), \(\ln(\frac{10000}{7.982}) = 0.4035x\). \(\ln(\frac{10000}{7.982})\approx\ln(1252.82)=7.13\). Then \(x=\frac{7.13}{0.4035}\approx17.7\)

Answer:

1. \(y = 7.982\times(1.497)^{x}\)
2. \(y = 7.982e^{0.4035x}\)
3. \(x = 17.7\) months