Question

An investor is deciding between two options for a short-term investment. One option has a return \(R\), in dollars, \(t\) months after investment, and is modelled by the equation \(R=100\left(3^t\right)\). The other option has a return \(R\), in dollars, \(t\) months after investment, and is modeled by the equation \(R=350 t\). After 4 months, how much less is the return given by the linear model than the return given by the exponential model? A) \(\$ 1,400\) B) \(\$ 4,050\) C) \(\$ 6,700\) D) \(\$ 8,100\) $$ n-\sqrt{2 n+22}=1 $$

Short answer

The return is \$6,700 less for the linear model after 4 months. So, the correct answer is C) \( \$ 6,700\).

Step-by-step solution

Substitute t in the models

First, we need to find the return values for both models at 4 months (t=4). Let's substitute t=4 in both equations and find the values of R.

Calculate R for exponential model

For the exponential model, the equation is \(R = 100(3^t)\). Plug in t=4: \[R = 100(3^4) = 100(81) = 8100\]

Calculate R for linear model

For the linear model, the equation is \(R = 350t\). Plug in t=4: \[R = 350(4) = 1400\]

Find the difference in returns

Now that we have the returns for both models, we need to find the difference. We are asked to find how much less the return is given by the linear model than by the exponential model. The difference is: \[R_{difference} = R_{exponential} - R_{linear} = 8100 - 1400\]

Calculate the difference and choose the correct option

Calculate the difference in returns: \[R_{difference} = 8100 - 1400 = 6700\] The return is \$6,700 less for the linear model after 4 months. So, the correct answer is C) \( \$ 6,700\).