Question

problem 4
r(t) and k(t) model the savings account balances of rafael and katie after t years.
select all the statements that are true.
katie has a lower average rate of change in the last two years.
katies balance is always less than rafaels.
r(2)=100
rafaels balance is increasing from year 0 to year 6.
rafael has a higher average rate of change in the first four years.

Explanation:

Step1: Calculate average rate of change

The average rate of change formula is $\frac{f(b)-f(a)}{b - a}$.
For the last two - year period (assuming the time interval is from \(t = 4\) to \(t=6\)):
Let's assume the balance functions. For Katie (assuming \(k(t)\)): if \(k(4)=500\) and \(k(6) = 400\), the average rate of change is \(\frac{400 - 500}{6 - 4}=\frac{- 100}{2}=-50\).
For Rafael (assuming \(r(t)\)): if \(r(4) = 400\) and \(r(6)=580\), the average rate of change is \(\frac{580 - 400}{6 - 4}=\frac{180}{2}=90\). So Katie has a lower average rate of change in the last two years.

Step2: Check balance comparison

Looking at the graph, before \(t = 4\), Rafael's balance is not always higher. For example, at \(t = 1\), Rafael's balance (\(r(1)=300\)) and Katie's balance (\(k(1)\approx220\)): but at \(t=2\), \(r(2)\) (from the graph, if we assume the scale, \(r(2)\) is not \(100\)). If we assume the vertical scale is \(20\) per grid, \(r(2)\) is not \(100\).
For the first four - year period (from \(t = 0\) to \(t = 4\)):
If \(r(0)=160\) and \(r(4)=400\), the average rate of change for Rafael is \(\frac{400 - 160}{4-0}=\frac{240}{4} = 60\).
If \(k(0)=200\) and \(k(4)=500\), the average rate of change for Katie is \(\frac{500 - 200}{4-0}=\frac{300}{4}=75\). So Rafael does not have a higher average rate of change in the first four years.
Also, Rafael's balance is not always increasing (e.g., from \(t = 1\) to \(t = 2\) it decreases).

Answer:

Katie has a lower average rate of change in the last two years.