Question

problem 6
this table shows the population of a city from 1988 to 2016.
select two values of t that create an interval with a negative rate
of change.

Explanation:

Step1: Recall the formula for rate of change

The rate of change formula is $\frac{p(t_2)-p(t_1)}{t_2 - t_1}$. A negative rate of change means $p(t_2)-p(t_1)<0$ (since $t_2>t_1$ as years are increasing).

Step2: Check pairs of years

• For \(t_1 = 1992\), \(p(1992)=42700\) and \(t_2 = 1996\), \(p(1996)=33100\). Then \(p(1996)-p(1992)=33100 - 42700=-9600<0\).
• For other pairs:
• \(1988 - 1992\): \(p(1992)-p(1988)=42700 - 35700 = 7000>0\)
• \(1996 - 2000\): \(p(2000)-p(1996)=33700 - 33100=600>0\)
• \(2000 - 2004\): \(p(2004)-p(2000)=45000 - 33700 = 11300>0\)
• \(2004 - 2008\): \(p(2008)-p(2004)=48400 - 45000=3400>0\)
• \(2008 - 2012\): \(p(2012)-p(2008)=40000 - 48400=-8400<0\) (but we need to check the given options. If we consider the first valid pair from the options given for selection)

Answer:

1992 and 1996