Question

the population of a city was about 170 thousand in 1980 and has decreased by about 2.7 thousand per year since then. a. complete the table to help find an expression that stands for the population of this city (in thousands) at t years since 1980. b. evaluate the expression that you found in part (a) for t = 26. what does your result mean in this situation? a. complete the table below with the appropriate expressions years since 1980 expression for the population (thousands) 0 1 2 3 4 t (do not simplify. use integers or decimals for any numbers in the expressions )

Explanation:

Step1: Initial population value

The initial population in 1980 (when $t = 0$) is 170 thousand. So for $t=0$, the population expression is 170.

Step2: Population after 1 - year

Since the population decreases by 2.7 thousand per year, after 1 year ($t = 1$), the population is $170-2.7\times1=170 - 2.7$.

Step3: Population after 2 - years

After 2 years ($t = 2$), the population is $170-2.7\times2=170 - 5.4$.

Step4: Population after 3 - years

After 3 years ($t = 3$), the population is $170-2.7\times3=170 - 8.1$.

Step5: Population after 4 - years

After 4 years ($t = 4$), the population is $170-2.7\times4=170 - 10.8$.

Step6: General expression for population

For $t$ years since 1980, the expression for the population (in thousands) is $170-2.7t$.

Step7: Evaluate for $t = 26$

Substitute $t = 26$ into the expression $170-2.7t$. We get $170-2.7\times26=170 - 70.2 = 99.8$. This means that 26 years after 1980 (i.e., in 2006), the population of the city is about 99.8 thousand.

Answer:

Years Since 1980	Expression for the Population (thousands)
1	$170 - 2.7$
2	$170 - 5.4$
3	$170 - 8.1$
4	$170 - 10.8$
$t$	$170-2.7t$

For part b, when $t = 26$, the population is 99.8 thousand, which means the population of the city in 2006 is about 99.8 thousand.