A city is growing at the rate of 0.8% annually. If there were 2,632,000 residents in the city in 1993, find how many (to the nearest ten-thousand) are living in that city in 2000. Use [tex]y = 2,632,000(2.7)^{0.008t}[/tex]

The population of the city in 2000 is approximately 3,010,000 residents. This was calculated using the growth formula provided and the initial population from 1993. The calculation involves evaluating the growth rate over 7 years.
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Answered by Anonymous | 2025-07-15

To solve this problem, we need to calculate the population of the city in the year 2000, given its population growth rate. We will use the formula for exponential growth:
y = y 0 ​ × ( 1 + r ) t
Where:

y 0 ​ is the initial population (in this case, 2,632,000).
r is the growth rate (0.8% per year, or 0.008 as a decimal).
t is the number of years since the starting year (1993).
y is the population after t years.

First, calculate t :
The year 2000 is 7 years after 1993, so t = 7 .
Next, substitute the values into the formula:
y = 2 , 632 , 000 × ( 1 + 0.008 ) 7
This simplifies to:
y = 2 , 632 , 000 × 1.00 8 7
Using a calculator, calculate 1.00 8 7 :
1.00 8 7 ≈ 1.0587
Now, multiply the initial population by this growth factor:
y ≈ 2 , 632 , 000 × 1.0587
y ≈ 2 , 782 , 234
Rounding this to the nearest ten-thousand gives:
y ≈ 2 , 780 , 000
Therefore, the estimated population of the city in 2000 is approximately 2,780,000 residents.

Answered by AvaCharlotteMiller | 2025-07-22