Question

guided practice
exponential models
in 1990, the population of washington d.c. was about 604,000 people. since then, the population has decreased about 1.8% each year. what was the population 4 years later?
a. 593,128
b. 582,452
c. 561,672
d. 571,968

Explanation:

Step1: Identify the formula for exponential decay

The formula for exponential decay is \( P(t) = P_0(1 - r)^t \), where \( P_0 \) is the initial population, \( r \) is the rate of decay (as a decimal), and \( t \) is the time in years.

Step2: Plug in the values

Here, \( P_0 = 604000 \), \( r = 0.018 \) (since 1.8% = 0.018), and \( t = 4 \). So we calculate \( P(4)=604000\times(1 - 0.018)^4 \).

Step3: Calculate \( (1 - 0.018) \)

First, \( 1- 0.018=0.982 \).

Step4: Calculate \( 0.982^4 \)

\( 0.982^4=0.982\times0.982\times0.982\times0.982\approx0.9271 \) (using a calculator for the exponentiation).

Step5: Multiply by the initial population

\( P(4)=604000\times0.9271\approx604000\times0.9271 = 604000\times(0.9 + 0.02+0.007 + 0.0001)=604000\times0.9+604000\times0.02 + 604000\times0.007+604000\times0.0001=543600+12080 + 4228+60.4 = 543600+12080=555680; 555680 + 4228 = 559908; 559908+60.4 = 559968.4\) (this is a rough calculation, a more accurate calculation using a calculator for \( 604000\times0.982^4 \)):
\( 0.982^4\approx0.9271\), and \( 604000\times0.9271 = 604000\times0.9271 = 604000\times\frac{9271}{10000}=\frac{604000\times9271}{10000}=\frac{604\times9271}{10}=\frac{604\times(9000 + 271)}{10}=\frac{(600\times9000+600\times271 + 4\times9000+4\times271)}{10}=\frac{(5400000+162600+36000 + 1084)}{10}=\frac{5400000+162600 = 5562600; 5562600+36000=5598600; 5598600 + 1084 = 5599684}{10}=559968.4\). Wait, this is incorrect. Wait, maybe I made a mistake in the exponentiation. Let's calculate \( 0.982^2 = 0.982\times0.982 = 0.964324 \), then \( 0.964324\times0.982^2=0.964324\times0.964324\approx0.9299 \) (wait, no, \( t = 4 \), so it's \( (0.982)^4=(0.982^2)^2 \). \( 0.982^2 = 0.964324 \), then \( 0.964324^2=(0.96 + 0.004324)^2=0.96^2+2\times0.96\times0.004324+0.004324^2=0.9216+0.0083328+0.0000187=0.9299515 \). Then \( 604000\times0.9299515\approx604000\times0.9299515 = 604000\times(0.93 - 0.0000485)=604000\times0.93-604000\times0.0000485 = 561720-29.294 = 561720 - 29.294 = 561690.706\). Wait, but the options are A. 593,128, B. 582,452, C. 561,672, D. 571,968. Wait, maybe my initial rate is wrong? Wait, the problem says "decreased about 1.8% each year", so the decay factor is \( 1 - 0.018 = 0.982 \), that's correct. Wait, maybe I miscalculated \( 0.982^4 \). Let's use a calculator for \( 0.982^4 \):

\( 0.982\times0.982 = 0.964324 \)

\( 0.964324\times0.982 = 0.964324\times(1 - 0.018)=0.964324-0.964324\times0.018=0.964324 - 0.017357832 = 0.946966168 \)

\( 0.946966168\times0.982=0.946966168\times(1 - 0.018)=0.946966168-0.946966168\times0.018=0.946966168 - 0.017045391 = 0.929920777 \)

Then \( 604000\times0.929920777\approx604000\times0.929920777 \)

\( 604000\times0.929920777 = 604000\times0.9 + 604000\times0.02+604000\times0.009+604000\times0.000920777 \)

\( 604000\times0.9 = 543600 \)

\( 604000\times0.02 = 12080 \)

\( 604000\times0.009 = 5436 \)

\( 604000\times0.000920777\approx604000\times0.000920777\approx556.15 \)

Adding them up: \( 543600+12080 = 555680; 555680+5436 = 561116; 561116 + 556.15 = 561672.15 \), which is approximately 561,672, which is option C. Wait, but earlier I had a miscalculation. So the correct calculation gives us approximately 561,672, so the answer is C. 561,672. (I think I made a mistake in the exponentiation step earlier, the correct \( 0.982^4\approx0.9299 \), and \( 604000\times0.9299\approx561672 \))

Answer:

D. 571,968