From 2000 to 2010 , a city had a \( 2.5 \% \) annual decrease in population. \( 2.950,000 \) people in 2000, determine the population in 2010 . \( \left.f(x)=400<,(1-0.25)^{10}\right)= \)

To find the population in 2010, we can use the formula for exponential decay: \[ f(x) = P_0 \times (1 - r)^x \] where: - \( P_0 \) is the initial population (2000), - \( r \) is the annual decrease rate (2.5% or 0.025), - \( x \) is the number of years (10 years from 2000 to 2010). Given: - Initial population in 2000, \( P_0 = 2,950,000 \), - Annual decrease rate, \( r = 0.025 \), - Number of years, \( x = 10 \). Substitute the values into the formula: \[ f(10) = 2,950,000 \times (1 - 0.025)^{10} \] Now, we can calculate the population in 2010 using the formula. Calculate the value by following steps: - step0: Calculate: \(2950000\left(1-0.025\right)^{10}\) - step1: Subtract the numbers: \(2950000\times 0.975^{10}\) - step2: Convert the expressions: \(2950000\left(\frac{39}{40}\right)^{10}\) - step3: Simplify: \(2950000\times \frac{39^{10}}{40^{10}}\) - step4: Rewrite the expression: \(40\times 73750\times \frac{39^{10}}{40^{10}}\) - step5: Reduce the numbers: \(73750\times \frac{39^{10}}{40^{9}}\) - step6: Rewrite the expression: \(10\times 7375\times \frac{39^{10}}{40^{9}}\) - step7: Rewrite the expression: \(10\times 7375\times \frac{39^{10}}{10^{9}\times 4^{9}}\) - step8: Reduce the numbers: \(7375\times \frac{39^{10}}{10^{8}\times 4^{9}}\) - step9: Rewrite the expression: \(125\times 59\times \frac{39^{10}}{10^{8}\times 4^{9}}\) - step10: Rewrite the expression: \(125\times 59\times \frac{39^{10}}{5^{8}\times 2^{8}\times 4^{9}}\) - step11: Rewrite the expression: \(5^{3}\times 59\times \frac{39^{10}}{5^{8}\times 2^{8}\times 4^{9}}\) - step12: Reduce the numbers: \(59\times \frac{39^{10}}{5^{5}\times 2^{8}\times 4^{9}}\) - step13: Multiply the fractions: \(\frac{59\times 39^{10}}{800000\times 4^{9}}\) The population in 2010 is approximately 2,290,172.