Worked solutions with a scoring guide for every question. Accept answers that differ in the last decimal because of rounding, award partial credit for a correct setup with an arithmetic slip, and total the score out of 20 points.
Question 1: Growth model (2 points)
- Initial population 18,000; growth 4.5% per year.
- \(P(8)=18000(1.045)^{8}\approx\) 25,598.
Scoring: 1 point for each part.
Question 2: Compound interest (2 points)
- \(A=2400(1+\frac{0.032}{12})^{60}\).
- The balance is $2,815.83.
Scoring: 1 point for the correct formula with n = 12 and nt = 60; 1 point for the correct balance.
Question 3: Continuous versus yearly (2 points)
- \(A=1800e^{0.045\cdot 12}=1800e^{0.54}\approx\) $3,088.81.
- \(A=1800(1.045)^{12}\approx\) $3,052.59, which is $36.23 less (using unrounded values).
Scoring: 1 point for each balance method with the matching difference in part (b).
Question 4: Model from a graph (3 points)
- \(a=f(0)=5\) and \(b=\frac{15}{5}=3\). Check: \(\frac{45}{15}=3\).
- \(f(x)=5\cdot 3^{x}\).
- \(f(4)=5\cdot 81=\)405.
Scoring: 1 point for a and b, 1 point for the equation, 1 point for f(4) = 405.
Question 5: Solve for time (3 points)
- \(1.04^{t}=1.75\), so \(t=\frac{\ln 1.75}{\ln 1.04}\).
- \(t\approx\) 14.27.
- After 14 years \(A(14)\approx\) $3,463.35 is below $3,500, and after 15 years \(A(15)\approx\) $3,601.89 is above. The answer is 15 years.
Scoring: 1 point for the exact logarithm form, 1 point for 14.27, 1 point for checking years 14 and 15.
Question 6: Doubling time (2 points)
Solve \(1.06^{t}=2\): \(t=\frac{\ln 2}{\ln 1.06}\approx\) 11.90 years.
Scoring: 1 point for the equation 1.06^t = 2 with a logarithm step; 1 point for the correct value.
Question 7: Half-life (3 points)
- The mass halves from 240 to 120 in 5 days: 5 days.
- \(A(t)=240(\frac{1}{2})^{t/5}\).
- \(A(12)=240(\frac{1}{2})^{2.4}\approx\) 45.5 mg.
Scoring: 1 point for the half-life, 1 point for the model, 1 point for the value.
Question 8: Equivalent forms (3 points)
- Since \(b=1.12\), the growth rate is 12% per year.
- \(e^{k}=1.12\), so \(k=\ln 1.12\approx\) 0.1133, giving \(f(t)=500e^{0.1133t}\).
- \(b_{\text{month}}=1.12^{1/12}\approx\) 1.00949, so the monthly rate is about 0.95%.
Scoring: 1 point for each part.
20 points
See every chapter on the Algebra 2 math page.


