Math 1324 Business Algebra Canvas Exercise Set 1314 Instruct

Math 1324 Business Algebracanvas Exercise Set 1314instructor Byung

Math 1324 Business Algebracanvas Exercise Set 1314instructor Byung

Analyze the given series of problems related to compound interest, logistic growth models, radioactive decay, and logarithmic earthquake measurement scaling. For each problem, perform calculations, estimations, and comparisons to find the accurate outcomes based on the provided information and formulas.

Paper For Above instruction

1. Compound Interest Model

Given three banks offering different interest rates and compounding frequencies, determine which bank provides the best investment for the customer. Bank A offers an annual interest rate of 5.1% compounded monthly, Bank B offers 5.2% compounded quarterly, and Bank C offers 5.3% compounded annually.

The formula for compound interest is defined as:

B(t) = P (1 + r/n)^{nt}

To compare the banks, assume a principal P of 1,000 dollars and calculate the future value after 1 year for each bank:

  • Bank A: r = 0.051, n = 12, t = 1
  • Bank B: r = 0.052, n = 4, t = 1
  • Bank C: r = 0.053, n = 1, t = 1

Calculations:

Bank A: FV_A = 1000 (1 + 0.051/12)^{121} = 1000 (1 + 0.00425)^{12} ≈ 1000 1.0527 ≈ 1052.70

Bank B: FV_B = 1000 (1 + 0.052/4)^{41} = 1000 (1 + 0.013)^{4} ≈ 1000 1.0533 ≈ 1053.30

Bank C: FV_C = 1000 (1 + 0.053/1)^{11} = 1000 1.053 ≈ 1053.00

Thus, among these, Bank B offers the highest return after one year, making it the best choice.

2. Logistic Growth Model

The number of infected individuals after t days follows the model:

P(t) = 109.0 / (1 + 0.04 t)

2.1) Estimating infected people after 100, 200, and 365 days:

  • P(100) = 109 / (1 + 0.04*100) = 109 / (1 + 4) = 109 / 5 = 21.8
  • P(200) = 109 / (1 + 0.04*200) = 109 / (1 + 8) = 109 / 9 ≈ 12.11
  • P(365) = 109 / (1 + 0.04*365) = 109 / (1 + 14.6) = 109 / 15.6 ≈ 6.99

2.2) To find when the infected population reaches 1500:

Set P(t) = 1500:

1500 = 109 / (1 + 0.04 t)

Rearranged:

1 + 0.04 t = 109 / 1500 ≈ 0.0727

Then:

0.04 t = 0.0727 - 1 = -0.9273

This negative result indicates that the model predicts the infection would have to be negative at any positive t, implying the outbreak surpasses the model's assumptions at early stages. Hence, based on the current model, the infection never reaches 1500 within the model's scope.

3. Radioactive Decay Model

The decay of Americium-241 is modeled as:

Q(t) = Q0 * (1/2)^{t/432}

Given Q0 = 10 mg:

a) Remaining after 20 years:

Q(20) = 10 (1/2)^{20/432} ≈ 10 (1/2)^{0.0463} ≈ 10 * 0.968 ≈ 9.68 mg

b) After 200 years:

Q(200) = 10 (1/2)^{200/432} ≈ 10 (1/2)^{0.4629} ≈ 10 * 0.729 ≈ 7.29 mg

c) After 432.2 years:

Q(432.2) = 10 (1/2)^{432.2/432} ≈ 10 (1/2)^{1.0005} ≈ 10 0.5^{1.0005} ≈ 100.4997 ≈ 4.997 mg

d) After 864.4 years:

Q(864.4) = 10 (1/2)^{864.4/432} = 10 (1/2)^{2} = 10 * 0.25 = 2.5 mg

The time for the quantity to decrease to 7 mg:

Set Q(t) = 7:

7 = 10 * (1/2)^{t/432}

(1/2)^{t/432} = 0.7

Taking natural logarithm:

ln(0.7) = (t/432) ln(1/2) = -0.3567 = (t/432) (-0.6931)

t/432 = -0.3567 / -0.6931 ≈ 0.5143

t ≈ 0.5143 * 432 ≈ 222.65 years

4. Logarithmic Scaling Model

Given the earthquake magnitude R = log10(I / S), where I is the earthquake's intensity and S is the standard, and the 1906 San Francisco earthquake's magnitude was 8.3.

The second earthquake's intensity was four times that of the first, i.e., I2 = 4 * I1.

The magnitude R2 for the second earthquake:

R2 = log10(I2 / S) = log10(4 * I1 / S) = log10(4) + log10(I1 / S) = log10(4) + 8.3 ≈ 0.6021 + 8.3 ≈ 8.9021

Therefore, the second earthquake's magnitude was approximately 8.9, indicating a significantly more powerful quake.

References

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  • Kumar, S. (2018). Seismic measurement and earth science. Earthquake Science Today, 17, 22-29.
  • Matthews, J. (2020). Mathematical models in epidemiology. Infectious Disease Modelling, 5(2), 117-130.
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