QuestionApril 16, 2026

38. The rate of the spread of a communicable viral infection among the population of Eulerville is modeled by R(t)=530e^0.18t persons per day. If there is no cure in sight and 725 people now have the viral infection, what is the projected number of people who will have the viral infection when t=6 (six days from now)? (A) 2.286 (B) 5,726 (C) 6,451 (D) 7,287

38. The rate of the spread of a communicable viral infection among the population of Eulerville is modeled by R(t)=530e^0.18t persons per day. If there is no cure in sight and 725 people now have the viral infection, what is the projected number of people who will have the viral infection when t=6 (six days from now)? (A) 2.286 (B) 5,726 (C) 6,451 (D) 7,287
38. The rate of the spread of a communicable viral infection among the population of Eulerville is modeled by R(t)=530e^0.18t persons per day. If there is no cure in sight and 725 people now have the viral infection, what is the projected number of people who will have the viral infection when t=6 (six days from now)? (A) 2.286 (B) 5,726 (C) 6,451 (D) 7,287

Solution
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Answer

6,451 Explanation 1. Interpret the rate function R(t) is the rate of change of infected people. To find total infected at t=6, integrate R(t) from t=0 to t=6 and add current infected (725). 2. Integrate the rate function \int_{0}^{6} 530e^{0.18t} \, dt = \frac{530}{0.18} \left[e^{0.18t}\right]_{0}^{6} = \frac{530}{0.18} \left(e^{1.08} - e^{0}\right) 3. Compute values e^{1.08} \approx 2.944 and e^0 = 1 \frac{530}{0.18} \approx 2944.44 Total increase = 2944.44 \times (2.944 - 1) \approx 2944.44 \times 1.944 \approx 5710.6 4. Add initial infected count 725 + 5710.6 \approx 6435.6

Explanation

1. Interpret the rate function<br /> $R(t)$ is the rate of change of infected people. To find total infected at $t=6$, integrate $R(t)$ from $t=0$ to $t=6$ and add current infected (725).<br />2. Integrate the rate function<br /> $\int_{0}^{6} 530e^{0.18t} \, dt = \frac{530}{0.18} \left[e^{0.18t}\right]_{0}^{6}$<br /> $= \frac{530}{0.18} \left(e^{1.08} - e^{0}\right)$<br />3. Compute values<br /> $e^{1.08} \approx 2.944$ and $e^0 = 1$<br /> $\frac{530}{0.18} \approx 2944.44$<br /> Total increase = $2944.44 \times (2.944 - 1) \approx 2944.44 \times 1.944 \approx 5710.6$<br />4. Add initial infected count<br /> $725 + 5710.6 \approx 6435.6$
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