Chapter 5 Challenge Problems In an isolated town in Mongolia of 49,454 people th

Question

Chapter 5 Challenge Problems In an isolated town in Mongolia of 49,454 people there is an outbreak of flu that is extremely virulent. The anticipated population that is infected is given by the formula 1. 49454 1+312 0.75r Pt P(t) noticed is the population, and t is the time elapsed in weeks after the infection was What is the starting infected population? How long is it before 90% of the population has been infected, and possibly gotten better? How many people are infected after 1 month? a. b. c. We want to buy a house for $193,000, with a $35,000 down payment, and you get a loan for 3.2% per year. 2.

Explanation / Answer

1. a. since t = 0 at the time when the infection was noticed, hence the starting infected population is P(0) = 49454/[1+312*e0] =49454/313 = 158.

b. Let t weeks be the time elapsed when 90 % of Mongolia’s population got infected. Then P(t) = 90% of 49454 = 44509 ( on rounding off to the nearest whole number). Also, P(t) =49454/[1+312*e-0.75t]   so that 49454/[1+312*e-0.75t]=44509 or, [1+312*e-0.75t]=49454/44509=1.111101126 or,312*e-0.75t = 0.111101126 or, e-0.75t = 0.111101126 /312 or, e0.75t = 312/0.111101126 = 2808.252366. Now, on taking natural log of both the sides, we get 0.75t = ln 2808.252366 = 7.940317635 so that t = 7.940317635 /0.75 = 10.58709018. Thus, 90 % of Mongolia’s population got infected in the 11th week.

c. Since 1 month = 4 weeks, the number of people infected after 1 month = 49454/[1+312*e-0.75*4] = 49454/[1+312*e-3] = 49454/[1+312*0.049787068] = 49454/16.53356533 = 2991 ( on rounding off to the nearest whole number).

Related Questions

Navigate

• Browse (All)
• Browse C
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.