Question

how fast would a zombie outbreak take over the world? in a secret government facility, a biological weapons test goes horribly wrong, releasing two strains of a zombie virus that will turn subjects into the walking dead. strain a - runners when the strain is initially released, 10 people are infected, turning them into zombies. each hour, the original zombies \infect\ 3 new people, but the new zombies cannot infect anyone. strain b - stumblers when the strain is initially released, 1 person is infected, turning them into a zombie. each hour, every zombie is able to infect 1 person. new zombies are able to infect people. which strain do you think will have infected more after 15 hours? strain b because

Explanation:

for Strain A:

Step1: Identify the pattern for Strain A

Strain A starts with 10 zombies. Each hour, only the original 10 zombies infect 3 new people each. So the number of new infections per hour is constant: \( 10\times3 = 30 \) new infections per hour. But wait, actually, the initial 10 zombies, and each hour they infect 3, but new zombies can't infect. So the total number of infected after \( t \) hours is initial + (infections per hour)×\( t \). Wait, no: initial infected is 10 (zombies). Then each hour, 10 zombies infect 3 each, so 30 new infections per hour. So total infected \( I_A(t)=10 + 30t \)? Wait, no, wait: the initial 10 are infected (zombies). Then each hour, the original 10 zombies infect 3 new people each. So at hour 1: 10 + 10×3 = 40? Wait, no, the problem says "when the strain is initially released, 10 people are infected, turning them into zombies. Each hour, the original zombies 'infect' 3 new people, but the new zombies cannot infect anyone." So original zombies are always 10? Wait, no, the original zombies are the initial 10. So each hour, the 10 original zombies infect 3 new people each. So number of new infections per hour is \( 10\times3 = 30 \). So total infected after \( t \) hours: initial (10) + 30×\( t \). Wait, but at \( t = 0 \), it's 10. At \( t = 1 \), 10 + 30×1 = 40. At \( t = 2 \), 10 + 30×2 = 70, etc.

Step2: Calculate for Strain A at \( t = 15 \)

\( I_A(15)=10 + 30\times15 \)
\( I_A(15)=10 + 450 = 460 \)

for Strain B:

Step1: Identify the pattern for Strain B

Strain B starts with 1 zombie. Each hour, every zombie (including new ones) infects 1 person. So this is exponential growth. The formula for exponential growth is \( I_B(t)=I_0\times r^t \), where \( I_0 = 1 \) (initial infected), and \( r = 2 \) (since each zombie becomes 2 zombies: itself + 1 infected). Wait, because each zombie infects 1, so each zombie produces 1 new zombie. So the number of zombies doubles each hour? Wait, initial: 1. After 1 hour: 1 + 1×1 = 2. After 2 hours: 2 + 2×1 = 4. After 3 hours: 4 + 4×1 = 8. So it's \( I_B(t)=2^t \)? Wait, no: initial \( t = 0 \): 1. \( t = 1 \): 2. \( t = 2 \): 4. So \( I_B(t)=2^t \)? Wait, no, wait: at \( t = 0 \), 1. At \( t = 1 \), 1 (original) + 1×1 (infected) = 2. At \( t = 2 \), 2 (zombies) + 2×1 = 4. So yes, \( I_B(t)=2^t \)? Wait, no, wait: the number of zombies at time \( t \) is \( I_B(t) \), where \( I_B(0)=1 \), and \( I_B(t + 1)=I_B(t)+I_B(t)\times1 = 2\times I_B(t) \). So it's a geometric sequence with \( a = 1 \), \( r = 2 \). So \( I_B(t)=1\times2^t \).

Step2: Calculate for Strain B at \( t = 15 \)

\( I_B(15)=2^{15} \)
\( 2^{10}=1024 \), \( 2^{15}=2^5\times2^{10}=32\times1024 = 32768 \)

Comparison:

Strain A at 15 hours: 460. Strain B at 15 hours: 32768. So Strain B has more.

for the Reason:
Strain A has linear growth (constant rate of 30 new infections per hour from original 10 zombies), while Strain B has exponential growth (each zombie, including new ones, infects 1 person per hour, leading to doubling each hour). Exponential growth outpaces linear growth over time, especially after 15 hours, so Strain B (Stumblers) infects far more people.

Answer:

Strain B (Stumblers) will have infected more after 15 hours because Strain A has linear growth (total infected \( I_A(t) = 10 + 30t \)) while Strain B has exponential growth (total infected \( I_B(t)=2^t \)), and exponential growth outpaces linear growth over 15 hours ( \( I_A(15)=460 \), \( I_B(15)=32768 \) ).