Question

a scientist is studying how bacteria thrive under different conditions of temperature and pressure. the scientist develops bacteria population models for the different conditions, as shown below. the models predict the bacteria population, p, after t hours.
condition 1: $p(t)=500(0.3)^t$
condition 2: $p(t)=500(1.65)^t$
condition 3: $p(t)=500(0.95)^{2t}$
condition 4: $p(t)=500(1.2)^{0.5t}$
drag and drop each population model under the column that describes the change in bacteria population over time.
population growth|population decay

Explanation:

Step1: Recall exponential - growth and decay rules

For an exponential function of the form $P(t)=a(b)^t$ (or $P(t)=a(b)^{kt}$ where $a> 0$), if $b > 1$, it represents population growth; if $0 < b<1$, it represents population decay.

Step2: Analyze Condition 1

For $P(t)=500(0.3)^t$, since $0 < 0.3<1$, it is population decay.

Step3: Analyze Condition 2

For $P(t)=500(1.65)^t$, since $1.65>1$, it is population growth.

Step4: Analyze Condition 3

For $P(t)=500(0.95)^{2t}$, let $u = 2t$, then the function is $P(u)=500(0.95)^u$. Since $0 < 0.95<1$, it is population decay.

Step5: Analyze Condition 4

For $P(t)=500(1.2)^{0.5t}$, let $v = 0.5t$, then the function is $P(v)=500(1.2)^v$. Since $1.2>1$, it is population growth.

Answer:

Population Growth	Population Decay