Question

the graph shows the population of a bacteria in an experiment, measured every hour. which function represents the population of the bacteria after t hours? f(t) = 10(1 4)^t f(t) = 10(2 0)^t f(t) = 14(1 4)^t f(t) = 14(2 0)^t bacteria growth population time (hours)

Explanation:

Step1: Recall the general form of an exponential - growth function

The general form of an exponential - growth function is $f(t)=a\cdot b^{t}$, where $a$ is the initial amount and $b$ is the growth factor. When $t = 0$, the population is the initial population. Looking at the graph, when $t = 0$, the population $f(0)$ is 10. So, $a = 10$.

Step2: Check the growth factor

We can take two points on the graph, say $(0,10)$ and $(1, 20)$. Substitute $t = 1$ into the function $f(t)=a\cdot b^{t}$. We know $a = 10$ and $f(1)=20$. So, $f(1)=10\cdot b^{1}=20$. Solving for $b$ gives $b = 2$.

Answer:

$f(t)=10(2)^{t}$ (There is no correct option among the given ones in the problem - statement as written. But if we assume there are typos and the intended functions are exponential growth functions with the correct form based on the steps above, the correct function should be $f(t)=10(2)^{t}$). If we assume the closest form among the given options considering the correct initial - value and growth - factor concept, we note that the initial value $a$ should be 10 and the growth factor $b$ should be close to 2. The closest option is $f(t)=10(2.0)^{t}$). So the answer is $f(t)=10(2.0)^{t}$.