Question

multiple choice 20 points a particular type of cell doubles in number every hour. which function can be used to find the number of cells present at the end of h hours if there are initially 4 of these cells? n = 4(2)^h n = 4+(2)^h n = 4+(1/2)^h n = 4(1/2)^h

Explanation:

Step1: Recall exponential - growth formula

The general formula for exponential growth is $n = n_0(a)^h$, where $n_0$ is the initial amount, $a$ is the growth factor, and $h$ is the number of time - periods.

Step2: Identify initial amount and growth factor

We are given that the initial number of cells $n_0 = 4$, and the number of cells doubles every hour. So the growth factor $a = 2$ and the number of hours is $h$.

Step3: Substitute values into formula

Substituting $n_0 = 4$ and $a = 2$ into the formula $n=n_0(a)^h$, we get $n = 4(2)^h$.

Answer:

$n = 4(2)^h$ (corresponding to the third option in the multiple - choice question)