Choice C is correct. It’s given that on April $1$ , there were $233$ views of the advertisement. It’s also given that every $2$ days after April $1$ , the number of views of the advertisement had increased by $70 percent sign$ of the number of views $2$ days earlier. This situation can be represented by an exponential function of the form $f (x) = a {(1 + \frac{r}{100})}^{\frac{x}{c}}$ , where $a$ is the number of views on April $1$ and every $c$ days after April $1$ , the number of views had increased by $r %$ of the number of views $c$ days earlier. It follows that $a equals 233$ , $r equals 70$ , and $c equals 2$ . Substituting $233$ for $a$ , $70$ for $r$ , and $2$ for $c$ in the equation $f (x) = a {(1 + \frac{r}{100})}^{\frac{x}{c}}$ yields $f (x) = 233 {(1 + \frac{70}{100})}^{\frac{x}{2}}$ , or $f (x) = 233 {(1.70)}^{\frac{x}{2}}$ .

Choice A is incorrect. This function gives the predicted number of views for an advertisement for which every $2$ days, the number of views was $70 percent sign$ , rather than increased by $70 percent sign$ , of the number of views $2$ days earlier.

Choice B is incorrect. This function gives the predicted number of views for an advertisement for which every $\frac{1}{2}$ days, the number of views was $70 percent sign$ of the number of views $\frac{1}{2}$ days earlier, rather than an advertisement for which every $2$ days, the number of views had increased by $70 percent sign$ of the number of views $2$ days earlier.

Choice D is incorrect. This function gives the predicted number of views for an advertisement for which every $\frac{1}{2}$ days, rather than every $2$ days, the number of views had increased by $70 percent sign$ of the number of views $\frac{1}{2}$ days earlier, rather than $2$ days earlier.