We are given a set of data. We can determine how much of a percentage the number of students in the first year increases with respect to the second year using the following relationship:
![1555+1555*(n)/(100)=1602](https://img.qammunity.org/2023/formulas/mathematics/college/sfwvyjswwkfvwy8pyumcsr567sbulpg663.png)
Where "n" is the percentage. WE solve for "n" first by subtracting 1555 to both sides:
![1555*(n)/(100)=1602-1555](https://img.qammunity.org/2023/formulas/mathematics/college/r94uiyk5lb203o683mh51p75n3kdt8xd3s.png)
Now we divide both sides by 1555:
![(n)/(100)=(1602-1555)/(1555)](https://img.qammunity.org/2023/formulas/mathematics/college/n5sqq19zk0fy029vc24036ps9xzkiic2w6.png)
Now we multiply both sides by 100:
![n=(1602-1555)/(1555)*100](https://img.qammunity.org/2023/formulas/mathematics/college/j9zg686qzucdf8v8cbvrop6plscz88xweu.png)
Solving the operations:
![\begin{gathered} n=(47)/(1555)*100 \\ n=0.003*100 \\ n=3 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/ghklt4rkorzibciscqdvapidxq7cpfnlv3.png)
Therefore, we have a 3 percent increase. Now we check if this percentage is fixed for the other years.
![1602+1602*(3)/(100)=1602+48.06=1650](https://img.qammunity.org/2023/formulas/mathematics/college/74veyfk16gsd27opwiujpvt5q711cxi38u.png)
For year 4:
![1650+1650*(3)/(100)=1650+49.5=1699\approx1700](https://img.qammunity.org/2023/formulas/mathematics/college/zdwqcg5e8czhq8y9gigurtbbey7v09898c.png)
For the fifth year:
![1700+1700*(3)/(100)=1700+51=1751](https://img.qammunity.org/2023/formulas/mathematics/college/8thp3jdg88yqkszi8vfxa1culwk801sapy.png)
Therefore, each value increases by a fixed percentage of 3 per cent, therefore, the right answer is option B, since this means that the function can be modelled by an exponential function.