We are given that the number of papers increases by a percentage of 3% annually. since we have a quantity that grows by a percentage we can use the following exponential function:
![f(t)=a(1+r)^t](https://img.qammunity.org/2023/formulas/mathematics/college/azg3wguunqg8z7h3scy2lcb979266raenp.png)
Where "r" is the rate of growth, "a" is the initial amount, and "t" is time. Replacing the values:
![f(t)=11610(1+0.03)^t](https://img.qammunity.org/2023/formulas/mathematics/college/2aav0hnjqcltciowvotsqeuhsjtssukvtt.png)
Simplifying:
![f(t)=11610(1.03)^t](https://img.qammunity.org/2023/formulas/mathematics/college/gmoyjeeai9g762m30904wwt42a5mh8qgnu.png)
Now we replace t = 8:
![f(t)=11610(1.03)^8](https://img.qammunity.org/2023/formulas/mathematics/college/qvw3ubcsxc5j36u43zoins4k3p9rmvttk9.png)
Solving the operations:
![f(t)=14707](https://img.qammunity.org/2023/formulas/mathematics/college/l7a1x2hwyv9dn5rsjehtmgldf67tn0vwzo.png)
Therefore, 14707 papers will be published.