Answer:
![y = a(b)^t](https://img.qammunity.org/2021/formulas/mathematics/middle-school/4yjt14t16sg9xs9sz1njrel70wk0p1g3ms.png)
Where a =1 represent the initial amount of bacteria and b =2 represent the growth factor, for this case since each hour we double the number of bacteria for this reason b =2. And t represent the number of hours after the first bacteria is founded.
So then our model would be given by:
![y = 1 (2)^t](https://img.qammunity.org/2021/formulas/mathematics/middle-school/lpp3m21q3qgvev6daigchtu4prh3dlhg69.png)
And since we want to find the number of bacteria at the end of one day, and we know that one day = 24 hours we can replace the value of t =24 into the model and we got:
![f(24) = 1 (2)^(24)=16777216](https://img.qammunity.org/2021/formulas/mathematics/middle-school/krdks62kow466xrbpvjjl6xmd5w40i4oey.png)
Then we can conclude that at the end of the day we would expect 16777216 bacteria
Explanation:
For this case we can use the exponential model given by this general expression:
![y = a(b)^t](https://img.qammunity.org/2021/formulas/mathematics/middle-school/4yjt14t16sg9xs9sz1njrel70wk0p1g3ms.png)
Where a =1 represent the initial amount of bacteria and b =2 represent the growth factor, for this case since each hour we double the number of bacteria for this reason b =2. And t represent the number of hours after the first bacteria is founded.
So then our model would be given by:
![y = 1 (2)^t](https://img.qammunity.org/2021/formulas/mathematics/middle-school/lpp3m21q3qgvev6daigchtu4prh3dlhg69.png)
And since we want to find the number of bacteria at the end of one day, and we know that one day = 24 hours we can replace the value of t =24 into the model and we got:
![f(24) = 1 (2)^(24)=16777216](https://img.qammunity.org/2021/formulas/mathematics/middle-school/krdks62kow466xrbpvjjl6xmd5w40i4oey.png)
Then we can conclude that at the end of the day we would expect 16777216 bacteria