1. **State the problem:** We are given a bacterial population at two different times: at $t=4$ hours, the population is 59 thousand, and at $t=15$ hours, the population is 99.7 thousand. We need to find the average rate of change of the population over the interval from $t=4$ to $t=15$ hours. 2. **Formula for average rate of change:** The average rate of change of a function $f(t)$ over the interval $[a,b]$ is given by: $$\text{Average rate of change} = \frac{f(b) - f(a)}{b - a}$$ 3. **Apply the formula:** Here, $a=4$, $b=15$, $f(4)=59$, and $f(15)=99.7$. $$\text{Average rate of change} = \frac{99.7 - 59}{15 - 4}$$ 4. **Calculate the numerator and denominator:** $$99.7 - 59 = 40.7$$ $$15 - 4 = 11$$ 5. **Substitute and simplify:** $$\frac{40.7}{11}$$ 6. **Divide to find the average rate:** $$\frac{40.7}{11} = 3.7$$ (rounded to three decimal places: $3.700$) 7. **Interpretation:** The average rate of change is $3.700$ thousand bacteria per hour over the time interval from 4 to 15 hours. **Final answer:** The average rate of change of the bacterial population from $t=4$ to $t=15$ is $3.700$ thousand bacteria per hour.