1. **Problem:** Determine the speeds of accessing files in the Finance bucket folders by year and by month after 4 days. 2. **Given functions:** $$v_1(t) = \frac{3t-4}{t+1}$$ (speed by year folders) $$v_2(t) = \frac{2t+1}{t+1}$$ (speed by month folders) 3. **Step 5.1: Compute speed by year folder after 4 days.** Substitute $t=4$ in $v_1(t)$: $$v_1(4) = \frac{3(4)-4}{4+1} = \frac{12 - 4}{5} = \frac{8}{5} = 1.6$$ So, the speed to access files by year after 4 days is $1.6$. 4. **Step 5.2: Compute speed by month folder after 4 days.** Substitute $t=4$ in $v_2(t)$: $$v_2(4) = \frac{2(4)+1}{4+1} = \frac{8 + 1}{5} = \frac{9}{5} = 1.8$$ So, the speed to access files by month after 4 days is $1.8$. 5. **Step 5.3: Find the composition $v_2(v_1(t))$ to calculate total speed to access file in month folder inside year folder.** The composition means substitute $v_1(t)$ into $v_2(t)$: $$v_2(v_1(t)) = \frac{2v_1(t) + 1}{v_1(t) + 1}$$ Substitute $v_1(t) = \frac{3t -4}{t+1}$: $$v_2(v_1(t)) = \frac{2\left(\frac{3t -4}{t+1}\right) + 1}{\frac{3t -4}{t+1} + 1}$$ Multiply numerator and denominator by $(t+1)$ to clear fractions: Numerator: $$2(3t -4) + (t+1) = 6t -8 + t +1 = 7t -7$$ Denominator: $$(3t -4) + (t +1) = 4t -3$$ So, $$v_2(v_1(t)) = \frac{7t -7}{4t -3}$$ This function gives the total speed to access a file stored by month inside a year folder. **Final answers:** - Speed by year folder at $t=4$ days: $1.6$ - Speed by month folder at $t=4$ days: $1.8$ - Composition function for total speed: $$v_2(v_1(t)) = \frac{7t -7}{4t -3}$$