Stop speaking French to me 😡😡😡😡redo question 6 in English

I apologize for the language mix-up. Here is Question 6 re-done in English: Question 6 a) The half-life (T_1/2) of a radioactive isotope is the time required for half of the radioactive nuclei in a sample to decay. It is also the time required for the activity of a radioactive sample to be reduced by half. b) (i) To determine the half-life of the isotope: Step 1: Use the radioactive decay formula A = A_0 ((1)/(2))^n, where n is the number of half-lives. Initial activity A_0 = 1.2 × 10^5 Bq. Activity after t = 12 days is A = 1.5 × 10^4 Bq. (A)/(A_0) = ((1)/(2))^n 1.5 × 10^4 Bq1.2 × 10^5 Bq = ((1)/(2))^n 0.125 = ((1)/(2))^n Step 2: Solve for n. We know that 0.125 = (1)/(8) = ((1)/(2))^3. Therefore, n = 3. Step 3: Calculate the half-life T_1/2 using n = (t)/(T_1/2). 3 = 12 daysT_1/2 T_1/2 = 12 days3 T_1/2 = 4 days The half-life of the isotope is 4 days. (ii) To calculate the activity of the sample after an additional 20 days: Step 1: Calculate the total time elapsed. Total time t_total = 12 days + 20 days = 32 days. Step 2: Calculate the total number of half-lives n' for this time. n' = t_totalT_1/2 = 32 days4 days n' = 8 Step 3: Calculate the final activity A'. A' = A_0 ((1)/(2))^n' A' = (1.2 × 10^5 Bq) × ((1)/(2))^8 A' = (1.2 × 10^5 Bq) × (1)/(256) A' ≈ 468.75 Bq A' ≈ 4.7 × 10^2 Bq The activity of the sample after an additional 20 days will be approximately 4.7 × 10^2 Bq. What's next? Send 'em!