A store had 50 bottles of olive oil. Each week, 40% of the olive oil bottles were sold and 20 new bottles arrived in shipments. Which recursive function best represents the number of bottles in the store, given that f(0) = 50?
f(n) = f(n − 1) ⋅ 0.6 + 20, n > 0
f(n) = 50 − f(n − 1) ⋅ 0.6 + 20, n > 0
f(n) = 50 − f(n − 1) ⋅ 0.4 + 20, n > 0
f(n) = f(n − 1) ⋅ 0.4 + 20, n > 0
f(1) = .6(50) + 20 = 30 + 20 = 50
f(2) = .6(50) + 20 = 50
f(3) = .6(50) + 20 = 50 forever and ever, good inventory planning
well try them, but you do not have to try many :)
Calculate the first few terms using your description.
(If 40% are sold, then 60% remain)
f(0) = 50
f(1) = .6(50) + 20 , looks like f(1) = .6(f(0)) + 20 = 50
Well, isn't that special, the number of bottles remains at 50
#1 and #3 yield the same result
The correct recursive function that represents the number of bottles in the store is:
f(n) = f(n − 1) ⋅ 0.4 + 20, n > 0
To figure out which recursive function best represents the number of bottles in the store, we need to analyze the given scenario step by step.
Let's break it down:
- The initial number of bottles in the store is 50, which is represented as f(0) = 50.
- Each week, 40% of the olive oil bottles are sold. This means that the remaining 60% of the bottles remain in the store.
- Additionally, 20 new bottles arrive in shipments every week.
- We need to find a recursive function that represents the number of bottles in the store based on the previous week's value.
Taking all of this into consideration, the correct recursive function would be:
f(n) = f(n − 1) ⋅ 0.6 + 20, n > 0
This function takes the previous week's number of bottles (f(n-1)), multiplies it by 0.6 (representing the 60% remaining), and then adds 20 (the new bottles that arrive in shipments).
So, for every week (n) greater than 0, we use the previous week's number of bottles, multiply it by 0.6, and add 20 to get the current week's number of bottles.
Therefore, the correct answer is:
f(n) = f(n − 1) ⋅ 0.6 + 20, n > 0.