Answer: f(n) = 0.9 × f(n − 1) + 10, f(0) = 175, n > 0
Step-by-step explanation:
Given: A store had 175 cell phones in the month of January.
Every month, 10% of the cell phones were sold and 10 new cell phones were stocked in the store.
Let n be the number of months
Then when n=0
f(0)=175
After first month , n=1
The number of cell phone in store=[tex]175-10\%\ of\ 175+10[/tex]
Thus [tex]f(1)=f(0)-10\%\ of\ f(0)+10[/tex]
[tex]=f(0)-0.1\timesf(0)+10\\=f(0)(1-0.1)+10\\=f(0)(0.9)+10\\[/tex]
Similarly we can do till n months, we get
f(n)=0.9×f(n−1)+10, f(0) = 175, n > 0
Answer:
[tex]f(n)=0.9\times f(n-1)+10, f(0) = 175, n > 0[/tex]
Step-by-step explanation:
Given :
A store had 175 cell phones in the month of January.
Every month, 10% of the cell phones were sold and 10 new cell phones were stocked in the store.
To Find:
Which recursive function best represents the number of cell phones in the store f(n) after n months?
Solution:
Let n be the number of months
So, when n=0
f(0)=175
Now, after 1 month i.e. At n=1
Now we are given that Every month, 10% of the cell phones were sold and 10 new cell phones were stocked in the store.
The number of cell phone in store after 1 month =
[tex]175-10\%\ of\ 175+10[/tex]
Thus [tex]f(1)=f(0)-10\%\ of\ f(0)+10[/tex]
[tex]f(1)=f(0)-0.1\times f(0)+10[/tex]
[tex]f(1) =f(0)(1-0.1)+10[/tex]
[tex]f(1) =f(0)(0.9)+10[/tex]
Similarly we can do till n months,Thus , we get
[tex]f(n)=0.9\times f(n-1)+10, f(0) = 175, n > 0[/tex]
Hence Option 4 is correct.
