Question

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. Which recursive function best represents the number of cell phones in the store f(n) after n months? f(n) = 175 − 0.9 × f(n − 1) + 10, f(0) = 175, n > 0 f(n) = 0.1 × f(n − 1) + 10, f(0) = 175, n > 0 f(n) = 175 + 0.9 × f(n − 1) + 10, f(0) = 175, n > 0 f(n) = 0.9 × f(n − 1) + 10, f(0) = 175, n > 0
99 POINTS

Answer 1

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=$175-10\%\ of\ 175+10$

Thus $f(1)=f(0)-10\%\ of\ f(0)+10$

$=f(0)-0.1\timesf(0)+10\\=f(0)(1-0.1)+10\\=f(0)(0.9)+10$

Similarly we can do till n months, we get

f(n)=0.9×f(n−1)+10, f(0) = 175, n > 0

Answer 2

Answer:

$f(n)=0.9\times f(n-1)+10, f(0) = 175, n > 0$

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.  

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 =

$175-10\%\ of\ 175+10$

Thus  $f(1)=f(0)-10\%\ of\ f(0)+10$

$f(1)=f(0)-0.1\times f(0)+10$

$f(1) =f(0)(1-0.1)+10$

$f(1) =f(0)(0.9)+10$

Similarly we can do till n months,Thus , we get

$f(n)=0.9\times f(n-1)+10, f(0) = 175, n > 0$

Hence Option 4 is correct.