Answer:
745 books are expected to be donated to the library at the end of Month 10
Step-by-step explanation:
There were 80 books donated at the library:
No=80
The number of books donated to a library is increasing by 25% each month:
There were 100 books donatd at the end of Month 1:
N1=100
N1=No+25%No
N1=No+(25/100)No
N1=No+0.25 No
N1=(1+0.25) No
N1=(1.25) No
(1) N1=(1.25)^1 No
N1=(1.25)^1 80
N1=(1.25) 80
N1=100
N2=N1+25%N1
N2=N1+(25/100)N1
N2=N1+0.25 N1
N2=(1+0.25) N1
N2=(1.25) N1
Replacing equation (1) in the equation above:
N2=(1.25) (1.25)^1 No
N2=(1.25)^(1+1) No
(2) N2=(1.25)^2 No
N3=N2+25%N2
N3=N2+(25/100)N2
N3=N2+0.25 N2
N3=(1+0.25) N2
N3=(1.25) N2
Replacing equation (2) in the equation above:
N3=(1.25) (1.25)^2 No
N3=(1.25)^(1+2) No
(3) N3=(1.25)^3 No
In general:
Nt=N(t)=(1.25)^t No
N(t)=(1.25)^t (80)
N(t)=80 (1.25)^t
where t is the number of the month
How many books are expected to be donated to the library at the end of Month 10?
t=10
N(10)=80 (1.25)^10
N(10)=80 (9.313225746)
N(10)=745.0580596
Rounding to the nearest whole number:
N(10)=745
Answer:
745 books.
Step by step explanation:
We have been given that the number of books donated to a library is increasing by 25% each month. Initially, there were 80 books donated to the library.
We can see that number of books donated to library are increasing exponentially.
Since an exponential growth function is in form: [tex]f(x)=a\cdot (1+r)^x[/tex], where,
f(x) = Future value of function after x units of time.
a = Initial value of function.
r= rate in decimal form.
We have been given the growth rate in percentage. Let us convert growth rate in decimal from.
[tex]25\%=\frac{25}{100}=0.25[/tex]
Upon substituting a = 80, r = 0.25 and x = 10 in exponential function we will get,
[tex]y=80\cdot (1+0.25)^{10}[/tex]
[tex]y=80\cdot (1.25)^{10}[/tex]
[tex]y=80\cdot 9.3132257461547852[/tex]
[tex]y=745.058059692382816\approx 745[/tex]
Therefore, 745 books are expected to be donated to the library at the end of Month 10.
