Respuesta :
Answer:
1. [tex]A=800(1+0.05)^{n-1}[/tex] and $972.4
Step-by-step explanation:
We are given that,
Investment in the first year = $800
Rate of interest = 5% = 0.05
As we know, the compound interest is given by [tex]A=P(1+r)^n[/tex], where P= initial amount, r= rate of interest and n= time period.
Since, she has $800 in the 1st year. So, when n= 1, the compound value = $800.
So, from these conditions, we get,
Compound interest is [tex]A=800(1+0.05)^{n-1}[/tex].
Further, when n= 5, we have,
[tex]A=800(1+0.05)^{5-1}[/tex].
i.e. [tex]A=800(1.05)^4[/tex].
i.e. [tex]A=800\times 1.2155[/tex].
i.e. A= 972.4 dollars
Thus, the compounded value at the beginning of 5th year is $972.4
Hence, option 1 is correct.
Answer:
[tex]A(n)=800(1+0.05)^{n-1}; 972.41[/tex]
Step-by-step explanation:
Principal = $800
Rate of interest = 5% = 0.05
Since we are given that At the beginning of year 1, Amada invests $800 at an annual compound interest rate of 5%.
Formula :
[tex]A=P(1+r)^n[/tex]
A is the amount
P is the principal
r = rate of interest in decimal
n = times period
Since we are given that we need to find the amount at the beginning of year 5
So, we need to replace n with n-1
So, formula becomes :[tex]A=P(1+r)^{n-1}[/tex]
Substitute the values
[tex]A(n)=800(1+0.05)^{n-1}[/tex]
[tex]A(n)=800(1+0.05)^{5-1}[/tex]
[tex]A(n)=972.41[/tex]
Hence Option A is correct.
[tex]A(n)=800(1+0.05)^{n-1}; 972.41[/tex]
Thus explicit formula can be used to find the account’s balance at the beginning of year 5 is [tex]A(n)=800(1+0.05)^{n-1}[/tex]
Balance at the beginning of year 5 is $972.41
