Respuesta :
The answer is A.
A is the amount after t years, P is the amount originally deposited, r is the interest rate, and n is how often the interest is compounded per t.
Plug in what we know and solve for P:
[tex]7160.06=P(1+ \frac{0.026}{4})^{(4)(4)} [/tex]
P = [tex] \frac{7160.06}{(1+ \frac{0.026}{4} )^{16}} = 6454.999 [/tex]
A is the amount after t years, P is the amount originally deposited, r is the interest rate, and n is how often the interest is compounded per t.
Plug in what we know and solve for P:
[tex]7160.06=P(1+ \frac{0.026}{4})^{(4)(4)} [/tex]
P = [tex] \frac{7160.06}{(1+ \frac{0.026}{4} )^{16}} = 6454.999 [/tex]
Answer:
Option a. [tex]\$6,455[/tex]
Step-by-step explanation:
we know that
The compound interest formula is equal to
[tex]A=P(1+\frac{r}{n})^{nt}[/tex]
where
A is the Final Investment Value
P is the Principal amount of money to be invested
r is the rate of interest in decimal
t is Number of Time Periods
n is the number of times interest is compounded per year
in this problem we have
[tex]t=4\ years\\ A=\$7,160.06\\ r=0.026\\n=4[/tex]
substitute in the formula above and solve for P
[tex]7,160.06=P(1+\frac{0.026}{4})^{4*4}[/tex]
[tex]7,160.06=P(1+\frac{0.026}{4})^{4*4}[/tex]
[tex]P=7,160.06/1.109227=\$6,455[/tex]