Respuesta :
You will want to use the formula for continuously compounded interest:
[tex]A = Pe^{rt}[/tex]
where [tex]A[/tex] is the resulting amount, [tex]P[/tex] is the initial amount, [tex]e[/tex] is the mathematical constant (2.718...), [tex]r[/tex] is the interest (in percentage), and [tex]t[/tex] is the time in years. Plugging these numbers into the equation gives the following:
[tex]8000 = Pe^{0.052 * 5}[/tex]
Solving for [tex]P[/tex] will give you the initial amount that should be put into the account.
[tex]A = Pe^{rt}[/tex]
where [tex]A[/tex] is the resulting amount, [tex]P[/tex] is the initial amount, [tex]e[/tex] is the mathematical constant (2.718...), [tex]r[/tex] is the interest (in percentage), and [tex]t[/tex] is the time in years. Plugging these numbers into the equation gives the following:
[tex]8000 = Pe^{0.052 * 5}[/tex]
Solving for [tex]P[/tex] will give you the initial amount that should be put into the account.
you use this formula
amount=principal(1+r%)^n
where
amount= the needed money to have been saved after five years for the student to join college=$8000
principal=the amount needed to be saved/invested in the account to give amount $8000 at the end of fifth year.
r=the rate of interest
therefore
$8000=p(1+5.2/100)^5
$8000=p(1+0.052)^5
$8000=p(1.052)^5
$8000=p(1.288483018284032)
$8000/(1.288483018284042)=p
$6208.851716691=p
p=$6208.8517
amount=principal(1+r%)^n
where
amount= the needed money to have been saved after five years for the student to join college=$8000
principal=the amount needed to be saved/invested in the account to give amount $8000 at the end of fifth year.
r=the rate of interest
therefore
$8000=p(1+5.2/100)^5
$8000=p(1+0.052)^5
$8000=p(1.052)^5
$8000=p(1.288483018284032)
$8000/(1.288483018284042)=p
$6208.851716691=p
p=$6208.8517