contestada

opal deposited 2925.90$ into a savings account with an interest rate of 3.9% compounded twice a year. about how long will it take for the account to be worth 6000$

Respuesta :

altavistard
Use the compound amount formula:

A = P (1 + r/n)^( r/n )

Here,

A $6000 = $2925.90 ( 1 + 0.039/2 )^(2t).  We must solve for t.

    2.051  = ( 1 + 0.0195 )^(2t)

Take the natural log of both sides:

ln 2.051 = (2t) ln 1.0195    leads to     (2t) ( 0.0193 ) = 0.7183

                                                                         0.7183
Then                                                      2t = -------------- = 37.219
                                                                        0.0193
                   37.219
Finally, t = ------------- = 18.6 years
                         2

This is reasonable, because we're going from $2925.90 to more than twice that, or $6000, at the relatively low interest rate of 3.9%.

Answer:

It will take 18.55 years.

Step-by-step explanation:

The formula of Compound Interest is:

[tex]A = P(1+\frac{r}{n})^{nt}[/tex]

where A = Amount

P = Principle

r = rate

n = Number of Compounding per year

t = total number of year

Here, P = 2925.9, r = 3.9% = 0.039, n = 2, A = 6000, and t = ?.

Putting all these values in above formula:

[tex]6000 = 2925.9(1+\frac{0.039}{2})^{2\times t}[/tex]

⇒ [tex]6000 = 2925.9(\frac{2.039}{2})^{2t}[/tex]

⇒ [tex]6000 = 2925.9(1.0195)^{2t}[/tex]

⇒ [tex]6000 ÷ 2925.9 = (1.0195)^{2t}[/tex]

⇒ [tex]2.05 = (1.0195)^{2t}[/tex]

Taking log on both side

log(2.05) = 2t × log(1.0195)

⇒ 0.3117 = t × 0.0167

⇒ t = 18.55

Hence, Opal will get $6000 after 18.55 years.

Otras preguntas