Question

Marcus has opened a savings account where the yearly interest rate is 10%. He deposits $1,000 to start the account. After t years, the amount of money in the account is modeled by the function An=1,0001+0.10t. Which functions below would be an approximate equivalent function?


Select all that apply.


A. An=1,000(1.10)t

B. An=1,000(1+0.008)^12t

C. An=1,000(1+0.46)^4t

D. An=1,000(1.008)^t/12

E. An=1,000(1.46)^t/4

Answer 1

Answer:

Option A and option B applies

Step-by-step explanation:

The compound interest formula is given by:

$A(t) = P(1 + \frac{r}{n})^{nt}$

Where A(t) is the amount of money after t years, P is the principal(the initial sum of money), r is the interest rate(as a decimal value), n is the number of times that interest is compounded per year and t is the time in years for which the money is invested or borrowed.

Marcus has opened a savings account where the yearly interest rate is 10%. He deposits $1,000 to start the account.

This means, respectively, that: $r = 0.1, P = 1000$. So

$A(t) = 1000(1 + \frac{0.1}{n})^{nt}$

Option A:

$n = 1$. So

$A(t) = 1000(1 + \frac{0.1}{1})^{t}$

$A(t) = 1000(1.1)^{t}$

So option A applies

Option B:

$n = 12$. So

$A(t) = 1000(1 + \frac{0.1}{12})^{12t}$

0.1/12 = 0.008. So

$A(t) = 1000(1 + 0.008)^{12t}$

$A(t) = 1000(1.008)^{12t}$

So option B also applies.

The other options will not apply.

Answer 2

Answer:

A and B

Step-by-step explanation: