Question

HELP with 4 and 6 please

Answer 1

4.) a.) 2.01 b.) 404.01

6.) a.) 152.17 b.) 18412.17 c.) 153.14 d.) 18565.60

Explanation:

To solve, use the exponential growth formula
`A_t=A_0(1+(r)/(p))^t` where

`A_t` = Amount as a function of time

`A_0` = Original amount

`r` = rate

`p` = period

`t` = time

For question 4, let

`A_0=400\\r=0.06\\p=12\\t=2`

So,

`A_t=400(1+(0.06)/(12))^2\\A_t=400((201)/(200))^2\\A_t=400(1.01)\\A_t=404.01`

So, the second-period interest would be 404.01-402=2.01

Question 6 can be solved using the same formula, but let

`A_0=18260\\r=0.025\\p=(12)/(4)=3\\t_1=1\\t_2=2`

So,

`A_t=18260(1+(0.025)/(3))^1\\A_t=18260((121)/(120))^1\\A_t=18412.17`

The first-period interest would be 18412.17-18260=152.17

For the second-period, use
`t_2`:

`A_t=18260(1+(0.025)/(3))^2\\A_t=18260((121)/(120))^2\\A_t=18260(1.02)\\A_t=18565.60`

The second-period interest would be 18565.60-18412.17=153.43