Question

Help plzzz thank you !

Answer 1

Explanation:

a. Ameribank-$15,157.50

b. Capital Two-$4,646.25

a. Tad's savings is $15,000, we calculate his total amount at the end of the year for each bank:

Answer 2

a. Ameribank-$15,157.50

b. Capital Two-$4,646.25

Explanation:

a. Tad's savings is $15,000, we calculate his total amount at the end of the year for each bank:

#Ameribank

`A=P+I=P+PRT\\\\=15000+15000* 0.0105* 1\\\\=\$15,157.50`

#Huffington( we use the effective rate to calculate the compound amount):

`i_m=(1+i/m)^m-1\\\\=(1+0.0095/12)^[12}-1=0.009541\\\\A=P(1+i_m)^n\\\\=15000(1.009541)^1\\\\=\$15,143.12`

#Sixth-Third, Take 1 yrs=52 weeks:

`i_m=(1+i/m)^m-1\\\\=(1+0.01/52)^(52)-1=0.01005\\\\A=15000(1.01005)^1\\\\=\$15,150.74`

#Hence, Ameribank is the best option as his money grows to $15,157.50 which is greater than all the remaining two options.

b. We use the compound interest formula
`A=P(1+r/n)^(nt)` to determine which bank gives the best option:

#Capital Two. r=3.75%, n=12,t=4

`A=P(1+r/n)^(nt)\\\\=4000(1+0.0375/12)^(12*4)\\\\=\$4,646.25`

#J.C Morgan, t=2, r=3.55% n=12

`A=P(1+r/n)^(nt)\\\\=4000(1+0.0355/12)^(12* 2)\\\\=\$4,293.87`

#Silverman Slacks, n=12,t=3, r=3.65%

`A=P(1+r/n)^(nt)\\\\=4000(1+0.0365/12)^(12*3)\\\\=\$4,462.14`

We compare the investment amounts after t years:

`Capital>Silver>Morgan=4646.25>4462.14>4293.87`

Hence, Capital two is the best option with an investment amount of $4,646.25