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2 years ago

What is the answer to this particular question?

Mathematics

1 answer:

STatiana [176]2 years ago

5 0

The anwser is c
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VashaNatasha [74]

Answer:

a. Ameribank-$15,157.50

b. Capital Two-$4,646.25

Step-by-step explanation:

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

#Ameribank

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

#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

$A=P(1+r/n)^{nt}\\\\=4000(1+0.0375/12)^{12\times4}\\\\=\$4,646.25$

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

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

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

$A=P(1+r/n)^{nt}\\\\=4000(1+0.0365/12)^{12\times3}\\\\=\$4,462.14$

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

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