Melanie invested $8,600 in an account paying an interest rate of 2\tfrac{7}{8}2 8 7 ​ % compounded monthly. Oliver invested $8,600 in an account paying an interest rate of 2\tfrac{3}{4}2 4 3 ​ % compounded daily. After 10 years, how much more money would Melanie have in her account than Oliver, to the nearest dollar?

$139

Step-by-step explanation:

The formula for compound interest =

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

Where

P = Principal

r = Interest rate

n = Compounding frequency

t = time in years

A = Amount after t years

For Melanie

Melanie invested $8,600 in an account paying an interest rate of 2 7/8% compounded monthly.

Rate = 2 7/8% = 23/8% = 2.875%

Time = 10 years

n = monthly = 12

First, convert R percent to r a decimal

r = R/100

r = 2.875%/100

r = 0.02875 per year,

Then, solve our equation for A

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

A = 8,600.00(1 + 0.02875/12)^(12)(10

A = 8,600.00(1 + 0.002395833)^(120)

A = $11,460.64

For Oliver

Oliver invested $8,600 in an account paying an interest rate of 2 3/4 ​ % compounded daily.

Rate = 2 3/4% = 11/4% = 2.75%

Time = 10 years

n = daily = 360 days

First, convert R percent to r a decimal

r = R/100

r = 1.375%/100

r = 0.01375 per year,

Then, solve our equation for A

First, convert R percent to r a decimal

r = R/100

r = 2.75%/100

r = 0.0275 per year,

Then, solve our equation for A

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

A = 8,600.00(1 + 0.0275/360)^(360)(10)

A = 8,600.00(1 + 7.6389E-5)^(3600)

A = $11,322.05

After 10 years, how much more money would Melanie have in her account than Oliver, to the nearest dollar?

This is calculated as

$11,460.64 - $11,322.05

=$ 138.59

Approximately to the nearest dollar = $139

explanation:

previous steps have made pa≅ pb ≅ st ≅ su. setting the compass width to ab allows you to make tu the same length. then δstu ≅ δpab by the sss congruence postulate, and ∠p ≅ ∠s by the reason cpctc.