Question

Kyleigh invested $2800 in an account paying an interest rate of 5 1/2% compounded annually Ariana invested $2800 in an account paying in interest rate of 5 3/4%compounded continuously after 12 years how much more money with Ariana having her account then Kylie to the nearest dollar?

Answer 1

Answer:

Step-by-step explanation:

Kyleigh invested $2800 in an account paying an interest rate of 5 1/2% compounded annually Ariana invested $2800 in an account paying in interest rate of 5 3/4%compounded continuously after 12 years how much more money with Ariana having her account then Kylie to the nearest dollar?

Answer 2

Answer:

$259

Step-by-step explanation:

\text{Rate 1: }5\tfrac{1}{2}\%=5+1/2=

Rate 1: 5  

2

1

​  

%=5+1/2=

\,\,5.5\%\rightarrow 0.055

5.5%→0.055

\text{Rate 2: }5\tfrac{3}{4}\%=5+3/4=

Rate 2: 5  

4

3

​  

%=5+3/4=

\,\,5.75\%\rightarrow 0.0575

5.75%→0.0575

\text{Calculate Final Amount for Kylie}

Calculate Final Amount for Kylie

\overline{\phantom{\text{Calculate Final Amount for Kylie}}}

Calculate Final Amount for Kylie

​  

 

\text{Compounded Annually:}

Compounded Annually:

A=P(1+r)^t

A=P(1+r)  

t

 

P=2800\hspace{35px}r=0.055\hspace{35px}t=12

P=2800r=0.055t=12

Given values

A=2800(1+0.055)^{12}

A=2800(1+0.055)  

12

 

Plug in values

A=2800(1.055)^{12}

A=2800(1.055)  

12

 

Add

A=5323.381

A=5323.381

Use calculator

\text{Calculate Final Amount for Ariana}

Calculate Final Amount for Ariana

\overline{\phantom{\text{Calculate Final Amount for Ariana}}}

Calculate Final Amount for Ariana

 

\text{Compounded Continuously:}

Compounded Continuously:

A=Pe^{rt}

A=Pe  

rt

 

P=2800\hspace{35px}r=0.0575\hspace{35px}t=12

P=2800r=0.0575t=12

Given values

A=2800e^{0.0575(12)}

A=2800e  

0.0575(12)

 

Plug in

A=2800e^{0.69}

A=2800e  

0.69

 

Multiply

A=5582.4035

A=5582.4035

Use calculator (with e button)

\text{How much more money Ariana has:}

How much more money Ariana has:

5582.4035-5323.381

5582.4035−5323.381

259.0225

259.0225