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NO LINKS!! Fredrico's Bank will you decide how often your interest will be computed, but with certain restrictions. Read the following options. What is the best deal after one year? Justify your answer.​

NO LINKS Fredricos Bank will you decide how often your interest will be computed but with certain restrictions Read the following options What is the best deal class=

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Answer:  Option B

Compounded quarterly; annual interest rate is 7.875%

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Explanation:

Let's say we are depositing P = 100 dollars, and we leave it in the account for a timespan of t = 1 year.

This will apply for options A through E.

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If we go with option A, then we would have n = 1 and r = 0.08

Use the compound interest formula to get the following:

A = P*(1+r/n)^(n*t)

A = 100*(1+0.08/1)^(1*1)

A = 108

We'll have $108 dollars in the account at the end of 1 year if we go with option A. The amount of interest only is A-P = 108-100 = 8 dollars.

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For option B, we have: n = 4 and r = 0.07875

A = P*(1+r/n)^(n*t)

A = 100*(1+0.07875/4)^(4*1)

A = 108.110625948487

A = 108.11

So far option B is slightly better since you've earned an extra $0.11, aka 11 cents.

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For option C, we have: n = 4 and r = 0.07875

A = P*(1+r/n)^(n*t)

A = 100*(1+0.0775/12)^(12*1)

A = 108.031299777277

A = 108.03

This amount is less compared to the $108.11 with option B. Therefore, option B is still the best so far.

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For option D, we have: n = 52 and r = 0.07625

A = P*(1+r/n)^(n*t)

A = 100*(1+0.07625/52)^(52*1)

A = 107.917207523206

A = 107.92

This is worse compared to option C.

Option B is still the best so far.

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For option E, we have: n = 365 and r = 0.075

A = P*(1+r/n)^(n*t)

A = 100*(1+0.075/365)^(365*1)

A = 107.787584644

A = 107.79

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Here's a table comparing the balance after 1 year for options A through E. Those balances all have a starting point of $100 as the deposit.

[tex]\begin{array}{|c|c|} \cline{1-2}\text{Option} & \text{Balance after 1 year}\\\cline{1-2}A & \$108\\\cline{1-2}\boldsymbol{B} & \boldsymbol{\$108.11}\\\cline{1-2}C & \$108.03\\\cline{1-2}D & \$107.92\\\cline{1-2}E & \$107.79\\\cline{1-2}\end{array}[/tex]

Option B is the best choice earning the most ($108.11) if you deposit $100.

There's nothing special about the $100, so feel free to change it to something else to try other starting deposits. You'll find that option B will still provide the highest return.

semsee45

Answer:

B

Step-by-step explanation:

Compound Interest Formula

[tex]\large \text{$ \sf A=P\left(1+\frac{r}{n}\right)^{nt} $}[/tex]

where:

  • A = final amount
  • P = principal amount
  • r = interest rate (in decimal form)
  • n = number of times interest applied per time period
  • t = number of time periods elapsed

Use the Compound Interest Formula with the given values for each option to calculate a final equation where A equals a multiple of P.  

The option with the greatest coefficient of P is the account which yields the highest interest after one year, and is therefore the best deal.

Option A

Given:

  • r = 8% = 0.08
  • n = 1 (annually)
  • t = 1 year

Substitute the given values into the formula:

[tex]\implies \sf A=P\left(1+\dfrac{0.08}{1}\right)^1[/tex]

[tex]\implies \sf A=1.08P[/tex]

Option B

Given:

  • r = 7.875% = 0.07875
  • n = 4 (quarterly)
  • t = 1 year

Substitute the given values into the formula:

[tex]\implies \sf A=P\left(1+\dfrac{0.07875}{4}\right)^{4 \cdot 1}[/tex]

[tex]\implies \sf A=P\left(1.0196875)^4[/tex]

[tex]\implies \sf A=1.081106259P[/tex]

Option C

Given:

  • r = 7.75% = 0.0775
  • n = 12 (monthly)
  • t = 1 year

Substitute the given values into the formula:

[tex]\implies \sf A=P\left(1+\dfrac{0.0775}{12}\right)^{12\cdot 1}[/tex]

[tex]\implies \sf A=P\left(1.006458333\right)^{12}[/tex]

[tex]\implies \sf A=1.080312998P[/tex]

Option D

Given:

  • r = 7.625% = 0.07625
  • n = 52 (weekly)
  • t = 1 year

Substitute the given values into the formula:

[tex]\implies \sf A=P\left(1+\dfrac{0.0765}{52}\right)^{52\cdot 1}[/tex]

[tex]\implies \sf A=P\left(1.001471154\right)^{52}[/tex]

[tex]\implies \sf A=1.079441506P[/tex]

Option E

Given:

  • r = 7.5% = 0.075
  • n = 365 (daily)
  • t = 1 year

Substitute the given values into the formula:

[tex]\implies \sf A=P\left(1+\dfrac{0.075}{365}\right)^{365\cdot 1}[/tex]

[tex]\implies \sf A=P\left(1.000205479\right)^{365}[/tex]

[tex]\implies \sf A=1.077875846P[/tex]

Summary

[tex]\begin{array}{|c|l|}\cline{1-2} \sf Answer\;Option & \sf Coefficient\;of\;P\\\cline{1-2} A & 1.08\\\cline{1-2} B & 1.081106259\\\cline{1-2} C & 1.080312998\\\cline{1-2} D & 1.079441506\\\cline{1-2} E & 1.077875846\\\cline{1-2} \end{array}[/tex]

Therefore, the option with the greatest coefficient of P and therefore the best deal is option B.

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