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alina1380 [7]
3 years ago
13

Use the compound interest formulas A= P (1 + r/n) ^nt and A= Pe ^ rt to solve the problem given. Round to the nearest cent. Find

the accumulated value of an investment of $25,000 for 4 years at an interest rate of 5% if the money is 1. compounded semiannually; 2. compounded quarterly; 3. compounded monthly; 4. compounded continuously. (If possible, please explain)
Mathematics
1 answer:
Dennis_Churaev [7]3 years ago
7 0

Answer:

  1. \$30,460
  2. \$30,497
  3. \$30,522
  4. \$30,535

Step-by-step explanation:

We know that,

A=P\left (1+\dfrac{r}{n}\right )^{n\cdot t}

where,

A = Amount after time t,

P = Principle amount,

r = Rate of interest,

n = Number of times interest is compounded per year,

t = time period in year.

Investment of $25,000 for 4 years at an interest rate of 5% if the money is compounded semiannually

Here,

P = $25,000

r = 5% = 0.05

n = 2 (as compounded semiannually)

t = 4 years

Putting the values,

A=25000\left (1+\dfrac{0.05}{2}\right )^{2\times 4}

=25000\left (1+0.025\right )^{8}

=25000\left (1.025\right )^{8}

=\$30,460

Investment of $25,000 for 4 years at an interest rate of 5% if the money is compounded quarterly.

Here,

P = $25,000

r = 5% = 0.05

n = 4 (as compounded quarterly)

t = 4 years

Putting the values,

A=25000\left (1+\dfrac{0.05}{4}\right )^{4\times 4}

=25000\left (1+0.0125\right )^{16}

=25000\left (1.0125\right )^{8}

=\$30,497

Investment of $25,000 for 4 years at an interest rate of 5% if the money is compounded monthly.

Here,

P = $25,000

r = 5% = 0.05

n = 12 (as compounded monthly)

t = 4 years

Putting the values,

A=25000\left (1+\dfrac{0.05}{12}\right )^{12\times 4}

A=25000\left (1+\dfrac{0.05}{12}\right )^{48}

=\$30,522

Investment of $25,000 for 4 years at an interest rate of 5% if the money is compounded continuously.

A= Pe^{rt}

where,

A = Amount after time t,

P = Principle amount,

r = Rate of interest,

t = time period in year.

Putting all the values,

A= 25000e^{0.05\times 4}=\$30,535

It can be observed that, the frequent we compound the amount, the more we get.

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2) For each two values of x, a value of y is increased by 10.

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