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shtirl [24]
3 years ago
13

A man deposits $ 14,850 into a bank, which pays 4% interest that is compounded

Mathematics
1 answer:
Ganezh [65]3 years ago
6 0

Given:

Principal = $14850

Rate of interest = 4% compounded semiannually.

Time = 3 years

To find:

The amount after 3 years.

Solution:

Formula for amount is:

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

Where, P is principal, r is the rate of interest in decimal, n is the number of times interest compounded and t is the number of years.

The interest is compounded semiannually, so n=2.

Putting P=14850, r=4, n=2, t=3 in the above formula, we get

A=14850\left(1+\dfrac{0.04}{2}\right)^{2(3)}

A=14850\left(1+0.02\right)^{6}

A=14850\left(1.02\right)^{6}

On further simplification, we get

A=14850(1.12616242)

A=16723.511937

A\approx 16723.51

Therefore, the amount in the account after three years is $16723.51.

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2 years ago
Please help me. This is real confusing. ​
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Answer:

1st problem: b) A=2500(1.01)^{12t}

2nd problem:  c) A=2500e^{.12t}

Step-by-step explanation:

1st problem:

The formula/equation you want to use is:

A=P(1+\frac{r}{n})^{nt}

where

t=number of years

A=amount he will owe in t years

P=principal (initial amount)

r=rate

n=number of times the interest is compounded per year t.

We are given:

P=2500

r=12%=.12

n=12 (since there are 12 months in a year and the interest is being compounded per month)

A=2500(1+\frac{.12}{12})^{12t}

Time to clean up the inside of the ( ).

A=2500(1+.01)^{12t}

A=2500(1.01)^{12t}

----------------------------------------------------

2nd Problem:

Compounded continuously problems use base as e.

A=Pe^{rt}

P is still the principal

r is still the rate

t is still the number of years

A is still the amount.

You are given:

P=2500

r=12%=.12

Let's plug that information in:

A=2500e^{.12t}.

6 0
3 years ago
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