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3 years ago

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$765.13 is deposited at the end of each month for 3 years in an account paying 2% Interest compounded monthly. Find the final am

ount of the account. Round to the nearest
cent
O A. $27,598.32
OB. $28,363,45
OC. $29,128.58
OD. $26,787.27

1 answer:

Crank3 years ago

8 0

Answer:

Option B.

Step-by-step explanation:

The future value formula, for an annuity, is:

$FV = \frac{P((1+r)^{n} - 1)}{r}$

An annuity means that a number of payments happen during the period(an year, for example).

P is the value of the deposit, r is the interest rate, as a decimal, and n is the number of deposits.

In this question:

Deposits of $765.13, so $P = 765.13$

Each month, for 3 years. An year has twelve months, so $n = 3*12 = 36$

2% Interest a year. An year has 12 months, so $r = \frac{0.02}{12} = 0.00167$

Find the final amount of the account.

$FV = \frac{765.13*((1 + 0.00167)^{36} - 1)}{0.00167} = 28,363.46$

The final amount of the account will be $28,363.46, which is option B.

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Step-by-step explanation:

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First, convert $4^{x-3}$ to base 2:

$4^{x-3}$ = $(2^{2})^{x-3}$

$\frac{1}{2} ^{x-4}$ - 3 = $(2^{2})^{x-3}$ - 2

Next, convert $\frac{1}{2} ^{x-4}$ to base 2:

$\frac{1}{2} ^{x-4}$ = $(2^{-1})^{x-4}$

$(2^{-1})^{x-4}$ - 3 = $(2^{2})^{x-3}$ - 2

Apply exponent rule: $(a^{b})^{c}$ = $a^{bc}$ :

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Apply exponent rule: $(a^{b})^{c}$ = $a^{bc}$ :

$(2^{2})^{x-3}$ = $2^{2(x-3)}$

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Apply exponent rule: $a^{b+c}$ = $a^{b}$ $a^{c}$ :

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$2^{-1x}$ = $(2^{x})^{-1}$ , $2^{2x}$ = $(2^{x})^{2}$

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Rewrite the equation with $2^{x} = u$ :

$(u)^{-1}$ * $2^{4}$ - 3 = $(u)^{2}$ * $2^{-6}$ - 2

Solve $u^{-1}$ * $2^{4}$ - 3 = $u^{2}$ * $2^{-6}$ - 2:

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Refine:

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Add 3 to both sides:

$\frac{16}{u}$ - 3 + 3 = $\frac{1}{64}$ $u^{2}$ - 2 + 3

Simplify:

$\frac{16}{u}$ = $\frac{1}{64}$ $u^{2}$ + 1

Multiply by the Least Common Multiplier (64u):

$\frac{16}{u}$ * 64u = $\frac{1}{64}$ $u^{2}$ + 1 * 64u

Simplify:

$\frac{16}{u}$ * 64u = $\frac{1}{64}$ $u^{2}$ + 1 * 64u

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1024

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Substitute:

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u = 8

Substitute back u = $2^{x}$ :

8 = $2^{x}$

Solve for x:

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