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prisoha [69]
2 years ago
9

Whitney deposits $9,000 for two years. She compares two different banks. State Banks will pay her 4.1% interest, compounded mont

hly. Kings Savings will pay her 4.01% interest, compounded continuously.
Mathematics
1 answer:
dlinn [17]2 years ago
3 0

Answer:

1)state bank

2)$16.21

Step-by-step explanation:

CHECK THE COMPLETE QUESTION BELOW

Whitney deposits $9,000 for two years. She compares two different banks. State Banks will pay her 4.1% interest, compounded monthly. Kings Savings will pay her 4.01% interest, compounded continuously.Which bank pays higher interest? How much higher?

The Amount in the formula for compound interest is

A= P(1+r)^2

P = principal amount,

r = rate per periods,

t = number of periods,

To know the interest she got from the State bank, we say

P = $ 9000, r = 4.1% = 0.041, t = 2 years,

A= 9000(1+0.041)^2

A= 9000(1.041)^2

= 9767.74

Then interest earned = A - P

= 9767.74 - 9000

= $ 767.74

To know the interest she got from the State bank, we say

P = $ 9000, r = 4.01% = 0.0401, t = 2 years,

A= 9000(1+0.0401)^2

A= 9000(1.0401)^2

= 9751.53

Then interest earned = A - P

= 9767.74 - 9000

= $ 751.53

The difference between the two interest is ($ 767.74 - $ 751.53)

=$16.21

Hence state bank will have $16.21 higher interest

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Answer:

\frac{9}{3x + 2} = 1 - \frac{1}{3}(x - \frac{7}{3}) + \frac{1}{9}(x - \frac{7}{3})^2 - \frac{1}{27}(x - \frac{7}{3})^3 ........

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Step-by-step explanation:

Given

f(x)= \frac{9}{3x+ 2}

c = 6

The geometric series centered at c is of the form:

\frac{a}{1 - (r - c)} = \sum\limits^{\infty}_{n=0}a(r - c)^n, |r - c| < 1.

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a \to first term

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We have to write

f(x)= \frac{9}{3x+ 2}

In the following form:

\frac{a}{1 - r}

So, we have:

f(x)= \frac{9}{3x+ 2}

Rewrite as:

f(x) = \frac{9}{3x - 18 + 18 +2}

f(x) = \frac{9}{3x - 18 + 20}

Factorize

f(x) = \frac{1}{\frac{1}{9}(3x + 2)}

Open bracket

f(x) = \frac{1}{\frac{1}{3}x + \frac{2}{9}}

Rewrite as:

f(x) = \frac{1}{1- 1 + \frac{1}{3}x + \frac{2}{9}}

Collect like terms

f(x) = \frac{1}{1 + \frac{1}{3}x + \frac{2}{9}- 1}

Take LCM

f(x) = \frac{1}{1 + \frac{1}{3}x + \frac{2-9}{9}}

f(x) = \frac{1}{1 + \frac{1}{3}x - \frac{7}{9}}

So, we have:

f(x) = \frac{1}{1 -(- \frac{1}{3}x + \frac{7}{9})}

By comparison with: \frac{a}{1 - r}

a = 1

r = -\frac{1}{3}x + \frac{7}{9}

r = -\frac{1}{3}(x - \frac{7}{3})

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r = -\frac{1}{3}(x - \frac{7}{3}+6-6)

Take LCM

r = -\frac{1}{3}(x + \frac{-7+18}{3}+6-6)

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So, the power series becomes:

\frac{9}{3x + 2} =  \sum\limits^{\infty}_{n=0}ar^n

Substitute 1 for a

\frac{9}{3x + 2} =  \sum\limits^{\infty}_{n=0}1*r^n

\frac{9}{3x + 2} =  \sum\limits^{\infty}_{n=0}r^n

Substitute the expression for r

\frac{9}{3x + 2} =  \sum\limits^{\infty}_{n=0}(-\frac{1}{3}(x - \frac{7}{3}))^n

Expand

\frac{9}{3x + 2} =  \sum\limits^{\infty}_{n=0}[(-\frac{1}{3})^n* (x - \frac{7}{3})^n]

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\frac{1}{3}|x - \frac{7}{3}| < 1

Multiply both sides by 3

|x - \frac{7}{3}|

Expand the absolute inequality

-3 < x - \frac{7}{3}

Solve for x

\frac{7}{3}  -3 < x

Take LCM

\frac{7-9}{3} < x

-\frac{2}{3} < x

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