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prohojiy [21]
3 years ago
6

Please help!!! ASAP someone please

Mathematics
1 answer:
murzikaleks [220]3 years ago
3 0

Answer:

0.31 yr

Step-by-step explanation:

The formula for interest compounded continuously is

FV = PVe^{rt}

FV = future value, and

PV = present value

If FV is twice the PV, we can calculate the doubling time, t

\begin{array}{rcl}2 & = & e^{rt}\\\ln 2 & = & rt\\t & = & \dfrac{\ln 2}{r} \\\end{array}

1. Brianna's doubling time

\begin{array}{rcl}t & = & \dfrac{\ln 2}{0.065}\\\\& = & \textbf{10.663 yr}\\\end{array}

2. Adam's doubling time

The formula for interest compounded periodically is

FV = PV\left (1 + \dfrac{r}{n} \right )^{nt}

where

n = the number of payments per year

If FV is twice the PV, we can calculate the doubling time.

\begin{array}{rcl}2 & = & \left (1 + \dfrac{0.0675}{4} \right )^{4t}\\\\&= & (1 + 0.016875 )^{4t}\\& = & 1.016875^{4t}\\\ln 2& = & 4 (\ln 1.01688)\times t \\& = & 0.066937t\\t& = & \dfrac{\ln 2}{0.066937}\\\\& = & \textbf{10.355 yr}\\\end{array}

3. Brianna's doubling time vs Adam's

10.663 - 10.355 = 0.31 yr

It would take 0.31 yr longer for Brianna's money to double than Adam's.

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Please check for me and if any are incorrect please explain!
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Answer:  (1) 14.4%   (2) 3.78%   (3) $222.48     (4) 21,176.47    

               (5) $425    (6) 14.3 cents per mile

<u>Step-by-step explanation:</u>

1) This is a calculator question. I = 0.144    --> 14.4%

A=P\bigg(1+\dfrac{r}{n}\bigg)^{nt}\\\\\bullet \text{A = Accrued amount (total amount paid)}\\\bullet \text{P = Principal (initial cost of the car)}\\\bullet \text{r = rate (interest rate in decimal form)}\\\bullet \text{n = number of times in a year (number of months)}\\\bullet \text{t = number of years}\\\\A = m \times nt\\\bullet \text{m = monthly payment amount}\\\bullet \text{nt = number of payments made}\\\implies m\times nt=P\bigg(1+\dfrac{r}{n}\bigg)^{nt}

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m\times nt=P\bigg(1+\dfrac{r}{n}\bigg)^{nt}\\\\\\93.33\times 36=3000\bigg(1+\dfrac{r}{12}\bigg)^{36}\\\\\\3359.88=3000\bigg(1+\dfrac{r}{12}\bigg)^{36}\\\\\\1.11996=\bigg(1+\dfrac{r}{12}\bigg)^{36}\\\\\\1.00315198=1+\dfrac{r}{12}\\\\\\0.00315198=\dfrac{r}{12}\\\\\\0.0378=r\\\\\\\large\boxed{3.78\%}=r

3) <em>same equation as #2 but deduct the down payment from the Principal</em>

   nt = 42

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   r = 18% --> 0.18

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m\times nt=P\bigg(1+\dfrac{r}{n}\bigg)^{nt}\\\\\\m\times 42=(5,555-555)\bigg(1+\dfrac{.18}{12}\bigg)^{42}\\\\\\42m=5000(1.015)^{42}\\\\\\42m=5000(1.868847)\\\\\\42m=9344.23557\\\\\\m=\large\boxed{222.48}

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   1000 + 800 = 0.085x

               1800 = 0.085x

        \large\boxed{21,176.47}=x

5)  \text{Annual Depreciation}=\dfrac{\text{Cost of car - Trade-in value}}{\text{Years driven}}

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=\dfrac{(5000-3800)+750+250+300}{17,500}\\\\\\=\dfrac{2500}{17,500}\\\\\\=0.142857\\\\=\large\boxed{14.3\ cents\ per\ mile}

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