MATHEMATICS COLLEGE

You purchase a car in 2010 for $25,000. The value of the car decreases by 14% annually. What would the value of the car be in 2020?

Answers

Answer 1

Answer:

The value of the car in 2020 is $5532.53

What is exponential decay?

A quantity is subject to exponential decay if it decreases at a rate proportional to its current value.

Given that, You purchase a car in 2010 for $25,000. The value of the car decreases by 14% annually.

The exponential decay is given by =

$A = P(1-r)^t$

A = final amount

P = principal amount

r = rate of decrease.

t = 10

Therefore,

A = 25000(1-0.14)¹⁰

A = 25000×0.86¹⁰

A = 5532.53

Hence, the value of the car in 2020 is $5532.53

For more references on exponential decay, click;

brainly.com/question/2193799

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Answer 2

Answer:

Answer:

42,690

Step-by-step explanation:

it is the rule of LONG numbers

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Answers

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Answers

Answer:

It should be 1/10

Step-by-step explanation:

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Kelly flies a distance of 2,100 miles. the trip takes 4 2/3 hourswhat is the airplanes unit rate of speed in miles per hour?

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Answers

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Step-by-step explanation:

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Answers

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Step-by-step explanation:

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Answers

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Trisha Long wants to buy a boat in five years. She estimates the boat will cost $15,000 at that time. What must Trisha deposit today in an account earning 5% annually to have enough to buy the boat in five years?

Answers

We use the formula for compound growth to figure this problem out. Formula is:

$P=P_(0)(1+(r)/(n))^(nt)$

Where,

P is the future value
$P_(0)$ is the initial deposite
r is the rate of interest annually
n is the number of times compounding occurs (n=1 for annual compounding, n=2 for semiannual compounding etc.)
t is time

Given P=15,000, r=5%=0.05 (in decimal), n=1 (since annual compounding), and t=5 years, we can solve:

$15000=P_(0)(1+(0.05)/(1) )^((1)(5))\n15000=P_(0)(1+0.05)^(5)\nP_(0)=(15000)/((1+0.05)^(5))\nP_(0)=(15000)/(1.05^(5))\nP_(0)=11,752.89$

So, Trisha Long needs to deposit $11,752.89 today in the account.

ANSWER: $11,752.89

$\bf \qquad \textit{Compound Interest Earned Amount}\n\nA=P\left(1+(r)/(n)\right)^(nt)\qquad \begin{cases}A=\textit{compounded amount}\to &15000\nP=\textit{original amount deposited}\nr=rate\to 5\%\to (5)/(100)\to &0.05\nn=\begin{array}{llll}\textit{times it compounds per year}\n\textit{annually means, once}\end{array}\to &1\nt=years\to &5\end{cases}$

solve for "P", to see how much Principal she should deposit today

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