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VARVARA [1.3K]
3 years ago
11

You are debating about whether to buy a new computer for $800.00 or a refurbished computer with the same equipment for $640.00.

If a savings account earns 4.5% APR interest, how much do you really save with a refurbished computer if you put the difference into the savings account for a year?
Mathematics
1 answer:
Misha Larkins [42]3 years ago
5 0

Answer:

$167.2

Step-by-step explanation:

It has been given that the price of a new computer is $800.00 and refurbished computer with the same equipment has a  price of $640.00.

Let us find the amount that we will save with a refurbished computer if we put the difference into the savings account for a year using simple interest formula.  

A=P(1+rt), where A= Amount after t years, P=principal amount, r = interest rate (decimal form) and t=time.

Our principal amount will be the difference of prices of new computer and refurbished computer.

\text{Principal amount}=800-640=160

\text{Interest rate}=\frac{4.5}{100}=0.045  

Upon substituting our given values in above formula we will get,

\text{Amount we will save with a refurbished computer}=160(1+0.045\times 1)

\text{Amount we will save with a refurbished computer}=160(1+0.045)  

\text{Amount we will save with a refurbished computer}=160(1.045)

\text{Amount we will save with a refurbished computer}=167.2

Therefore, we will save $167.2 with refurbished computer when we put the difference into the savings account for a year.


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Answer:

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Step-by-step explanation:

A student invests $500 in a savings account

Principal = $500

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We are supposed to find equation can be used to determine the amount of money S(t)  that her savings account has after t years

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Substitute the value in the formula :

So, S(t)=500(1+0.04)^t

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Hence The equation can be used to determine the amount of money S(t)  that her savings account has after t years is S(t)=500(1.04)^t

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\bold{\text{Answer:  b.}\quad \dfrac{y^2}{100}+\dfrac{x^2}{49}=1}

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

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\dfrac{(x-0)^2}{7^2}+\dfrac{(y-0)^2}{10^2}=1\\\\\\\dfrac{x^2}{49}+\dfrac{y^2}{100}=1\\\\\\\longrightarrow \dfrac{y^2}{100}+\dfrac{x^2}{49}=1

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