Question

HELP PLEASE I BEG YOU

Answer 1

Part A: Find the third terms in the table, A and B, using a linear model (simple interest) and an exponential model (compound interest).
Part B: Assuming it has the same annual interest rate rounded to the nearest cent, how much more is the investment worth using an exponential model versus a linear model?
Part C: How would a rate of decay or decay factor affect the outcome in Parts A and B? Explain.

Answer 2

you typically need a period too, to calc the interest in each case

Answer 3

I'm not sure I know what that is, sorry. I was just given this table (which i attracted) and that's it. I:

Answer 4

Thank you :)

Answer 5

@PickleLover ... Looking at the chart. what is the principle?

Answer 6

NP Always happy to help if i can :D

Answer 7

Uh, would it be the price?

Answer 8

... The initial base price @ year 0 yep

Answer 9

So: the question says that it is the same annual interest rate.
We can solve for R using the simple interest formula:

\[$32,000.00=$20,000.00(r)(10)\]

Answer 10

So I would need to calculate that?

Answer 11

yep... so what is r? aka the annual interest rate

Answer 12

.16

Answer 13

Yep. so...
\[Interest=($20,000.00)(.16)(20years)\]
What is the amount of $ @ 20 years? using the simple interest formula

Answer 14

64000

Answer 15

yep!(:

Answer 16

Awesome! :) Would I have to do the same thing for other missing part of the table?

Answer 17

Hmm... I think so. Let's look at it

Answer 18

\[$35,816.95=$20,000(1+\frac{ r }{ 100 })^{10}\]

--> Solving for annual rate

\[\frac{$35,816.95}{$20,000.00}=(1+\frac{ r }{ 100 })^{10}\]

Answer 19

r = 60000 ?

Answer 20

*Take tenth root of both sides
\[1.059=1+\frac{ r }{ 100 }\]
Multiply everything by 100
\[105.9=100+r\]
\[r=5.9\]

Answer 21

Ohhh, okay. I see my mistake

Answer 22

Yep... but the rate of interest is basically 6%. if you plug it into the calculator it is 5.99999
so basically 6 (:

\[A=$20,000.00(1+\frac{ 6.00 }{ 100 })^n\]

Answer 23

What is the amount at 20 years? of B

Answer 24

I calculated it and got \[\frac{ 53 }{ 50 }\] ?

Answer 25

Hmm... be careful with how you enter this into your calculator.

It should be
$64,142.71

Answer 26

Okay, got it :)

Answer 27

Okay awesome. So part A is done.
Part B is asking you what the price difference of the model is at 20 years... comparing the simple interest & compounding interest.

AKA
What is $64,142.71-$64,000.00? & How much more $ do you get if you use the compounding interest rate?

Answer 28

Okay, so I would have to subtract the two?

Answer 29

Hmm.. I would say so. Looks like you are just comparing the two interest rates at year 20.

Answer 30

Ah okay :)
Would it be 142.71 then?

Answer 31

Yep... I believe so.
I'm not sure if this question is asking you to compare the two rates though...
So simple interest: (0.16)
Compounding (1.06)
Anyways... moving on to part C.
What is a decay factor?

Answer 32

Decay factor is when the original amount declines

Answer 33

Yep.
So it looks like both equations on have a growth factor.. aka positive rate.
Or more importantly:

\[A=A _{0}b^t\]
When b<0<1 --> exponential decay

Answer 34

But looks like you got the right idea. If the amount compounded had a decay factor... 0<b<1*** sorry typo
Than the amount compounded decreases in parts A, B @ year 20.

Answer 35

So both amounts had a decay or had a growth?

Answer 36

Nevermind I got it haha, thank you so much for your help and being so patient! You rock! <3

Answer 37

NP :D I am just happy to help!