My TI-83 calculator was worth $75 after I owned it for 2 years. Since the new and improved TI-84 came out, the price of the calculator keeps going down. After 8 years, it is now worth only $10. Write an equation to model this situation and use it to predict when the calculator will be worth $2. Show all work.

It will take approximately 13 years for the calculator to be $2

Step-by-step explanation:

We need to set up a depreciation equation for the cost of the calculator per year.

Let the initial cost be I, the present value be P and the depreciation percentage be r%

Thus;

P = I( 1 - r%)^t

r% is same as r/100

For the two years initially;

75 =I( 1 - r/100)^2•••••••••(i)

For the 8 years, we have the equation as

10 = I(1 - r/100)^8 ••••••••(ii)

Now, simply divide the second equation by the first directly; we have;

10/75 = (1 - r/100)^8 / (1 - r/100)^2

2/15 = (1-r/100)^6

Take the log of both sides

ln(2/15) = ln(1-r/100)^6

ln(2/15) = 6 ln (1-r/100)

-2.015 = 6 ln (1 - r/100)

divide both sides by 6

-0.34 = ln (1-r/100)

1-r/100 = e^(-0.34)

1-r/100 = 0.715

Multiply through by 100

100 - r = 71.5

r = 100-71.5

r = 28.5%

Let’s now calculate the initial value;

P = I( 1 - r/100)^t

Let’s use the two years time frame

75 = I( 1 - 28.5/100)^2

75 = I(0.715)^2

I = 75/(0.715)^2

I = 146.71 which is approximately $147

Now, the time it will take for the calculator to be $2

P = I(1-r/100)^t

2 = 147 ( 1 - r/100)^t

divide both sides by 147

0.014 = (0.715)^t

ln 0.014 = ln (0.715)^t

t ln 0.715 = ln 0.014

t = ln 0.014/ln 0.715

t = 12.72

t is approximately 13 years

the answer is a

step-by-step explanation:

39.00 x 24 = total installment price

carrying installment price - cash price

needed to save the .00/