2. Mary deposited $8000 in an account that pays 6% annual interest for 5 years. Using A=p(1+r)^t, what is the balance at the end of the 5 years?
Question 2 options:

a. 10706
b. 0.006
c. 134456000
d. 83886

3. Mary deposited $8000 in an account that pays 6% annual interest compounded quarterly for 5 years. What is the balance at the end of the 5 years? Use A = p(1 + rn)n*t

a. 10706
b. 10775
c. 12806
d. 8652

4. A $15,000 car depreciates at 20% each year. If you were to write an exponential model in the form , which value represents B in the function? (which is the growth factor)

a. 15,000
b. 0.2
c. 0.8
d. 0.02

Respuesta :

Hi there

1)
A=8,000×(1+0.06)^(5)
A=10,705.8 round your answer to get
A=10706

2)
A=8,000×(1+0.06÷4)^(4×5)
A=10,774.8 round to get
A=10775

3) with depreciation the formula is
V (t)=Vb^t
Where b=(1-r) (r rate of depreciation)
B=1-0.2=0.8

Good luck!

Answer:

2) the correct option is A) 10706

3) the correct option is A) 10775

4) the correct option is C) 0.8

Step-by-step explanation:

2)

Given:- amount is $8000, annual interest is 6% and time for 5 years

Formula used :- [tex]A=p(1+r)^{t}[/tex] Where p is principle amount , r is interest rate (in decimal) and t is the total number of years.

We need to find the balance at the end of the 5 years

Put given data in provided formula

[tex]A=p(1+r)^{t}[/tex]  

[tex]A=8000\times (1+0.06)^{5}[/tex]

[tex]A=10,705.8[/tex]

Round answer to get A=10706

Therefore, the correct option is A) 10706

3)

Given:- amount is $8000, annual interest is 6% and time for 5 years

Formula used :- [tex]A=p(1+\frac{r}{n})^{nt}[/tex]  Where p is the principal amount , r is the interest rate (in decimal), n is the number of compounding year and t is the total number of years.

We need to find the balance at the end of the 5 years

Put given data in provided formula

[tex]A=p(1+\frac{r}{n})^{nt}[/tex]

[tex]A=8000(1+\frac{0.06}{4})^{4\times 5}[/tex]

[tex]A=10774.8[/tex]

Round answer to get A=10775

Therefore, the correct option is A) 10775

4)

Given:- amount is $15000, annual interest is 20% and time for 5 years

We need to find the which value represents B in the function

depreciation the formula is

[tex]V(t)=Vb^{t}[/tex]

Where b=(1-r) (r rate of depreciation)

b = 1 - 0.2 = 0.8

Therefore, the correct option is C) 0.8