The technology of a school was valued at $153,500 three years ago. It is now valued at $130,570.80. What was the depreciation percentage (%) rate over those years? (A = C(1 - r)t) A. 0.525% B. 5.20% C. 5.25% D. 5.30%

Question

The technology of a school was valued at $153,500 three years ago. It is now valued at $130,570.80. What was the depreciation percentage (%) rate over those years? (A = C(1 - r)t) 	A.	0.525% 	B.	5.20% 	C.	5.25% 	D.	5.30%

aum · Answer

$\Large A = C(1-r)^t $
A = $130,570.80
C = $153,500
t = 3 years
Substitute and solve for r.

anonymous · Answer

Thanks!

anonymous · Answer

May have another for you :)

aum · Answer

You are welcome.  I can try one more.

anonymous · Answer

what is the common ratio of this geometric sequence? (following are in 2 column chart)

1 - 27
2 - 18
3 - 12
4 - 8
5 - 16/3

Answers possibilities:
-9
2/3
3/2
9

aum · Answer

For the first one, when you solve for r it will be in decimal.  Multiply it by 100 to convert to percentage.

anonymous · Answer

got it.
any advice on the other question? thanks!

aum · Answer

To find the common ratio, take any term and divide it by the previous term.
Example, second term / first term = 18/27 = 2/3
third term / second term = 12/18 = 2/3

anonymous · Answer

thanks!