Question

Sara's new car depreciates 8% each year. 1. A= P(1 - 8/100 x 12)^12t = P(298/300)^12t Per month. 2. A= P(1 - 8/100 x 52)^12t = P(298/300)^12t Per week. 3. A= P(1-3/100 x 365)^12t = P(298/300)^12t Per day. What relationship is there between the amount of depreciation and the time interval measured in these exponential functions?

Answer 1

i think for 2.
\[A=P \left( 1-\frac{ 8 }{ 100 \times 52 } \right)^{52t}\] is per week.
for 3.
\[A=P \left( 1-\frac{ 8 }{ 100 \times 365 } \right)^{365t}\] per day

Answer 2

Let us find amount for one year.
for 1.
\[A=P \left( 1-\frac{ 8 }{ 100\times 12 } \right)^{12\times 1}=P \left( \frac{ 298 }{ 300 } \right)^{12} =0.9229P\]

Answer 3

\[for ~2. A=P \left( 1-\frac{ 8 }{ 100 \times52 } \right)^{52 \times 1}=P \left( \frac{ 1298 }{ 1300 } \right)^{52}=0.9231P\]
price after one year when depreciation rate is every week.

Answer 4

What I need to find out is between
1. A= P(1 - 8/100 x 12)^12t = P(298/300)^12t Per month.

2. A= P(1 - 8/100 x 52)^52t = P(298/300)^52t Per week.

3. A= P(1-3/100 x 365)^365t = P(298/300)^365t Per day.

What is the relation between the amount of depreciation, and the time intervals recorded.