If Rick has $500 In his account after 4 years investing at 2.5% per year , how much money did he deposit initially ? Use =A =P (1+ r/n)
2). Your new laptop cost $1500 but depreciates in value by about 21% each year.
a) How,much will your laptop be worth in 6 years ? Use A= Pe . ( Remember; Is r posting or negative if it is depreciating ? )
b) About how long will it take before your laptop is worth only $60?
2. V = Vo(1-r)^n.
a. V = 1500(1-0.21)^6 = $364.63
b. V = 1500(1-0.21)^n = 60.
1500(0.79)^n = 60,
0.79^n = 60/1500 = 0.04, n*Log0.79 = Log0.04,
n = Log0.04/Log0.79 = 13.7 Years.
To solve both of these questions, we can use the formula for compound interest:
A = P(1 + r/n)^(nt)
Where:
A = the final amount
P = the initial amount (principal)
r = annual interest rate (as a decimal)
n = number of times interest is compounded per year
t = number of years
Let's go through each question step by step:
1) If Rick has $500 in his account after 4 years investing at 2.5% per year, how much money did he deposit initially?
First, let's rearrange the formula to solve for P:
P = A / (1 + r/n)^(nt)
Now, we substitute the given values into the formula:
A = $500
r = 2.5% = 0.025 (as a decimal)
n = 1 (compounded once per year)
t = 4
P = $500 / (1 + 0.025/1)^(1*4)
P = $500 / (1.025)^4
P ≈ $452.89
Therefore, Rick initially deposited approximately $452.89.
2a) If your new laptop costs $1500 but depreciates in value by about 21% each year, how much will it be worth in 6 years?
Since the laptop price is decreasing each year, we need to use a negative interest rate (or a negative value for r).
A = P(1 + r/n)^(nt)
Substituting the given values:
P = $1500
r = -21% = -0.21 (as a decimal)
n = 1 (compounded once per year)
t = 6
A = $1500(1 - 0.21/1)^(1*6)
A = $1500(0.79)^6
A ≈ $488.54
Therefore, the laptop will be worth approximately $488.54 in 6 years.
2b) About how long will it take before your laptop is worth only $60?
To determine the number of years, we need to rearrange the formula to solve for t:
t = log(A/P) / (n * log(1 + r/n))
Substituting the given values:
A = $60
P = $1500
r = -21% = -0.21 (as a decimal)
n = 1 (compounded once per year)
t = log($60/$1500) / (1 * log(1 - 0.21/1))
t ≈ log(0.04) / (1 * log(0.79))
t ≈ -2.255 / (1 * -0.204)
t ≈ 11.05
Therefore, it will take approximately 11.05 years before the laptop is worth only $60.