1. Traci purchases a car for $16,000. If its depreciates by 12% per year, what will be the car's value in 5 years?
2. The population of flies in Mr. Bunn's classroom doubles every 5 days. If there where 10 flies on the first day of school, when will the population reach 2,000,000?
2 answers
I don't understand how to do it .Is there a formula that I just, && what is this called? decay?
Yes, the formula is:
A(t) = A(0)*R^t
where
A(t) is the quantity at time t,
A(0) is the initial quantity
R=compounding rate (R>1), or decay rate(R<1)
t = time in appropriate units.
Take
1a.
Traci purchases a car for $10,000. If its depreciates by 11% per year, what will be the car's value in 5 years?
A(0)=10000,
t=5 years
R=1.0 - 0.11= 0.89
A(t)=A(5)=10000*0.89^5=10000*0.5584=$5584
2a.
The population of flies in Mr. Bunn's classroom doubles every 6 days. If there were 100 flies on the first day of school, when will the population reach 2,000,000?
A(0)=100
A(t)=2,000,000
R=2^(1/6)=1.12246 (because it doubles every 6 days).
Then
2000000=100*1.12246^t
1.12246^t=2000000/100=20000
t*log(1.12246)=log(20000)
t=log(20000)/log(1.12246)
=85.7 days
Post if you need further help.
A(t) = A(0)*R^t
where
A(t) is the quantity at time t,
A(0) is the initial quantity
R=compounding rate (R>1), or decay rate(R<1)
t = time in appropriate units.
Take
1a.
Traci purchases a car for $10,000. If its depreciates by 11% per year, what will be the car's value in 5 years?
A(0)=10000,
t=5 years
R=1.0 - 0.11= 0.89
A(t)=A(5)=10000*0.89^5=10000*0.5584=$5584
2a.
The population of flies in Mr. Bunn's classroom doubles every 6 days. If there were 100 flies on the first day of school, when will the population reach 2,000,000?
A(0)=100
A(t)=2,000,000
R=2^(1/6)=1.12246 (because it doubles every 6 days).
Then
2000000=100*1.12246^t
1.12246^t=2000000/100=20000
t*log(1.12246)=log(20000)
t=log(20000)/log(1.12246)
=85.7 days
Post if you need further help.
5 answers