SOLUTION: Can you help with this word problem. Thanks 1.) Suppose that $4,000 is invested at 7% compounded weekly. How much money will be in the account in 10 years? (show equation us

Algebra ->  Percentage-and-ratio-word-problems -> SOLUTION: Can you help with this word problem. Thanks 1.) Suppose that $4,000 is invested at 7% compounded weekly. How much money will be in the account in 10 years? (show equation us      Log On


   



Question 64010: Can you help with this word problem. Thanks
1.)
Suppose that $4,000 is invested at 7% compounded weekly. How much money will be in the account in 10 years? (show equation used to find your answer)
2.)
In 2002, the estimated population of Ethiopia was 68 million with an annual growth of 2.8% compounded continuously. Based on this information, what will the population of Ethiopia be in 2020? (show equation used to find your answer)

Found 2 solutions by Edwin McCravy, josmiceli:
Answer by Edwin McCravy(20056) About Me  (Show Source):
You can put this solution on YOUR website!
Can you help with this word problem. Thanks 
1.) 
Suppose that $4,000 is invested at 7% compounded weekly. How much money will be
in the account in 10 years? (show equation used to find your answer) 

The formula is

A = P(1 + r/n)nt

where P = 4000, r = .07, n = 52, t = 10

A = 4000(1 + .07/52)(52)(10)

A = 4000(1 + .0013461538)520

A = 4000(1.0013461538)520

A = 8051.219976, or $8051.22

Notice that if you used the formula for continuous
compounding, as you have to do in the next problem,
you get:

A = Pert 

A = 4000e(.07)(10)

A = 8055.01083 or $8055.01

which is very close to what you get compounding 
weekly.

2.)
In 2002, the estimated population of Ethiopia was 68 million with an annual
growth of 2.8% compounded continuously. Based on this information, what will
the population of Ethiopia be in 2020? (show equation used to find your answer)

A = Pert

where P = 68, r = .028, t = 18 (18 years 
from 2002 till 2020) 

A = 68e(.028)(18)

A = 68e.504

A = 112.5623967 or about 112.6 million

Edwin

Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
My book tells me A+=+P%281+%2B+r%2Fk%29%5Ekt where
P = 4000
r = .07
k = 52
t = 10
A+=+4%2A10%5E3%281+%2B+.07%2F52%29%5E%2852%2A10%29
log%28A%29+=+log%284%2A10%5E3%29+%2B+log%28%281+%2B+.07%2F52%29%5E%2852%2A10%29%29
log%28A%29+=+log%284%29+%2B+3+%2B+520%2Alog%281.001346%29
log%28A%29+=+.602+%2B+3+%2B+520%2Alog%281.001346%29
log%28A%29+=+3.602+%2B+520%2A.000584
log%28A%29+=+3.602+%2B+.3037
log%28A%29+=+3.906
A+=+8047.85
There will be about $8048 in the account after 10 years
---------------------
A+=+Pe%5E%28rt%29
A+=+6.8%2A10%5E7%2Ae%5E%28.028%2A18%29
log%28A%29+=+log%286.8%29+%2B+7+%2B+log%28e%5E.504%29
log%28A%29+=+.833+%2B+7+%2B+.504%2Alog%28e%29
log%28A%29+=+7.833+%2B+.504%2A.434
log%28A%29+=+7.833+%2B+.219
log%28A%29+=+8.052
A+=+112.690%2A10%5E6
The population will be 112,690,000 by 2020