The number of fishes in a lake is expected to increase at a rate of 8% per year. How many fishes will be in the lake in 5 years if 10,000 fishes are placed in the lake today?
If 8/100=1y,5*8/100=40/100 by 40/100=140/100*10,000=1.4*10,000 ans:140,000
To find out how many fishes will be in the lake in 5 years, we need to calculate the future value of the initial population of 10,000 fishes using the compound interest formula.
The compound interest formula is given by:
FV = PV * (1 + r)^n
Where:
FV is the future value
PV is the present value
r is the rate of increase (expressed as a decimal)
n is the number of years
In this case, the present value (PV) is 10,000 fishes, the rate of increase (r) is 8% or 0.08, and the number of years (n) is 5.
Substituting these values into the formula, we get:
FV = 10,000 * (1 + 0.08)^5
Now, let's simplify this calculation step by step:
Step 1: Calculate the rate of increase in parentheses (1 + 0.08):
1 + 0.08 = 1.08
Step 2: Raise this result to the power of 5:
1.08^5 ≈ 1.469
Step 3: Multiply the result by the present value of 10,000:
10,000 * 1.469 ≈ 14,690
Therefore, in 5 years, there will be approximately 14,690 fishes in the lake if the population is expected to increase at a rate of 8% per year.