Exponential growth of the form:
F=Ar^t
F=100(1.22)^t
F(5)=100(1.22)^5
F(5)=270.27
F(5)=270 to the nearest frog...
....
1.22=r^12
r=1.22^(1/12)
F=100(1.22^(1/12)^t)
F=100(1.22^(t/12)) now it will produce monthly populations...
Say we did the same as before but instead of 5 years you have 60 months...
F(60)=100(1.22^(60/12))
F(60)=270 to the nearest frog... :P
Answer:
D. It would be less steep
Step-by-step explanation:
The first graph moves at a rate of 5/1 which is a greater fraction than 3/4
- The second graph is shallow due to the close points in x and y that are able to be conducted
- The first Graph rapidly increases at a way higher rate making it VERY steep
- While both are linear the second strays away in terms of plot lines
Answer:
251.2
Step-by-step explanation:
V=πr²h
3.14×4²×5
Answer:
3.2%
Step-by-step explanation:
The computation of the annual rate of interest is shown below:
As we know that
Simple interest = Principal × rate of interest × time period
($1,857.60 - $1,800) = $1,800 × rate of interest × 1
$57.60 = $1,800 × rate of interest × 1
So, the rate of interest is
= $57.60 ÷ $1,800 × 100
= 3.2%
Hence, the rate of interest on annual basis is 3.2%
Answer:
3x-4y=16
Step-by-step explanation:
