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:
29.4
Step-by-step explanation:
42*0.7
Answer: x>18.64
Step-by-step explanation:
x+6.56-6.56>25.2-6.56
x>18.64
Answer:
C
Step-by-step explanation:
To solve, use inverse operations:
Now we solve for both +/- 7.
x+6=7
x=1
x+6=-7
x=-13
Answer:
E
Step-by-step explanation:
Confidence Interval = mean + or - error margin
Mean = 42,000, error margin = width of estimate of the parameter ÷ 2 = 175 ÷ 2 = 87.50
We can be 95% confident that the population mean is 42,000 plus or minus 87.50
