Respuesta :
This is an exponential equation of the form:
y=ab^x, where a=initial value and b=common ratio or "rate"
In this case a=9000 and b=(100-5)/100=0.95 so
f(x)=9000(0.95)^x
After two years...
f(2)=9000(0.95)^2=8122.5....so I guess that you could estimate that there are about 8120 bees left to the nearest 10 bees.
however f(4)≈7331, far more than 1800
In reality, the domain is not really limited to whole values, unless we are to believe that the 5% die off in an instant at the end of each year :)
The range is the number of bees, and as such, you cannot have fractional bees, thus the range, in reality, is limited to whole numbers.
y=ab^x, where a=initial value and b=common ratio or "rate"
In this case a=9000 and b=(100-5)/100=0.95 so
f(x)=9000(0.95)^x
After two years...
f(2)=9000(0.95)^2=8122.5....so I guess that you could estimate that there are about 8120 bees left to the nearest 10 bees.
however f(4)≈7331, far more than 1800
In reality, the domain is not really limited to whole values, unless we are to believe that the 5% die off in an instant at the end of each year :)
The range is the number of bees, and as such, you cannot have fractional bees, thus the range, in reality, is limited to whole numbers.
Answer:
Step-by-step explanation:
1. The function f(x) = 9,000(0.95)x represents the situation.
2. After 2 years, the farmer can estimate that there will be about 8,120 bees remaining.
3. The range values, in the context of the situation, are limited to whole numbers.