Please show the full question. I would love to help
Answer:
- The function f(x) = 9,000(0.95)^x represents the situation.
- After 2 years, the farmer can estimate that there will be about 8,120 bees remaining.
- The range values, in the context of the situation, are limited to whole number
Step-by-step explanation:
The "growth" rate is -5%, so the growth factor, the base in the exponential equation, is 1.00-5% =0.95.
Using x=2, we find the population in 2 years is expected to be about ...
f(2) = 9000·0.95^2 ≈ 8123 . . . . about 8120
Using x=4, we find the population in 4 years is expected to be about ...
f(4) = 9000·0.95^4 ≈ 7331 . . . . about 7330
Since population is whole numbers of bees, the range of the function is limited to whole numbers.
The domain of the function is numbers of years. Years can be divided into fractions as small as you want, so the domain is not limited to whole numbers.
The choices listed above are applicable to the situation described.
Answer:
n=12
Step-by-step explanation:
5/6n=10
*6 *6
5n=60
n=12
Answer:
<u><em>Least Amount Likley: 7800 </em></u>
<u><em>Favorable:8200 </em></u>
Step-by-step explanation:
<u><em>Favorable 8000+2.5% </em></u>
<u><em>Least 8000-2.5% this is the range</em></u>
<u><em /></u>
Unfortunately I can not answer this question because there are no underlined digits in your question
