Answer:
5,850 - 6,150/ 5,850≤x≤6,150
Step-by-step explanation:
All we really need to see here is what is 2.5% of 6,000, that is 150,
Now, we just add/ subtract that from 6,000 to get our range
5,850 - 6,150/ 5,850≤x≤6,150
Answer:
because its have √-1 so it is irrational
Answer:
The population that gives the maximum sustainable yield is 45000 swordfishes.
The maximum sustainable yield is 202500 swordfishes.
Step-by-step explanation:
Let be
, the maximum sustainable yield can be found by using first and second derivatives of the given function: (First and Second Derivative Tests)
First Derivative Test
![f'(p) = -0.02\cdot p +9](https://tex.z-dn.net/?f=f%27%28p%29%20%3D%20-0.02%5Ccdot%20p%20%2B9)
Let equalize the resulting expression to zero and solve afterwards:
![-0.02\cdot p + 9 = 0](https://tex.z-dn.net/?f=-0.02%5Ccdot%20p%20%2B%209%20%3D%200)
![p = 450](https://tex.z-dn.net/?f=p%20%3D%20450)
Second Derivative Test
![f''(p) = -0.02](https://tex.z-dn.net/?f=f%27%27%28p%29%20%3D%20-0.02)
This means that result on previous part leads to an absolute maximum.
The population that gives the maximum sustainable yield is 45000 swordfishes.
The maximum sustainable yield is:
![f(450) = -0.01\cdot (450)^{2}+9\cdot (450)](https://tex.z-dn.net/?f=f%28450%29%20%3D%20-0.01%5Ccdot%20%28450%29%5E%7B2%7D%2B9%5Ccdot%20%28450%29)
![f(450) =2025](https://tex.z-dn.net/?f=f%28450%29%20%3D2025)
The maximum sustainable yield is 202500 swordfishes.
Let X be the number of lightning strikes in a year at the top of particular mountain.
X follows Poisson distribution with mean μ = 3.8
We have to find here the probability that in randomly selected year the number of lightning strikes is 0
The Poisson probability is given by,
P(X=k) = ![\frac{e^{-mean} mean^{x}}{x!}](https://tex.z-dn.net/?f=%20%5Cfrac%7Be%5E%7B-mean%7D%20mean%5E%7Bx%7D%7D%7Bx%21%7D%20%20%20%20)
Here we have X=0, mean =3.8
Hence probability that X=0 is given by
P(X=0) = ![\frac{e^{-3.8} 3.8^{0}}{0!}](https://tex.z-dn.net/?f=%20%5Cfrac%7Be%5E%7B-3.8%7D%203.8%5E%7B0%7D%7D%7B0%21%7D%20%20%20%20)
P(X=0) = ![\frac{0.02237 * 1}{1}](https://tex.z-dn.net/?f=%20%5Cfrac%7B0.02237%20%2A%201%7D%7B1%7D%20%20)
P(X=0) = 0.0224
The probability that in a randomly selected year, the number of lightning strikes is 0 is 0.0224
21x³ - 18x²y + 24xy²
Simplify!
7x³ - 6x²y + 8xy²