Answer:
The hens are the independent variable
Step-by-step explanation:
An independent variable is something that isn't dependent on anything else. Whose laying the eggs? The hens right? So the number of eggs laid is dependent on the number of hens laying them. Basically the hens are the independent variable because the number of eggs is relying on the number of hens laying them.
D I think so, I might be wrong tho
Answer:
x = 60
1) 60
2) 360
3) 160
60 + 360 + 160 = 580
Step-by-step explanation:
1) x
2) 6x
3) x + 100
x + 6x + x + 100 = 580
8x = 580 - 100
x = 480/8
x = 60
Answer: 1.8
Step-by-step explanation:
Answer:
(a) The probability of the event (<em>X</em> > 84) is 0.007.
(b) The probability of the event (<em>X</em> < 64) is 0.483.
Step-by-step explanation:
The random variable <em>X</em> follows a Poisson distribution with parameter <em>λ</em> = 64.
The probability mass function of a Poisson distribution is:

(a)
Compute the probability of the event (<em>X</em> > 84) as follows:
P (X > 84) = 1 - P (X ≤ 84)
![=1-\sum _{x=0}^{x=84}\frac{e^{-64}(64)^{x}}{x!}\\=1-[e^{-64}\sum _{x=0}^{x=84}\frac{(64)^{x}}{x!}]\\=1-[e^{-64}[\frac{(64)^{0}}{0!}+\frac{(64)^{1}}{1!}+\frac{(64)^{2}}{2!}+...+\frac{(64)^{84}}{84!}]]\\=1-0.99308\\=0.00692\\\approx0.007](https://tex.z-dn.net/?f=%3D1-%5Csum%20_%7Bx%3D0%7D%5E%7Bx%3D84%7D%5Cfrac%7Be%5E%7B-64%7D%2864%29%5E%7Bx%7D%7D%7Bx%21%7D%5C%5C%3D1-%5Be%5E%7B-64%7D%5Csum%20_%7Bx%3D0%7D%5E%7Bx%3D84%7D%5Cfrac%7B%2864%29%5E%7Bx%7D%7D%7Bx%21%7D%5D%5C%5C%3D1-%5Be%5E%7B-64%7D%5B%5Cfrac%7B%2864%29%5E%7B0%7D%7D%7B0%21%7D%2B%5Cfrac%7B%2864%29%5E%7B1%7D%7D%7B1%21%7D%2B%5Cfrac%7B%2864%29%5E%7B2%7D%7D%7B2%21%7D%2B...%2B%5Cfrac%7B%2864%29%5E%7B84%7D%7D%7B84%21%7D%5D%5D%5C%5C%3D1-0.99308%5C%5C%3D0.00692%5C%5C%5Capprox0.007)
Thus, the probability of the event (<em>X</em> > 84) is 0.007.
(b)
Compute the probability of the event (<em>X</em> < 64) as follows:
P (X < 64) = P (X = 0) + P (X = 1) + P (X = 2) + ... + P (X = 63)
![=\sum _{x=0}^{x=63}\frac{e^{-64}(64)^{x}}{x!}\\=e^{-64}\sum _{x=0}^{x=63}\frac{(64)^{x}}{x!}\\=e^{-64}[\frac{(64)^{0}}{0!}+\frac{(64)^{1}}{1!}+\frac{(64)^{2}}{2!}+...+\frac{(64)^{63}}{63!}]\\=0.48338\\\approx0.483](https://tex.z-dn.net/?f=%3D%5Csum%20_%7Bx%3D0%7D%5E%7Bx%3D63%7D%5Cfrac%7Be%5E%7B-64%7D%2864%29%5E%7Bx%7D%7D%7Bx%21%7D%5C%5C%3De%5E%7B-64%7D%5Csum%20_%7Bx%3D0%7D%5E%7Bx%3D63%7D%5Cfrac%7B%2864%29%5E%7Bx%7D%7D%7Bx%21%7D%5C%5C%3De%5E%7B-64%7D%5B%5Cfrac%7B%2864%29%5E%7B0%7D%7D%7B0%21%7D%2B%5Cfrac%7B%2864%29%5E%7B1%7D%7D%7B1%21%7D%2B%5Cfrac%7B%2864%29%5E%7B2%7D%7D%7B2%21%7D%2B...%2B%5Cfrac%7B%2864%29%5E%7B63%7D%7D%7B63%21%7D%5D%5C%5C%3D0.48338%5C%5C%5Capprox0.483)
Thus, the probability of the event (<em>X</em> < 64) is 0.483.