It is probably 0,2,4,6 because other ones go "out of the sequence".
Hope this helps. :)
The law of supply<span> states that the quantity of a good </span>supplied<span> (i.e., the amount owners or producers offer for sale) rises as the market price rises, and falls as the price falls. Conversely, the </span>law<span> of </span>demand<span> (see </span>demand<span>) says that the quantity of a good demanded falls as the price rises, and vice versa.</span>
Hey!
Weight of small box = 5 lb. x no. of small boxes total weight = 5*x
Weight of Large boxes = 12 lb. y no. of large boxes total weight = 12*y
Total weight = 5x + 12y.
Since total weight carrying capacity of shelf is 80, thus the inequality becomes:
5x + 12y Less than or equal to 80.
(a) satisfies the inequality.
(b) satisfies the inequality.
Hope my answer helps!
Using the Poisson distribution, there is a 0.8335 = 83.35% probability that 2 or fewer will be stolen.
<h3>What is the Poisson distribution?</h3>
In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by:
![P(X = x) = \frac{e^{-\mu}\mu^{x}}{(x)!}](https://tex.z-dn.net/?f=P%28X%20%3D%20x%29%20%3D%20%5Cfrac%7Be%5E%7B-%5Cmu%7D%5Cmu%5E%7Bx%7D%7D%7B%28x%29%21%7D)
The parameters are:
- x is the number of successes
- e = 2.71828 is the Euler number
is the mean in the given interval.
The probability that a rental car will be stolen is 0.0004, hence, for 3500 cars, the mean is:
![\mu = 3500 \times 0.0004 = 1.4](https://tex.z-dn.net/?f=%5Cmu%20%3D%203500%20%5Ctimes%200.0004%20%3D%201.4)
The probability that 2 or fewer cars will be stolen is:
![P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2)](https://tex.z-dn.net/?f=P%28X%20%5Cleq%202%29%20%3D%20P%28X%20%3D%200%29%20%2B%20P%28X%20%3D%201%29%20%2B%20P%28X%20%3D%202%29)
In which:
![P(X = x) = \frac{e^{-\mu}\mu^{x}}{(x)!}](https://tex.z-dn.net/?f=P%28X%20%3D%20x%29%20%3D%20%5Cfrac%7Be%5E%7B-%5Cmu%7D%5Cmu%5E%7Bx%7D%7D%7B%28x%29%21%7D)
![P(X = 0) = \frac{e^{-1.4}1.4^{0}}{(0)!} = 0.2466](https://tex.z-dn.net/?f=P%28X%20%3D%200%29%20%3D%20%5Cfrac%7Be%5E%7B-1.4%7D1.4%5E%7B0%7D%7D%7B%280%29%21%7D%20%3D%200.2466)
![P(X = 1) = \frac{e^{-1.4}1.4^{1}}{(1)!} = 0.3452](https://tex.z-dn.net/?f=P%28X%20%3D%201%29%20%3D%20%5Cfrac%7Be%5E%7B-1.4%7D1.4%5E%7B1%7D%7D%7B%281%29%21%7D%20%3D%200.3452)
![P(X = 2) = \frac{e^{-1.4}1.4^{2}}{(2)!} = 0.2417](https://tex.z-dn.net/?f=P%28X%20%3D%202%29%20%3D%20%5Cfrac%7Be%5E%7B-1.4%7D1.4%5E%7B2%7D%7D%7B%282%29%21%7D%20%3D%200.2417)
Then:
![P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) = 0.2466 + 0.3452 + 0.2417 = 0.8335](https://tex.z-dn.net/?f=P%28X%20%5Cleq%202%29%20%3D%20P%28X%20%3D%200%29%20%2B%20P%28X%20%3D%201%29%20%2B%20P%28X%20%3D%202%29%20%3D%200.2466%20%2B%200.3452%20%2B%200.2417%20%3D%200.8335)
0.8335 = 83.35% probability that 2 or fewer will be stolen.
More can be learned about the Poisson distribution at brainly.com/question/13971530
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