Okay. so we need 15 2/5 pound bags. We'll calculate how many pounds that actually comes to. Our problem is 15x2/5. Using cross simplification, our problem becomes 3x2. 3x2=6. We need 6 pounds. That means that we must have 6 1-pound bags to fit with no leftovers.
Answer:
Step-by-step explanation:
x^2 + 24x + 144 is a perfect square: (x + 12)². The square root of this is
±(x + 12).
g(x) = square root of x^3 -216 = √(x^3 - 216), or
√(x³ - 6³). x³ - 6³ is not a perfect square, altho' it can be factored.
Answer:
The Product of Powers property allows you to rewrite
.
Step-by-step explanation:
We are given that,
The property allows us to write, ![x^a\times x^b=x^{a+b}](https://tex.z-dn.net/?f=x%5Ea%5Ctimes%20x%5Eb%3Dx%5E%7Ba%2Bb%7D)
It is required to know the corresponding property.
As, we see that,
<em>The expression involves the product of terms having same base resulting in the addition of the powers.</em>
So, we have that,
The corresponding property is 'Product of Powers'.
Hence, the complete statement is,
The Product of Powers property allows you to rewrite
.
Answer:
18.88% probability that three or four customers will arrive during the next 30 minutes
Step-by-step explanation:
In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:
![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%2A%5Cmu%5E%7Bx%7D%7D%7B%28x%29%21%7D)
In which
x is the number of sucesses
e = 2.71828 is the Euler number
is the mean in the given time interval.
Average rate of 6.4 per 30 minutes.
This means that ![\mu = 6.4](https://tex.z-dn.net/?f=%5Cmu%20%3D%206.4)
What is the probability that three or four customers will arrive during the next 30 minutes?
![P = P(X = 3) + P(X = 4)](https://tex.z-dn.net/?f=P%20%3D%20P%28X%20%3D%203%29%20%2B%20P%28X%20%3D%204%29)
![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%2A%5Cmu%5E%7Bx%7D%7D%7B%28x%29%21%7D)
![P(X = 3) = \frac{e^{-6.4}*(6.4)^{3}}{(3)!} = 0.0726](https://tex.z-dn.net/?f=P%28X%20%3D%203%29%20%3D%20%5Cfrac%7Be%5E%7B-6.4%7D%2A%286.4%29%5E%7B3%7D%7D%7B%283%29%21%7D%20%3D%200.0726)
![P(X = 4) = \frac{e^{-6.4}*(6.4)^{4}}{(4)!} = 0.1162](https://tex.z-dn.net/?f=P%28X%20%3D%204%29%20%3D%20%5Cfrac%7Be%5E%7B-6.4%7D%2A%286.4%29%5E%7B4%7D%7D%7B%284%29%21%7D%20%3D%200.1162)
![P = P(X = 3) + P(X = 4) = 0.0726 + 0.1162 = 0.1888](https://tex.z-dn.net/?f=P%20%3D%20P%28X%20%3D%203%29%20%2B%20P%28X%20%3D%204%29%20%3D%200.0726%20%2B%200.1162%20%3D%200.1888)
18.88% probability that three or four customers will arrive during the next 30 minutes