The solutions of the equations of the situation can be:
z = 1 , y = 5, x = 4
z= 2 , y = 3, x = 5
z=3 , y = 1, x = 6
z = 0 , y = 7, x =3
The question can be expressed as a equation
6 x + 8 y + 10 z = 84
also, x + y + z = 10
⇒ x = 10- y - z
Putting it in first equation,
6(10 - y - z ) +8y + 10z = 84
⇒ 60 +2y + 4z = 84
⇒2(y + 2z ) = 14
⇒ y + 2z = 7
Now putting
z = 1 , we get y = 5,
z= 2 , y = 3
z=3 , y = 1
z = 0 , y = 7
So, only 4 possible solutions.
Therefore, there can only be 4 possible solutions for the equations.
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Hello,
g_m(15,1215)=√(5*1215)=9√15≈77,942286340....
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Answer:

Step-by-step explanation:
We are given that:

And we want to find F'(0).
First, find F(x):
![\displaystyle F'(x) = \frac{d}{dx}\left[ f(3x)]](https://tex.z-dn.net/?f=%5Cdisplaystyle%20F%27%28x%29%20%3D%20%5Cfrac%7Bd%7D%7Bdx%7D%5Cleft%5B%20f%283x%29%5D)
From the chain rule:
![\displaystyle \begin{aligned} F'(x) &= f'(3x) \cdot \frac{d}{dx} \left[ 3x\right] \\ \\ &= 3f'(3x)\end{aligned}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cbegin%7Baligned%7D%20F%27%28x%29%20%26%3D%20f%27%283x%29%20%5Ccdot%20%5Cfrac%7Bd%7D%7Bdx%7D%20%5Cleft%5B%203x%5Cright%5D%20%5C%5C%20%5C%5C%20%26%3D%203f%27%283x%29%5Cend%7Baligned%7D)
Then:

In conclusion, F'(0) = 15.
Answer:
24.24%
Step-by-step explanation:
In other words we need to find the probability of getting one blue counter and another non-blue counter in the two picks. Based on the stats provided, there are a total of 12 counters (6 + 4 + 2), out of which only 4 are blue. This means that the probability for the first counter chosen being blue is 4/12
Since we do not replace the counter, we now have a total of 11 counters. Since the second counter cannot be blue, then we have 8 possible choices. This means that the probability of the second counter not being blue is 8/11. Now we need to multiply these two probabilities together to calculate the probability of choosing only one blue counter and one non-blue counter in two picks.
or 0.2424 or 24.24%
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