Answer:
It would be changing by which a person thinks.
But, I will pick 569.124.
It would be 5.69124e2. (or 5.6912 x 10^3)
Answer:
The answer is in the photo
Step-by-step explanation:
Answer:
![Probability = \frac{2}{35}](https://tex.z-dn.net/?f=Probability%20%3D%20%5Cfrac%7B2%7D%7B35%7D)
Step-by-step explanation:
Given
![Total = 70](https://tex.z-dn.net/?f=Total%20%3D%2070)
First, we need to list the multiples of 5
![M_5 = \{5,10,15,20,25,30,35,40,45,50,55,60,65,70\}](https://tex.z-dn.net/?f=M_5%20%3D%20%5C%7B5%2C10%2C15%2C20%2C25%2C30%2C35%2C40%2C45%2C50%2C55%2C60%2C65%2C70%5C%7D)
Then, multiples of 3![M_3 = \{3,6,9,12,15,18,21,24,27,30,33,36,39,42,45,48,51,54,57,60,63,66,69\}](https://tex.z-dn.net/?f=M_3%20%3D%20%5C%7B3%2C6%2C9%2C12%2C15%2C18%2C21%2C24%2C27%2C30%2C33%2C36%2C39%2C42%2C45%2C48%2C51%2C54%2C57%2C60%2C63%2C66%2C69%5C%7D)
Next, is to list out the common elements in both
![M_3\ n\ M_5 = \{15,30,45,60\}](https://tex.z-dn.net/?f=M_3%5C%20n%5C%20M_5%20%3D%20%5C%7B15%2C30%2C45%2C60%5C%7D)
![n(M_3\ n\ M_5) = 4](https://tex.z-dn.net/?f=n%28M_3%5C%20n%5C%20M_5%29%20%3D%204)
The required probability is then calculated as thus:
![Probability = \frac{n(M_3\ n\ M_5)}{Total}](https://tex.z-dn.net/?f=Probability%20%3D%20%5Cfrac%7Bn%28M_3%5C%20n%5C%20M_5%29%7D%7BTotal%7D)
![Probability = \frac{4}{70}](https://tex.z-dn.net/?f=Probability%20%3D%20%5Cfrac%7B4%7D%7B70%7D)
![Probability = \frac{2}{35}](https://tex.z-dn.net/?f=Probability%20%3D%20%5Cfrac%7B2%7D%7B35%7D)
Answer:
<em>A=3 and B=6</em>
Step-by-step explanation:
<u>Increasing and Decreasing Intervals of Functions</u>
Given f(x) as a real function and f'(x) its first derivative.
If f'(a)>0 the function is increasing in x=a
If f'(a)<0 the function is decreasing in x=a
If f'(a)=0 the function has a critical point in x=a
As we can see, the critical points may define open intervals where the function has different behaviors.
We have
![f ( x ) = - 2 x^3 + 27 x^2 - 108 x + 9](https://tex.z-dn.net/?f=f%20%28%20x%20%29%20%3D%20-%202%20x%5E3%20%2B%2027%20x%5E2%20-%20108%20x%20%2B%209)
Computing the first derivative:
![f' ( x ) = - 6 x^3 + 54 x - 108](https://tex.z-dn.net/?f=f%27%20%28%20x%20%29%20%3D%20-%206%20x%5E3%20%2B%2054%20x%20-%20108)
We find the critical points equating f'(x) to zero
![- 6 x^3 + 54 x - 108=0](https://tex.z-dn.net/?f=-%206%20x%5E3%20%2B%2054%20x%20-%20108%3D0)
Simplifying by -6
![x^2 -9 x +18=0](https://tex.z-dn.net/?f=x%5E2%20-9%20x%20%2B18%3D0)
We get the critical points
![x=3,\ x=6](https://tex.z-dn.net/?f=x%3D3%2C%5C%20x%3D6)
They define the following intervals
![(-\infty,3),\ (3,6),\ (6,+\infty)](https://tex.z-dn.net/?f=%28-%5Cinfty%2C3%29%2C%5C%20%283%2C6%29%2C%5C%20%286%2C%2B%5Cinfty%29)
Thus A=3 and B=6
Answer:
answer is attached on the photo