Answer:
The answer is "It would decrease, but not necessarily by 8%".
Step-by-step explanation:
They know that width of the confidence level is proportional to a confidence level. As just a result, reducing the confidence level decreases the width of a normal distribution, but not with the amount of variance in the confidence level. As just a result, when a person teaches a 90% standard deviation rather than a 98 percent normal distribution, the width of the duration narrows.
Answer:
5√3
Step-by-step explanation:
The <em>geometric mean</em> (GM) of two numbers x and y is the square root of their product.
Thus, the GM of x and y is √(xy).
The GM of 3√5 and 5√5
= √(3√5 × 5√5)
= √[3 × 5 × (√5)²]
= √(3 × 5 × 5)
= √(3 × 5²)
= 5√3
So, I came up with something like this. I didn't find the final equation algebraically, but simply "figured it out". And I'm not sure how much "correct" this solution is, but it seems to work.
![f(x)=\sin(\omega(x))\\\\f(\pi^n)=\sin(\omega(\pi^n))=0, n\in\mathbb{N}\\\\\\\sin x=0 \implies x=k\pi,k\in\mathbb{Z}\\\Downarrow\\\omega(\pi^n)=k\pi\\\\\boxed{\omega(x)=k\sqrt[\log_{\pi} x]{x},k\in\mathbb{Z}}](https://tex.z-dn.net/?f=f%28x%29%3D%5Csin%28%5Comega%28x%29%29%5C%5C%5C%5Cf%28%5Cpi%5En%29%3D%5Csin%28%5Comega%28%5Cpi%5En%29%29%3D0%2C%20n%5Cin%5Cmathbb%7BN%7D%5C%5C%5C%5C%5C%5C%5Csin%20x%3D0%20%5Cimplies%20x%3Dk%5Cpi%2Ck%5Cin%5Cmathbb%7BZ%7D%5C%5C%5CDownarrow%5C%5C%5Comega%28%5Cpi%5En%29%3Dk%5Cpi%5C%5C%5C%5C%5Cboxed%7B%5Comega%28x%29%3Dk%5Csqrt%5B%5Clog_%7B%5Cpi%7D%20x%5D%7Bx%7D%2Ck%5Cin%5Cmathbb%7BZ%7D%7D)
(8) if you mean -4 then it's (-8)
