Subjects algebra

Random Values

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1. The problem is to generate three new random values based on the given values $0.8551459$, $0.9577651$, and $1.07$ such that the new values do not exceed the original upper and lower limits and their average remains the same. 2. First, calculate the average of the original values: $$\text{average} = \frac{0.8551459 + 0.9577651 + 1.07}{3} = \frac{2.882911}{3} = 0.9609703333$$ 3. The lower limit is the minimum of the original values: $0.8551459$. 4. The upper limit is the maximum of the original values: $1.07$. 5. To generate new values $x$, $y$, and $z$ such that: - $0.8551459 \leq x,y,z \leq 1.07$ - $\frac{x + y + z}{3} = 0.9609703333$ 6. One approach is to randomly select $x$ and $y$ within the limits, then solve for $z$: $$z = 3 \times 0.9609703333 - x - y$$ 7. Ensure $z$ also lies within the limits. If not, repeat the random selection. 8. Example: - Let $x = 0.9$ - Let $y = 1.0$ Then: $$z = 3 \times 0.9609703333 - 0.9 - 1.0 = 2.882911 - 1.9 = 0.982911$$ 9. Check if $z$ is within limits: $$0.8551459 \leq 0.982911 \leq 1.07$$ which is true. 10. Therefore, new values can be: $$x = 0.9, y = 1.0, z = 0.982911$$ 11. Verify the average: $$\frac{0.9 + 1.0 + 0.982911}{3} = \frac{2.882911}{3} = 0.9609703333$$ which matches the original average. This method ensures the new values respect the original limits and maintain the same average.