Answer:
At a level of 95%, it is expected that the interval [0.45; 11.59] contains the value of the ductility in steel when its carbon content is 0.5%.
Step-by-step explanation:
Hello!
Considering the dependent variable:
Y: Ductility in steel.
And the independent variable:
X: Carbon content of the steel.
The linear regression was estimated and a prediction interval was calculated.
The prediction interval is calculated to predict a value that the variable Y (response variable) can take for a given value of the variable X (predictor variable) in the definition range of the linear regression line. Symbolically [Y/X=
]
In this case 95% prediction interval for Y/X=0.5
At a level of 95%, it is expected that the interval [0.45; 11.59] contains the value of the ductility in steel when its carbon content is 0.5%.
I hope it helps!
Step-by-step explanation:
![\sqrt[8]{ {x}^{2} {y}^{6} } \\ \\ = ( {x}^{2} {y}^{6} )^{ \frac{1}{8} } \\ \\ = {x}^{2 \times \frac{1}{8}} {y}^{6\times \frac{1}{8}} \\ \\ = x^{\frac{1}{4}} {y}^{3\times \frac{1}{4}} \\ \\ = x^{\frac{1}{4}} {y}^{\frac{3}{4}} \\ \\ = \sqrt[4]{x {y}^{3} } \\](https://tex.z-dn.net/?f=%20%5Csqrt%5B8%5D%7B%20%7Bx%7D%5E%7B2%7D%20%20%7By%7D%5E%7B6%7D%20%7D%20%20%5C%5C%20%20%5C%5C%20%20%3D%20%28%20%7Bx%7D%5E%7B2%7D%20%7By%7D%5E%7B6%7D%20%20%29%5E%7B%20%5Cfrac%7B1%7D%7B8%7D%20%7D%20%20%5C%5C%20%20%5C%5C%20%20%3D%20%20%7Bx%7D%5E%7B2%20%5Ctimes%20%5Cfrac%7B1%7D%7B8%7D%7D%20%7By%7D%5E%7B6%5Ctimes%20%5Cfrac%7B1%7D%7B8%7D%7D%20%20%5C%5C%20%20%5C%5C%20%20%3D%20x%5E%7B%5Cfrac%7B1%7D%7B4%7D%7D%20%7By%7D%5E%7B3%5Ctimes%20%5Cfrac%7B1%7D%7B4%7D%7D%20%20%5C%5C%20%20%5C%5C%20%20%20%3D%20x%5E%7B%5Cfrac%7B1%7D%7B4%7D%7D%20%7By%7D%5E%7B%5Cfrac%7B3%7D%7B4%7D%7D%20%20%5C%5C%20%20%5C%5C%20%20%20%3D%20%20%5Csqrt%5B4%5D%7Bx%20%7By%7D%5E%7B3%7D%20%7D%20%20%5C%5C%20)
Answer:
65.434
Step-by-step explanation:
-2 - 3 = -5
The answer is negative 5.