52? ahh thats what im thinking. Maybe wait for someone elses answer
        
                    
             
        
        
        
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!
 
        
             
        
        
        
Answer:
x=-2
Step-by-step explanation:
1/2(2-6x)-4(x+3/2)=-(x-3)+4
(1-3x)-4x-6=-x+3+4
-7x-5=-x+7
Tranpose
-7x+x=5+7
-6x=12
Divide by -6
x=-2
 
        
             
        
        
        
14/4 is a fraction greater than 1.