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!
y = mx + b
m = slope and b = y-intercept
We can arrange 6y = x - 12 in the form of y = mx + b
6y = x - 12
y = 1/6(x) - 2
Slope of y = 1/6(x) - 2 is 1/6. Taking the negative reciprocal of the slope we get the slope for the perpendicular line.
Negative reciprocal of 1/6 is -6.
The equation for the perpendicular line is
y = -6x + b
To find b we can plug in the x and y values of (4,-4) into it since it passes through those coordinates
-4 = -6(4) + b
b = -4 + 6(4)
b = -4 + 24
b = 20
So the equation for the perpendicular line is y = -6x + 20
Answer:
san po jan
<h3>step_by istep explanation_</h3>