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!
I think it’s b
Not for sure but it makes the most sense
Answer:
Its the last option
y = 1488.33x + 23800 ; $172,600.
Step-by-step explanation:
The slope of the trend line = (median value for 2000 - median value for 1940) / 60
= 89300 / 60
= 1488.33.
So the equation is 1488.33x + 23,800
Estimate for 2040
is 1483.33* 100 + 23800
= 172,600.
Answer:
I'm pretty sure it's A. It will help you get your equations set up correctly
Step-by-step explanation:
For anyone needing this answer in the future lol
Answer:
The blue one
4, 2
6, 3
12, 6
Step-by-step explanation: