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!
3/4 + 1/2
multiply 1/2 denominator and numerator by 2 to match 3/4
= 3/4 + 2/4 = 5/4 (copy same denominator add numerator)
2/6 + 1/3
divide 2/6 denominator and numerator by 2 to match 1/3
= 1/3 + 1/3 = 2/3 (copy same denominator add numerator)
5/9 + 2/3
multiply 2/3 denominator and numerator by 3 to match 5/9
= 5/9 + 6/9 = 11/9 (copy same denominator add numerator)
6/9 -1/5
cross multiply 6x5 - 1x9 for numerator
for denominator multiply 9x5
=30/45 - 9/45= 21/45
divide num and den by 3
=7/15
5/8-1/3
cross multiply 5x3-1x8 for numerator
multiply 8x3 for denominator
= 15/24 -8/24 =7/24
The picture is kind of blurry could you please take a clearer one?
Answer:
-2x-5
Step-by-step explanation: