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!
N + d = 20....n = 20 - d
0.05n + 0.10d = 1.35
0.05(20 - d) + 0.10n = 1.35
1 - 0.05d + 0.10d = 1.35
-0.05d + 0.10d = 1.35 - 1
0.05d = 0.35
d = 0.35/0.05
d = 7 <==== 7 dimes
n + d = 20
n + 7 = 20
n = 20 - 7
n = 13 <=== 13 nickels
Part A: After 9 days the radius of algae was approximately 12.81 mm. The reasonable domain to point is (0,9).
Part B: The 9 on the y-intercept represents the amount of algae the experiment started with.
Part C: (12.81-10.12)/7=0.38
Answer:
4. SR= 17 (opposite angle of parallogram are equal)