Simplify. 4−16÷4+42 −16 13 15 16PLEASE HELP
2 answers:
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Answer:
( 0 , 8)
Step-by-step explanation:
The y-intercept is when the graph intersects the y axis, in other words, when x is zero. Hence substitute 0 in for x
f ( x ) = 8
f ( 0 ) = 8
Any number to the power zero is one;
f ( 0 ) = 8 * 1
f(0) = 8
The y-intercept is 8.
Answer:
Step-by-step explanation:
1) The null hypothesis is,
H_0: The mean thickness of teh wafers for the five positions are equal
i.e,
2)
The alternative hypothesis is,
H_1: There is an evidence of a difference in the mean thickness of the wafers for the five positions
3)
Let us consider the level of significance
from the Minitab outout
One-way ANOVA:C1 versus C2
source DF SS MS F P
C2 4 1417.73 354.43 51.00 0.00
Error 145 1007.77 6.95
Total 14 2425.50
S = 2.636 R - S = 58.45% R - Sq(adj) = 57.31%
Individual 95% CIs For Mean Based on Pooled StDev
level N Mean StDev -,----------,----------,----------,----------
1 30 240.53 2.62 (--,--)
2 30 243.73 2.79 (--,--)
3 30 246.07 2.90 (--,--)
4 30 249.10 2.66 (--,--)
5 30 247.07 2.15 (--,--)
-,----------,----------,----------,----------
240.0 243.0 246.0 249.0
Pooled StDev = 2.64
4)
The test statistic is, F = 51
5)
The P-value is approximately 0
6)
Here, the P - value is less than the level of significance
So, we do not accept our null hypothesis H_0
7)
Therefore, we conclude that there is an evidence of a difference in the mean thickness of the wafers for the five positions at level of significance
b)
chek attachment
we observe that,
The mean thickness of the wafer for position 1 is significant with position 2,
position 18, position 19 and position 28.
The mean thickness of the wafer for position 2 is significant with position 18,
position 19 and position 28.
The mean thickness of the wafer for position 18 is significant with the
position 19.
But the mean thickness of the wafer for position18 is not significant with
position 28.
The mean thickness of the wafer for position 19 is significant with position 28.
