Answer:
No ;
Step-by-step explanation:
Given ;
s2 = 10 ; n2 = 10 ; s1 = 12 ; n1 = 12
Hypothesis :
H0 : σ1² = σ2²
H1 : σ1² ≠ σ2²
The test statistic :
Ftest = s1² / s2² = 12² / 10² = 144 / 100 = 1.44
The Pvalue using the Pvalue from Ftest calculator :
Numerator df = 12 - 1 = 11
Denominator df = 10 - 1 = 9
Pvalue = 0.2969
At α = 0.1
If Pvalue > α ; Fail to reject H0
0.2969 > 0.1 ; Hence, we fail to reject the Null
Hence, we cannot conclude that variation exists
Answer:
23
Step-by-step explanation:
-2(8p+2)-3(2-7p)-2(4+2p)=0
mutiply the first bracket by -2
(-2)(8p)= -16p
(-2)(+2)= -4
mutiply the second bracket by -3
(-3)(2)= -6
(-3)(-7p)= 21p
mutiply the third bracket by -2
(-2)(4)= -8
(-2)(+2p)= -4p
-16p-4-6+21p-8-4p= 0
-16p+21p-4p-4-6-8= 0 ( combine like terms)
5p-4p-4-6-8= 0
p-4-6-8= 0
p-10-8= 0
p-18= 0
move -18 to the other side to get p by itself
sign changes from -18 to +18
p-18+18= 0+18
p= 0+18
Answer: p= 18
