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
Tahmid has 96 figures, Sandeep has 1/3 of it so
96(1/3) = 96/3 = 32
Sandeep has 32 figures
Tahmid 96 ; Sandeep 32
How many must tahmid give so they have the same figures?
First take the difference of the two so we know how many tahmid can give.
96-32 = 64
now they each have 32 and 64 give or keep. for each one tahmid gives, he must keep one himself so to match the number they both own. so we divide the number of 2, one part to give, the other part to keep himself.
64/2 = 32
So, Tahmid must give Sandeep 32 figures to have equal number of figures of 64 each.
Answer:
x = -3
Step-by-step explanation:
Step 1: Simplify signs
3x + 7 = -2
Step 2: Subtract 7 on both sides
3x = -9
Step 3: Divide both sides by 3
x = -3
Answer:
f(x) = x + 7
or
y=x+7
Step-by-step explanation:
The term of the sequence is eleven 11
