Answer:
Step-by-step explanation:
Given the sample data
Pre-test... 12 14 11 12 13
Post-Test 15 17 11 13 12
The mean of pre-test
x = ΣX / n
x = (12+14+11+12+13) / 5
x = 12.4
The standard deviation of pre-test
S.D = √Σ(X-x)² / n
S.D = √[(12-12.4)²+(14-12.4)²+(11-12.4)²+(12-12.4)²+(13-12.4)² / 5]
S.D = √(5.2 / 5)
S.D = 1.02.
The mean of post-test
x' = ΣX / n
x' = (15+17+11+13+12) / 5
x' = 13.6
The standard deviation of post-test
S.D' = √Σ(X-x)² / n
S.D' = √[(15-13.6)²+(17-13.6)²+(11-13.6)²+(13-13.6)²+(12-13.6)² / 5]
S.D = √(23.2 / 5)
S.D = 2.15
Test value
t = (sample difference − hypothesized difference) / standard error of the difference
t = [(x-x') - (μ- μ')] / (S.D / n — S.D'/n)
t = (12.4-13.6) - (μ-μ')/ (1.02/5 - 2.15/5)
-1.5 = -1.2 - (μ-μ') / -0.226
-1.5 × -0.226 = -1.2 -(μ-μ')
0.339 = -1.2 - (μ-μ')
(μ-μ') = -1.2 -0.339
μ-μ' = -1.539
Then, μ ≠ μ'
We can calculate our P-value using table.
This is a two-sided test, so the P-value is the combined area in both scores.
The p-value is 0.172
The p value > 0.1
Answer: x = 13
Step-by-step explanation:
They are equal so set them equal and solve for x.
(4x+1) = 53
4x = 52
x = 13
Answer:
Step-by-step explanation:
Given is an algebraic polynomial of degree 5.
Here leading term is p=3 and constant term is q =12
Factors of p are ±1,±2,±3
Factors of q are 
Possible forms of p/q will be the same for any other polynomial of degree 5 with leading term =3 and constant term = 12
Hence any other polynomial
will have same possible zeroes of p/q, when a,b,c are rational.
Hence any polynomial of this type would have the same possible rational roots.
L=5w
96 =2(5w+w)
96=2(6w)
w is 8
So l=5w
D. (y-3)
I'm assuming that they want you to find the factors of the quadratic expression, which are (y-5)(y-3). (y-5) isn't up there, so (y-3) is the only solution that's really possible.
