Answer: The data is a random sample from the population of interest.
And
Individual observations can be considered independent.
Step-by-step explanation:
On khan academy
1) acceleration
2) speed
3) velocity
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
This whole ugly thing is just 6 numbers that are all multiplied.
Here they are:
(-7) · (x⁶) · (y⁻³) · (5) · (x⁻¹) · (y) .
To help you see what's going on, I'm going to rearrange them
and write them in a different order. (You'll remember that when
you multiply, the order of the numbers doesn't matter.)
(-7)·(5) · (x⁶)·(x⁻¹) · (y⁻³)·(y) .
I'll bet you can see it already. It's really starting to fall apart.
Remember that when you have to multiply the same base to
different powers, you just add the powers.
So here's the multiplication:
(-7)·(5) · (x⁶)·(x⁻¹) · (y⁻³)·(y) .
| | |
-35 · x⁵ · y⁻²
-35 x⁵ y⁻²
or you could write it as -35x⁵ / y² which is the same thing.
Answer:
yes its linear
Step-by-step explanation:
