Answer:
We accept H₀ we don´t have enough evidence to support that the mean thickness is greater than 41 mm
Step-by-step explanation:
Sample Information:
Results:
41.8
40.9
42.1
41.2
40.5
41.1
42.6
40.6
From the table we get:
sample mean : x = 41.35
sample standard deviation s = 0.698
Hypothesis Test:
Null Hypothesis H₀ x = 41
Alternative Hypothesis Hₐ x > 41
The test is a one-tail test
If significance level is 0.01 and n = 8 we need to use t-student distribution
From t-table α = 0.01 and degree of freedom df = n - 1 df = 8 - 1
df = 7 t(c) = 2.998
To calculate t(s) = ( x - 41 ) / s/√n
t(s) = ( 41.35 - 41 ) / 0.698/√8
t(s) = 0.35 * 2.83/ 0.698
t(s) = 1.419
Comparing t(s) and t(c)
t(s) < t(c)
t(s) is in the acceptance region we accept H₀
Answer:
5/6 liters.
Step-by-step explanation:
Given that:
Amount of juice in a punch bowl = 2 * 1/2 liters
= 5/2 liters.
Part of juice drank by Liam from the bowl = 1/3
Amount of juice drank by Liam =
= 5/2 * 1/3 = 5\6 liters.
Answer:
here my answer
Step-by-step explanation:
5+1=6
7-1=6
3x2=6
hope it helps :)
Domain: (-♾,2)U(2,♾), {x | x ≠2}
Range: (-♾,3) U(3,♾), {y | y ≠ 3}
Sorry that’s all I can help with
