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
(11/3-7/4)/[3.4-(1/2+2/3*3/2)]
First I converted fractions to mixed numbers.
11/3-7/4 / 3.4-3/2
Then, I solve the inside of the parentheses.
23/12 / 3.4-3/2
Continue to simplify
Convert to decimals
1.91667/1.9
Hope this helps
Answer:
1) binomial
2) monomial
Step-by-step explanation:
mono-think one
bi-think two
tri-think thee
poly-1 or more
So 1 goes with binomial
and 2 goes with monomial
See if you can figure out 3 and 4
I can check them.
The cost of 1 adult ticket in the off season is 42$, but it is asking for the child ticket's price so you would do 194-42 and you get 152. Then, you would divide 152/4 and you get 38 but this is during the off-season so you would double it and get 76.
The regular price for a child is 76$
But feel free to check my work :D
Where the heck did x come from!
