Our current list has 11!/2!11!/2! arrangements which we must divide into equivalence classes just as before, only this time the classes contain arrangements where only the two As are arranged, following this logic requires us to divide by arrangement of the 2 As giving (11!/2!)/2!=11!/(2!2)(11!/2!)/2!=11!/(2!2).
Repeating the process one last time for equivalence classes for arrangements of only T's leads us to divide the list once again by 2
Answer:
The required confidence inteval = 94.9%.
Step-by-step explanation:
Confidence interval: Mean ± Margin of error
Given: A confidence interval for the true mean diameter of all oak trees in the neighbourhood is calculated to be (36.191, 42.969).
i.e. Mean + Margin of error = 42.969 (i)
Mean - Margin of error = 36.191 (ii)
Adding (i) and (ii), we get

Margin of error = 42.969-39.58 [from (i)]
= 3.389
Margin of error = 
here n= 25 
i.e.

Using excel function 1-TDIST.2T(2.054,24)
The required confidence inteval = 94.9%.
For every value tht x increases, y decreases by 3
-3x
Then to find the y intercept, look at 0's corresponding value (8)
-3x+8
Answer:
3.5
Step-by-step explanation:
you will go like sum of frequency multiplied by score divided by sum of frequency
Answer:
C. -12
Step-by-step explanation:
divide each side by 2/3
2/3x ÷ 2/3 = x
-8 ÷ 2/3 = -12
x = -12