Answer:
Teachers in his school district have less than 5 yearſ of experience on average is true.
Step-by-step explanation:
Sample mean =
Sample standard deviation = s = 2
We are supposed to to test Ha :u= 5 versus H : u < 5 using a sample of 25 teachers
n = 25
Since n<30 and population standard deviation is unknown
So, we will use t test
Formula :
t=-2.5
Degree of freedom = n-1 = 25-1 = 24
Refer the t table
t critical > t calculated
So, We are failed to reject null hypothesis
So, teachers in his school district have less than 5 yearſ of experience on average is true.
Turn the 50 into a fraction (50/1) then cross multiply the fractions after that turn it into a equivalent fraction and you you get your answer
(I think)
Answer:
n=5
Step-by-step explanation:
Solve and isolate n.
8 x n = 5 x 8
8 x n = 40 Divide out the 8
n=5
One way: log(A)-log(B) = log(A/B) right?
So, log( x^2-9/(x+3))
Now x^2-9 = (x+3)*(x-3), and x^2-9/(x+3) = x-3,
so log(x^2-9)-log(x+3) = log(x-3),
Indeed for all x but for x = -3, but probably no one asks you about that detail?