Answer:
Step-by-step explain
Find the horizontal asymptote for f(x)=(3x^2-1)/(2x-1) :
A rational function will have a horizontal asymptote of y=0 if the degree of the numerator is less than the degree of the denominator. It will have a horizontal asymptote of y=a_n/b_n if the degree of the numerator is the same as the degree of the denominator (where a_n,b_n are the leading coefficients of the numerator and denominator respectively when both are in standard form.)
If a rational function has a numerator of greater degree than the denominator, there will be no horizontal asymptote. However, if the degrees are 1 apart, there will be an oblique (slant) asymptote.
For the given function, there is no horizontal asymptote.
We can find the slant asymptote by using long division:
(3x^2-1)/(2x-1)=(2x-1)(3/2x+3/4-(1/4)/(2x-1))
The slant asymptote is y=3/2x+3/4
Domain restrictions are what x values make the denominator zero.
a² + 5a - 36 = (a + 9)(a - 4)
a ≠ -9 ; a ≠ 4
The table is missing in the question. The table is provided here :
Group 1 Group 2
34.86 64.14 mean
21.99 20.46 standard deviation
7 7 n
Solution :
a). The IV or independent variable = Group 1
The DV or the dependent variable = Group 2
b).
Therefore, 
t = -2.579143
Now, 
df = 7 - 1
= 6
Therefore the value of p :
= 0.020908803
The p value is 0.0209
So we reject the null hypothesis and conclude that 
Step by step explanation
4n-8=5
4n= 5+8
4n=13
4/4 n = 13/4
n= 13/4
Answer:
80/81
Step-by-step explanation:
hope this helps!
