Answer:
y = 4,-6
Step-by-step explanation:
0 property states that 0 times any number is 0
applying that here, that means y+6 = 0 or y-4 = 0
0(y-4) = 0; (y+6)(0) = 0
y+ 6 = 0
y = -6
y-4 = 0
y = 4
You do 9x27=243 then take 2x31=62 add 243+62=305 then take 305 and subtract it from 28 305-28=277
so if i did it right you should get n=277
Answer:
31/40
Step-by-step explanation:
It cannot be reduced anymore. It is already in simplest form.
Intersection of the first two lines:
Multiply the first equation by 4 and the second by 5:
Subtract the two equations:
Plug this value for y in one of the equation, for example the first:
So, the first point of intersection is 
We can find the intersection of the other two lines in the same way: we start with
Use the fact that x and y are the same to rewrite the second equation as
And since x and y are the same, the second point is 
So, we're looking for a line passing through and . We may use the formula to find the equation of a line knowing two of its points, but in this case it is very clear that both points have the same coordinates, so the line must be 
In the attached figure, line 
is light green, line 
is dark green, and their intersection is point A.
Simiarly, line 
is red, line 
is orange, and their intersection is B.
As you can see, the line connecting A and B is the red line itself.
Answer:
D. There were no significant effects.
Step-by-step explanation:
The table below shows the representation of the significance level using the two-way between subjects ANOVA.
Source of Variation SS df MS F P-value
Factor A 10 1 10 0.21 0.660
Factor B 50 2 25 0.52 0.6235
A × B 40 2 20 0.42 0.6783
Error 240 5 48 - -
Total 340 10 - - -
From the table above , the SS(B) is determined as follows:
SS(B) = SS(Total)-SS(Error-(A×B)-A)
= 340-(240-40-10)
= 50
A researcher computes the following 2 x 3 between-subjects ANOVA;
k=2
n=3
N(total) = no of participants observed in each group =11
df for Factor A= (k-1)
=(2-1)
=1
df for Factor B = (n-1)
=(3-1)
=2
df for A × B
= 2 × 1
= 2
df factor for total
=(N-1)
=11-1
=10
MS = SS/df
Thus, from the table, the P-Value for all data is greater than 0.05, therefore we fail to reject H₀.
