50
Step-by-step explanation:
Since AE is the bisector of angle BAC,so angle BAE and Angle CAE is equal. Here Angle CAE and Angle EAC is same.
therefore,
m(Angle BAE)=m(Angle EAC)

or, 2x=40
or, x=20
Hence,
m(Angle EAC)=

=

The answer to your problem will be 1,220. hope this helps
First, I would distribute the 2 out to the (x+4).
2x+8.
Next, use the foil method to multiply together 2x+8 and (x+3).
2x^2 +14x+24.
Do the same process for the other side.
3(x+2)= 3x+6
(3x+6)(x-1)= 3x^2+3x-6
Set the remaining products of each side of the equation equal to each other.
2x^2 +14x+24=3x^2+3x-6.
Now you must cancel out one side to make the equation equal to zero. Do this by doing the inverse operation on one side of the equation (each value with their like term). I am going to subtract the left side values from the right:
This means:
3x^2 minus 2x^2 equals 1x^2 (or just x^2).
3x minus 14x equals -11x
-6 minus 24 equals -30
The equation should now look like this:
0=x^2-11x-30
Reverse the order to get zero on the right side:
x^2-11x-30=0
Hope this helps! Sorry if I made any careless mistakes (^^;)
Answer:
a) (iii) ANOVA
b) The ANOVA test is more powerful than the t test when we want to compare group of means.
Step-by-step explanation:
Previous concepts
Analysis of variance (ANOVA) "is used to analyze the differences among group means in a sample".
The sum of squares "is the sum of the square of variation, where variation is defined as the spread between each individual value and the grand mean"
If we assume that we have
groups and on each group from
we have
individuals on each group we can define the following formulas of variation:



And we have this property

Solution to the problem
Part a
(i) confidence interval
False since the confidence interval work just when we have just on parameter of interest, but for this case we have more than 1.
(ii) t-test
Can be a possibility but is not the best method since every time that we conduct a t-test we have a chance that we commit a Type I error.
(iii) ANOVA
This one is the best method when we want to compare more than 1 group of means.
(iv) Chi square
False for this case we don't want to analyze independence or goodness of fit, so this one is not the correct test.
Part b
The ANOVA test is more powerful than the t test when we want to compare group of means.