4. <br> HELP ME PLS W THE CORRECT ANSWER

For an analysis of variance, if the between-group estimate of the population variance is 30, and the within-group estimate is 25

borishaifa [10]

Answer:

1.2

Step-by-step explanation:

We have given the variance between - group estimate is = 30

And the variance within - group estimate is =25

Now we have to calculate the F ratio

F ratio is defined as the ratio of variance between the group and variance within the group

So F ratio $=\frac{30}{25}=1.2$

So for the given variance F ratio will be 1.2

8 0

3 years ago

3. If your siice a rectangular pyramid parallel to it base, what will be the shape of the cross section?

zubka84 [21]

Answer:

<em><u>1</u></em><em><u>.</u></em><em><u> </u></em><em><u>A cross section is the new face you see when you slice through a three-dimensional figure. For example, if you slice a rectangular pyramid parallel to the base, you get a smaller rectangle as the cross section.</u></em>

5 0

2 years ago

Factorise 2a – 4a3 + 6abc

Viefleur [7K]

The answer for this will be

2a(1-2a ²+3bc)

7 0

3 years ago

Line z has a slope of 2.5 and goes through point (4, 13). Which equation best represents line z?

AnnyKZ [126]

Answer:

y - 13 = (5/2)(x - 4)

Step-by-step explanation:

Here we know the slope and one point on the line. Use the point-slope formula:

y - k = m(x - h).

Substituting 13 for k, 4 for x and 2.5 for m, we get:

y - 13 = (5/2)(x - 4)

3 0

3 years ago

Let X be a set of size 20 and A CX be of size 10. (a) How many sets B are there that satisfy A Ç B Ç X? (b) How many sets B are

Svetlanka [38]

Answer:

(a) Number of sets B given that

A⊆B⊆C: 2¹⁰. (That is: A is a subset of B, B is a subset of C. B might be equal to C)
A⊂B⊂C: 2¹⁰ - 2. (That is: A is a proper subset of B, B is a proper subset of C. B≠C)

(b) Number of sets B given that set A and set B are disjoint, and that set B is a subset of set X: 2²⁰ - 2¹⁰.

Step-by-step explanation:

Let $x_1, x_2, \cdots, x_{20}$ denote the 20 elements of set X.

Let $x_1, x_2, \cdots, x_{10}$ denote elements of set X that are also part of set A.

For set A to be a subset of set B, each element in set A must also be present in set B. In other words, set B should also contain $x_1, x_2, \cdots, x_{10}$ .

For set B to be a subset of set C, all elements of set B also need to be in set C. In other words, all the elements of set B should come from $x_1, x_2, \cdots, x_{20}$ .

$\begin{array}{c|cccccccc}\text{Members of X} & x_1 & x_2 & \cdots & x_{10} & x_{11} & \cdots & x_{20}\\[0.5em]\displaystyle\text{Member of}\atop\displaystyle\text{Set A?} & \text{Yes}&\text{Yes}&\cdots &\text{Yes}& \text{No} & \cdots & \text{No}\\[0.5em]\displaystyle\text{Member of}\atop\displaystyle\text{Set B?}& \text{Yes}&\text{Yes}&\cdots &\text{Yes}& \text{Maybe} & \cdots & \text{Maybe}\end{array}$ .

For each element that might be in set B, there are two possibilities: either the element is in set B or it is not in set B. There are ten such elements. There are thus $2^{10} = 1024$ possibilities for set B.

In case the question connected set A and B, and set B and C using the symbol ⊂ (proper subset of) instead of ⊆, A ≠ B and B ≠ C. Two possibilities will need to be eliminated: B contains all ten "maybe" elements or B contains none of the ten "maybe" elements. That leaves $2^{10} -2 = 1024 - 2 = 1022$ possibilities.

Set A and set B are disjoint if none of the elements in set A are also in set B, and none of the elements in set B are in set A.

Start by considering the case when set A and set B are indeed disjoint.

$\begin{array}{c|cccccccc}\text{Members of X} & x_1 & x_2 & \cdots & x_{10} & x_{11} & \cdots & x_{20}\\[0.5em]\displaystyle\text{Member of}\atop\displaystyle\text{Set A?} & \text{Yes}&\text{Yes}&\cdots &\text{Yes}& \text{No} & \cdots & \text{No}\\[0.5em]\displaystyle\text{Member of}\atop\displaystyle\text{Set B?}& \text{No}&\text{No}&\cdots &\text{No}& \text{Maybe} & \cdots & \text{Maybe}\end{array}$ .

Set B might be an empty set. Once again, for each element that might be in set B, there are two possibilities: either the element is in set B or it is not in set B. There are ten such elements. There are thus $2^{10} = 1024$ possibilities for a set B that is disjoint with set A.

There are 20 elements in X so that's $2^{20} = 1048576$ possibilities for B ⊆ X if there's no restriction on B. However, since B cannot be disjoint with set A, there's only $2^{20} - 2^{10}$ possibilities left.

5 0

3 years ago

4. HELP ME PLS W THE CORRECT ANSWER