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Lostsunrise [7]
3 years ago
11

Statistics - Please Help!!! The question is in the screenshot.

Mathematics
1 answer:
arsen [322]3 years ago
4 0

Answer: See the table below (attached image)

  • SSb = 2345
  • SSt = 5639
  • DFb = 5
  • DFw = 54
  • MSb = 469

=============================================================

Explanation:

I'll start at the right side of the table and work my way toward the left side.

I recommend you have a one way ANOVA formula sheet handy. Such a formula sheet should be located somewhere in the ANOVA chapter of your stats textbook, or toward the appendix section. There are various free online resources to search out as well.

On that formula sheet you should find that

(MSb)/(MSw) = F

where MSb and MSw are the mean squares between and within respectively.

We can find the value of MSb like so

(MSb)/(MSw) = F

MSb = F*MSw

MSb = 7.682*61

MSb = 468.602

This rounds to 469, so we'll go with MSb = 469. The reason I'm rounding is that I've done a similar problem in the past, and the computer system wanted me to round to the nearest whole number. Also, note how every value of the table given to us consists of whole numbers. Of course, your teacher may want you to keep the decimal value instead. So I would ask for clarification if possible.

Since we're going with MSb = 469, this means 469 goes in the last box of row 1.

-----------------------------------------------

Now go back to your reference sheet to locate this formula

MSw = (SSw)/(DFw)

Rearrange this formula to get

DFw = (SSw)/(MSw)

Now plug in the values we have from the table to find,

DFw = (SSw)/(MSw)

DFw = (3294)/(61)

DFw = 54  which goes in the box found in the second row.

-----------------------------------------------

Now recall that

DFb + DFw = df

where DFb is the degrees of freedom between and df is the total degrees of freedom, which in this case is 59.

Using the table and the value we found earlier, we can see that,

DFb = df - DFw

DFb = 59 - 54

DFb = 5   this goes in the second box of row one (ie above the 54).

-----------------------------------------------

Also on your formula sheet, you should see something like

MSb = (SSb)/(DFb)

where each lowercase 'b' stands for "between", and the other abbreviations are what were discussed earlier.

That formula rearranges into

SSb = (DFb)*(MSb)

Then we can use the values we calculated earlier to find

SSb = (DFb)*(MSb)

SSb = (5)*(469)

SSb = 2345  this goes in the first box of row one.

----------------------------------------------

Lastly, computing the total sum of squares (SSt) will simply involve us adding the two SSb and SSw values to get

SSt = SSb + SSw

SSt = 2345 + 3294

SSt = 5639 this goes in the box of row three.

----------------------------------------------

To summarize, we have these values to fill into the table

  • SSb = 2345
  • SSt = 5639
  • DFb = 5
  • DFw = 54
  • MSb = 469

where we have these abbreviations

  • SS = sum of squares
  • MS = mean square
  • DF = degrees of freedom
  • b = between
  • w = within

Check out the diagram below for the filled in values.

To summarize the formulas used, we basically have the template of

MS = (SS)/(DF)

for both between and within. And we also have the formula F = (MSb)/(MSw). So effectively, the values are found through various division formulas or rearrangements of such. I'm referring to working along any given row. For any given column, we add up the values to get the totals, though that of course does not apply to MS or F.

Admittedly, ANOVA tables and concepts can be confusing because there are a lot of variables and steps going on. But once you get used to it, it's not too bad if you can see how the patterns are emerging (the division and additions I was referring to earlier).

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Option 4

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Here, b_1 and b_2 are the bases and 'h' is the height between the bases.

Area of the composite figure = Area of two trapezoids given

                                                = \frac{1}{2}(12+12)14+\frac{1}{2}(25+31)11

                                                = \frac{1}{2}(24)(14)+\frac{1}{2}(56)(11)

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3 0
3 years ago
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What do you do to the equation y = x to make its graph move up on the y-axis?
densk [106]

Recall that in Linear Functions, we wrote the equation for a linear function from a graph. Now we can extend what we know about graphing linear functions to analyze graphs a little more closely. Begin by taking a look at Figure 8. We can see right away that the graph crosses the y-axis at the point (0, 4) so this is the y-intercept.

Then we can calculate the slope by finding the rise and run. We can choose any two points, but let’s look at the point (–2, 0). To get from this point to the y-intercept, we must move up 4 units (rise) and to the right 2 units (run). So the slope must be

\displaystyle m=\frac{\text{rise}}{\text{run}}=\frac{4}{2}=2m=

​run

​

​rise

​​ =

​2

​

​4

​​ =2

Substituting the slope and y-intercept into the slope-intercept form of a line gives

\displaystyle y=2x+4y=2x+4

HOW TO: GIVEN A GRAPH OF LINEAR FUNCTION, FIND THE EQUATION TO DESCRIBE THE FUNCTION.

Identify the y-intercept of an equation.

Choose two points to determine the slope.

Substitute the y-intercept and slope into the slope-intercept form of a line.

EXAMPLE 4: MATCHING LINEAR FUNCTIONS TO THEIR GRAPHS

Match each equation of the linear functions with one of the lines in Figure 9.

\displaystyle f\left(x\right)=2x+3f(x)=2x+3

\displaystyle g\left(x\right)=2x - 3g(x)=2x−3

\displaystyle h\left(x\right)=-2x+3h(x)=−2x+3

\displaystyle j\left(x\right)=\frac{1}{2}x+3j(x)=

​2

​

​1

​​ x+3

Graph of three lines, line 1) passes through (0,3) and (-2, -1), line 2) passes through (0,3) and (-6,0), line 3) passes through (0,-3) and (2,1)

Figure 9

SOLUTION

Analyze the information for each function.

This function has a slope of 2 and a y-intercept of 3. It must pass through the point (0, 3) and slant upward from left to right. We can use two points to find the slope, or we can compare it with the other functions listed. Function g has the same slope, but a different y-intercept. Lines I and III have the same slant because they have the same slope. Line III does not pass through (0, 3) so f must be represented by line I.

This function also has a slope of 2, but a y-intercept of –3. It must pass through the point (0, –3) and slant upward from left to right. It must be represented by line III.

This function has a slope of –2 and a y-intercept of 3. This is the only function listed with a negative slope, so it must be represented by line IV because it slants downward from left to right.

This function has a slope of \displaystyle \frac{1}{2}

​2

​

​1

​​  and a y-intercept of 3. It must pass through the point (0, 3) and slant upward from left to right. Lines I and II pass through (0, 3), but the slope of j is less than the slope of f so the line for j must be flatter. This function is represented by Line II.

Now we can re-label the lines as in Figure 10.

Figure 10

Finding the x-intercept of a Line

So far, we have been finding the y-intercepts of a function: the point at which the graph of the function crosses the y-axis. A function may also have an x-intercept, which is the x-coordinate of the point where the graph of the function crosses the x-axis. In other words, it is the input value when the output value is zero.

To find the x-intercept, set a function f(x) equal to zero and solve for the value of x. For example, consider the function shown.

\displaystyle f\left(x\right)=3x - 6f(x)=3x−6

Set the function equal to 0 and solve for x.

⎧

⎪

⎪

⎨

⎪

⎪

⎩

0

=

3

x

−

6

6

=

3

x

2

=

x

x

=

2

The graph of the function crosses the x-axis at the point (2, 0).

Q & A

Do all linear functions have x-intercepts?

No. However, linear functions of the form y = c, where c is a nonzero real number are the only examples of linear functions with no x-intercept. For example, y = 5 is a horizontal line 5 units above the x-axis. This function has no x-intercepts.

Graph of y = 5.

Figure 11

A GENERAL NOTE: X-INTERCEPT

The x-intercept of the function is value of x when f(x) = 0. It can be solved by the equation 0 = mx + b.

EXAMPLE 5: FINDING AN X-INTERCEPT

Find the x-intercept of \displaystyle f\left(x\right)=\frac{1}{2}x - 3f(x)=

​2

​

​1

​​ x−3.

SOLUTION

Set the function equal to zero to solve for x.

\displaystyle \begin{cases}0=\frac{1}{2}x - 3\\ 3=\frac{1}{2}x\\ 6=x\\ x=6\end{cases}

​⎩

​⎪

​⎪

​⎪

​⎪

​⎪

​⎨

​⎪

​⎪

​⎪

​⎪

​⎪

​⎧

​​  

​0=

​2

​

​1

​​ x−3

​3=

​2

​

​1

​​ x

​6=x

​x=6

​​  

The graph crosses the x-axis at the point (6, 0).

Analysis of the Solution

A graph of the function is shown in Figure 12. We can see that the x-intercept is (6, 0) as we expected.

Figure 12. The graph of the linear function \displaystyle f\left(x\right)=\frac{1}{2}x - 3f(x)=

​2

​

​1

5 0
2 years ago
Which statements are true based on the diagram? Select three options.
Aneli [31]

Answer:

On Edge its:

Option A: Points N and K are on plane A and plane S.

Option C: Point P is the intersection of line n and line g.

Option D; Point's M, P, and Q are noncollinear.

Are Correct.

Step-by-step explanation:

5 0
3 years ago
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