Plotting graphs

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Probability

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I've put this post in this section because I'm unsure where to post it. I thought about calculus but then thought that section is too advanced for what I'm learning.

I'm asked to construct a table of values for a formula f = 1.8c+32. The table I've calculated is;

c = (- 20), (0), (20), (40) This is in degrees C

f = (-4), (32), (68), (104)

I now know that (-20) degrees C is equivalent to (-4) degrees F etc.

I was then asked to plot a graph using graph paper if I had any, so not having any I used the Excell spread sheet.

I was then asked to use the graph to make the following conversions.

10 degrees C to degrees F
-10 degrees C to degrees F
-10 degrees F to degrees C
0 degrees F to 0 degrees C

Looking at the table of data above and using the formula above I can work out the conversions. So lets assume for a minute that I don't have access to the formula and I don't know how to convert these temperatures. Can somebody please show me how using this graph I'm supposed to know what degrees F is equal to say 10 degrees C?Capture of temperature graph.PNG

Now I look to the graph to convert 10 degrees C to degrees F. What I see here is a very inaccurate graph. If I now draw a vertical line from the (-c) axis to the blue line, and then draw another line from the (-f) axis to meet the vertical line, then based on the graph I could say that the equivalent temperature in degrees F is approx 50. This is completely wrong whereas I know the temperature should be 44 degrees F. Graphically examining the data its not reliable.

Without having to hand draw every graph there must be a better way using software to ensure that graphs produced are accurate for their intended purposes?
 
Probability said:
I've put this post in this section because I'm unsure where to post it. I thought about calculus but then thought that section is too advanced for what I'm learning.

I'm asked to construct a table of values for a formula f = 1.8c+32. The table I've calculated is;

c = (- 20), (0), (20), (40) This is in degrees C

f = (-4), (32), (68), (104)

I now know that (-20) degrees C is equivalent to (-4) degrees F etc.

I was then asked to plot a graph using graph paper if I had any, so not having any I used the Excell spread sheet.

I was then asked to use the graph to make the following conversions.

10 degrees C to degrees F
-10 degrees C to degrees F
-10 degrees F to degrees C
0 degrees F to 0 degrees C

Looking at the table of data above and using the formula above I can work out the conversions. So lets assume for a minute that I don't have access to the formula and I don't know how to convert these temperatures. Can somebody please show me how using this graph I'm supposed to know what degrees F is equal to say 10 degrees C?View attachment 19474

Now I look to the graph to convert 10 degrees C to degrees F. What I see here is a very inaccurate graph. If I now draw a vertical line from the (-c) axis to the blue line, and then draw another line from the (-f) axis to meet the vertical line, then based on the graph I could say that the equivalent temperature in degrees F is approx 50. This is completely wrong whereas I know the temperature should be 44 degrees F. Graphically examining the data its not reliable.

Without having to hand draw every graph there must be a better way using software to ensure that graphs produced are accurate for their intended purposes?
Click to expand...
Can you add more intermediate values on the axes and display vertical grid lines?
 
For some unknown reason to me the software seems to regulate the values itself. I can put a grid on the graph yes which would help, again I forgot about that idea having been previously advised about it on here.

Maybe something you could help with regarding values used. If say I have some data values between say (-4) and (104) as in this graph, then is there a mathematical way of finding out quickly how to find the intermediate values between (-4) and (104) so that I can enter and plot them accurately?
 
Probability said:
I've put this post in this section because I'm unsure where to post it. I thought about calculus but then thought that section is too advanced for what I'm learning.

I'm asked to construct a table of values for a formula f = 1.8c+32. The table I've calculated is;

c = (- 20), (0), (20), (40) This is in degrees C

f = (-4), (32), (68), (104)

I now know that (-20) degrees C is equivalent to (-4) degrees F etc.

I was then asked to plot a graph using graph paper if I had any, so not having any I used the Excell spread sheet.

I was then asked to use the graph to make the following conversions.

10 degrees C to degrees F
-10 degrees C to degrees F
-10 degrees F to degrees C
0 degrees F to 0 degrees C

Looking at the table of data above and using the formula above I can work out the conversions. So lets assume for a minute that I don't have access to the formula and I don't know how to convert these temperatures. Can somebody please show me how using this graph I'm supposed to know what degrees F is equal to say 10 degrees C?View attachment 19474

Now I look to the graph to convert 10 degrees C to degrees F. What I see here is a very inaccurate graph. If I now draw a vertical line from the (-c) axis to the blue line, and then draw another line from the (-f) axis to meet the vertical line, then based on the graph I could say that the equivalent temperature in degrees F is approx 50. This is completely wrong whereas I know the temperature should be 44 degrees F. Graphically examining the data its not reliable.

Without having to hand draw every graph there must be a better way using software to ensure that graphs produced are accurate for their intended purposes?
Click to expand...
....I know the temperature should be 44 degrees F....

How did you acquire such knowledge?? That is incorrect!

10 oC is exactly equal to 50 oF!

Check it with the given equation:

f = 1.8*c+32
 
Probability said:
For some unknown reason to me the software seems to regulate the values itself. I can put a grid on the graph yes which would help, again I forgot about that idea having been previously advised about it on here.

Maybe something you could help with regarding values used. If say I have some data values between say (-4) and (104) as in this graph, then is there a mathematical way of finding out quickly how to find the intermediate values between (-4) and (104) so that I can enter and plot them accurately?
Click to expand...
Regardless of software used, you can make your graph as precise as you want. But it can't be as precise as a formula. I can take your graph and ask for y that corresponds to an x value between your grid lines. So any graph should come with a disclaimer - estimate at your own risk!
Regarding "intermediate values between (-4) and (104)", please clarify your question. Do you to find how many intermediate values you need so that they are at least X units apart?
 
Subhotosh Khan said:
....I know the temperature should be 44 degrees F....

How did you acquire such knowledge?? That is incorrect!

10 oC is exactly equal to 50 oF!

Check it with the given equation:

f = 1.8*c+32
Click to expand...
Your correct I've redone the calculations.
 
lev888 said:
Regardless of software used, you can make your graph as precise as you want. But it can't be as precise as a formula. I can take your graph and ask for y that corresponds to an x value between your grid lines. So any graph should come with a disclaimer - estimate at your own risk!
Regarding "intermediate values between (-4) and (104)", please clarify your question. Do you to find how many intermediate values you need so that they are at least X units apart?
Click to expand...
Yes.
 
lev888 said:
You are planting a line of trees 10m long. You want them 2m apart. How many trees will you plant?
Click to expand...
That seems a little too simplistic to be honest. (10)/(2) = 5

If I have data such as (-4) to (104) I think you are saying I should just divide it out and say the solution is (26).

So 26 - (-4) = 30 + 26= 56, 56 + 26 = 82, 82 + 26 = 108

OK I can see that would work, but the arithmetic is long and needs a formula, so a little guidance on that would be appreciated.

So mathematically for this example 26*4 = 104

But a formula would be good!
 
Probability said:
That seems a little too simplistic to be honest. (10)/(2) = 5

If I have data such as (-4) to (104) I think you are saying I should just divide it out and say the solution is (26).

So 26 - (-4) = 30 + 26= 56, 56 + 26 = 82, 82 + 26 = 108

OK I can see that would work, but the arithmetic is long and needs a formula, so a little guidance on that would be appreciated.

So mathematically for this example 26*4 = 104

But a formula would be good!
Click to expand...
Well, 5 is incorrect. Draw them.
 
Probability said:
I'm not sure what you mean!
Click to expand...
Your original subject line was "plotting graphs".

You used technology and plotted the graph of a given equation. However, you miscalculated the given equation (while checking) and thought that the plot was incorrect.

Then your question (OP) was:

Without having to hand draw every graph there must be a better way using software to ensure that graphs produced are accurate for their intended purposes?​

You found your mistake (in checking) and corrected it.

You now know that the plot is correct and know that excel did produce a graph that is accurate for its intended purposes. Your initial concern has been ameliorated.

So - what is your question/concern now?
 
Subhotosh Khan said:
Your original subject line was "plotting graphs".

You used technology and plotted the graph of a given equation. However, you miscalculated the given equation (while checking) and thought that the plot was incorrect.

Then your question (OP) was:

Without having to hand draw every graph there must be a better way using software to ensure that graphs produced are accurate for their intended purposes?​

You found your mistake (in checking) and corrected it.

You now know that the plot is correct and know that excel did produce a graph that is accurate for its intended purposes. Your initial concern has been ameliorated.

So - what is your question/concern now?
Click to expand...
The same. When I use Excell the software somehow calculates the scaling for the axes. I can enter my data for the scales but Excell changes it, so my concern is the same as before, how do I work out the scaling of the axes without pages and pages of trial and error?

The graph from Excell is correct but on a small laptop screen reading the graticule screen of the graph it can be seen as inaccurate but I accept mathematically its OK now.
 
lev888 said:
You are planting a line of trees 10m long. You want them 2m apart. How many trees will you plant?
Click to expand...
I have a line of trees 10m long. All I know here is that I have 10m length trees. Place them 2m apart.
When I place them 2m apart I can only do this until I reach 10m.
(2 x 5) = 10
lev888 said:
Well, 5 is incorrect. Draw them.
Click to expand...
I can't draw something I don't understand?
 
Probability said:
I have a line of trees 10m long. All I know here is that I have 10m length trees. Place them 2m apart.
When I place them 2m apart I can only do this until I reach 10m.
(2 x 5) = 10

I can't draw something I don't understand?
Click to expand...
Draw 5 trees 2m apart:

Tree__2m__Tree__2m__Tree__2m__Tree__2m__Tree

What's the total length?
 
lev888 said:
Draw 5 trees 2m apart:

Tree__2m__Tree__2m__Tree__2m__Tree__2m__Tree

What's the total length?
Click to expand...
Look at what you have written above. I can add the 2's together but it makes little sense in the context of my question..

Draw 5 trees 2m apart. We are talking about 5 trees. Tree (1)_2m, Tree (2)_2m, Tree (3)_2m, Tree (4)_2m, Tree (5)_2m = 2 + 2 + 2 + 2 + 2 = 10

It's too easy to misinterpret what you mean when we don't keep the discussion in context with the problem in hand. I get it that there is no 6th tree.

Now if we refer back to my discussion about scaling a graph, would it not be much more accurate to just say something along the lines of;

Look at the data you have calculated in your table. (-4) to (104). Add this data; 104 - (-4) = 108

108 is not really a good number so I can round up to either 110 or even 120 as the computer software did.

Now I should be looking at the grids of the graph paper. Look how many divisions there are in the graph paper. My software shows 6 horizontal divisions in one column on the graph. (120)/(6) = 20. This means that at each square box of the graph paper I can write; 0, 20, 40, 60, 80, 100, 120. I can then do likewise with the x-axis and hence I have calculated by scaling correctly.

When scaling graphs on graph paper the starting point should therefore be look how many squares are in each square of the graph paper, its either squared paper or graph paper. Typical numbers to use for scaling could be 2, 5, 8, 10 etc.

Lets please try and keep with the context of the subject then the learner gains an understanding.
 
Probability said:
Look at what you have written above. I can add the 2's together but it makes little sense in the context of my question..

Draw 5 trees 2m apart. We are talking about 5 trees. Tree (1)_2m, Tree (2)_2m, Tree (3)_2m, Tree (4)_2m, Tree (5)_2m = 2 + 2 + 2 + 2 + 2 = 10

It's too easy to misinterpret what you mean when we don't keep the discussion in context with the problem in hand. I get it that there is no 6th tree.

Now if we refer back to my discussion about scaling a graph, would it not be much more accurate to just say something along the lines of;

Look at the data you have calculated in your table. (-4) to (104). Add this data; 104 - (-4) = 108

108 is not really a good number so I can round up to either 110 or even 120 as the computer software did.

Now I should be looking at the grids of the graph paper. Look how many divisions there are in the graph paper. My software shows 6 horizontal divisions in one column on the graph. (120)/(6) = 20. This means that at each square box of the graph paper I can write; 0, 20, 40, 60, 80, 100, 120. I can then do likewise with the x-axis and hence I have calculated by scaling correctly.

When scaling graphs on graph paper the starting point should therefore be look how many squares are in each square of the graph paper, its either squared paper or graph paper. Typical numbers to use for scaling could be 2, 5, 8, 10 etc.

Lets please try and keep with the context of the subject then the learner gains an understanding.
Click to expand...
My point with the trees was that it's important to differentiate between trees and spaces between trees. You want 10m total length with 2m spacing. 10/2 = 5. That's the number of spaces, not trees. How many trees would give us 5 spaces? 6 trees.
Same with values on graph axes. Given total distance of 108 and spacing, let's say 4, how many lines do we need on the graph? 108/4 = 27 spaces or intervals. How many lines? 28. If we are talking about intermediate values, then it's 28-2=26.
 
Probability said:
The same. When I use Excell the software somehow calculates the scaling for the axes. I can enter my data for the scales but Excell changes it, so my concern is the same as before, how do I work out the scaling of the axes without pages and pages of trial and error?

The graph from Excell is correct but on a small laptop screen reading the graticule screen of the graph it can be seen as inaccurate but I accept mathematically its OK now.
Click to expand...
Read and Thou shall know.

Open excel users manual - even the help line in your computer - and that will explain

To change property of x or y axis in your chart - click on the axis you want to control.

A side window will open up with plethora of choices available for you.

Go at it......

By the way, if you read carefully, the laptop resolution is generally good enough for "government work".
 
Subhotosh Khan said:
Read and Thou shall know.

Open excel users manual - even the help line in your computer - and that will explain

To change property of x or y axis in your chart - click on the axis you want to control.

A side window will open up with plethora of choices available for you.

Go at it......

By the way, if you read carefully, the laptop resolution is generally good enough for "government work".
Click to expand...
I must have a very old version of excel as the guidance provided is not advising what I required. Researching the software I must have a very basic package because recent experiences using the software and trying to plot this graph;

Line A and B (-0.6, 2) (0.4,1)
Line A and D (-0.6, 2) (0,2, 5)
Line B and C (0.4, 1) (-0.4, -2)

Shows that the version of excel I have cannot plot it, and when I looked further into why it is because excel cannot do free form graphs, the templates in the software decide what type of graph can be drawn, which I can't do the above on my version of excel!!

I've completed the graphs by hand and concluded that their gradients are (- 1), and [MATH]\frac{15}{4}[/MATH]
I just wish I could have used the software to print the hard copies.
 
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