Hi can someone help me with this table and graph stuff for algebra?

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Hi can someone help me with this table and graph stuff for algebra?

Mathematics
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I guess it could be either linear or cubic. It kind of looks like there's an inflection point between 1974 and 1975. Do you have to find the equation by hand?
it doesn't really say, but it would probably be for the best.
ok well if you have to do it by hand, it's probably linear. least square regression?

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It doesn't say specifically but that should work.
you have to change your table to do this. The y-column is fine the way it is. The x-column needs to start at 0. So let 1885 be year 0 and let x be years after 1885. So like 1917 will be 32, 1919 will be 34, etc
okay wait I think I misinterpreted what you typed. I have to change the table to find an equation?
Yeah, we're looking for an equation in y = mx + b form. If you leave the table like that it will skew the y-intercept ( I believe)
right okay I think I follow you a little.
this is making the gears in my head spin lol I know the equation has to be y=mx + b
This is one I helped someone with last night. Scroll to about the middle of the page. Do the formulas for a and b look familiar? http://openstudy.com/users/peachpi#/updates/55d7dbd9e4b02663346b9fa8
oh yeah! I've seen those before. I mean I always get the numbers mixed up. I wouldn't know where to put what.
that's fine. Just start with adjusting the years and we'll go from there
I feel like that table on that page or whatever has specific numbers for the x's and y's. I wouldn't know which ones to apply for my situation.
Your y's are going to be the stamp's costs (0.02, 0.03, 0.02, 0.03, etc.) Your x's are going to be years since 1885 (0, 32, 34, 47, etc). Subtract 1885 from each year to get these
oh that's how you got those, ok.
If you have excel, google docs, or some other spreadsheet program it will do the calculations for you and you just have to arrange the table
I had excel but the trial ended. I don't really have any way to do this. I don't know I gotta think about this.
for that girl you did a table that was x|y|x^2|xy is that what I have to do?
yeah
oh man okay x would start with 0? I think you said. I'm sorry I'm so hopeless I'm getting myself confused all over again.
or would the x and y columns look exactly like my table? and then I would multiply x by itself and then I multiply x and y?
hold up. I don't think those are going to be the right equations anyway. I think this might be exponential. It just doesn't look so on your graph because the years aren't evenly spaced. This is what I get with an evenly spaces x-axis
|dw:1440245393715:dw|
so I did my graph wrong? Or that's a new graph over my original graph?
no, your graph is wrong. You used 1 square for each difference in years, but the differences aren't all the same
so do I need to redo the years on my graph? I actually wasnt sure how to space them. I only had like 25 spaces. Yeah the years aren't spaced correctly.
what spacing would you recommend I use?
I can use the spacing you used for the y axis, but I'm not sure for the x axis.
Ok so we're going from 1885 to 2012, that's 127 years. Divide by 25 to get 5.02, so make every space 6 years to have a little extra room on the right
Don't put the actual years on the graph. so start at 0, then 6, 12, 18 etc. Then use a label that says "Years since 1885"
I'm sorry for being so confused. I'm really trying to understand this
so i didn't label the years but started with 0 and ended with 138 (that's as far as my graph will let me go)
0 represents 1885. 6 represents 1891. 12 represents 1897. Even though 1891 won't be a point on the graph, we have to show it on the x-axis to show how long it's taking the price to rise at the beginning. 1917 is 32 years after 1885. So that point will be about 5 or 6 grid lines right of the first point
okay I'm gonna need a much bigger graph
No your graph is fine. 2012 is 127 years after 1885. Your graph is showing 138 years after, so we have a little extra. That's exactly what we want
okay
Have you done the thing where you subtract 1885 from all the years yet?
is the y axis okay? I went by 2's on there?
yes the y-axis is good
I wasn't sure what you meant by that.
like this 1885 - 1885 = 0 1917-1885 = 32 1919 - 1885 = 34
oh okay well I'll do all those and get back to you okay?
ok.
almost done...
ok
ok. At the bottom of the page are the equations for curve fitting an exponential function (lowercase a and b). http://mathworld.wolfram.com/LeastSquaresFittingExponential.html
then I need to square the x's and then multiply x and y
oh man yikes okay
this one isn't linear so the equations are different. Have you seen those before? I had a hard time even finding them, so I'm wondering if you're supposed to do this using technology as opposed to hand
No I haven't really seen them before (hence my yikes lol) This is on a worksheet so I just assumed it was by hand
I feel like my heads gonna explode.
If you have or want to create a google account, google docs will calculate the curve once you enter the data in their spreadsheet
I just have a really hard time thinking they want an algebra student to do this by hand :/
I've never used spreadsheets before. How do I see the curve in the spreadsheet?
you have to enter the table and then create the chart. I can walk you through it. Search for google docs then sign in to create a spreadsheet
okay I'm on the spreadsheet (that link you gave me that you did last night since her graph thing was linear and my is expo than how I find an equation is different?
Linear is y = mx + b Exponential is \(y=ab^x\)
okay gotcha well i'm on the spread sheet now
can you explain to me why it is expo? I don't think we went over that
ok. so now make columns for x and y. Put the x column to the left and enter the numbers you got by subtracting (0, 32, etc) and another for the prices
make the correct graph first. It will be easier to understand that way. You really wouldn't know whether it was linear or exponential once you see the graph.
*until you see the graph
so in the A column put 0 32 34 47 all the way ↓ to 127 In the B column 0.02 0.03 0.02 0.03 ↓ 0.45
okay let me do all of that
I labeled the two sides of the table (years were x) (y were costs)
okay now what do I do?
ok. now highlight the cells with numbers in them and go to the INSERT menu at the top and select CHART
stupid question - how do I highlight them w/o highlight the little grey column on the left?
That's going to highlight automatically so you know which cells are selected. It's fine. Once you pick Chart A box should pop up with named Chart Editor with Recommendations, Chart Types, and Customization tabs
I don't think it matters
ok
okay I did insert and selected chart
i'm in the chart editor
One of the recommended charts should a grid showing dot. Select that one and hit INSERT at the bottom
okay
done
do you want me to screencap and upload it to make sure it looks good?
yes please
not sure if that looks right
This is mine
1 Attachment
hm seems like I missed something
For the most part it looks right. Maybe one of us made a typo when entering the points
yeah could be. I really don't think they will notice too much. it's fine
I'll leave it to you to check, but we can go on so you know how to get the equation
okay
Do you notice how it almost flat toward the left and then increases steeply toward the right? That's how I know it's exponential
right
|dw:1440249587962:dw|
The graph on the right is linear functions. They have the same slope everywher
I see
To get the equation of your graph hit the little triangle on the top right corner of the graph and select ADVANCED EDIT
okay
Then go down to where it has TRENDLINE and select EXPONENTIAL For LABEL select USE EQUATION
okay
then input?
or update I guess
oh yeah hit update
ok
do you see the equation to the right of the graph?
how do I see all of it?
you can make the chart bigger. grab one of the handles on the corner to stretch it
so that's my equation then
yeah mine is 8.66 instead of 9.538 yeah that's the equation.
can you tell me your full equation just incase mine was the one that wasn't 100% correct?
Mine was (8.661E-3)e^(0.03x)
awesome thank you
You should probably write it in decimal form though. So yours would be 0.009538e^(-0.03x)
(move the decimal point left 3 times and drop the E)
math teachers don't generally like E format
don't put it in y=ab^x form?
It is. a = 0.009538 b = e^0.03
oh okay I see didn't know if y= is relevant or not.
It should be y = 0.009538e^(0.03x) as your final equation. (the - I put there was a mistake)
okay
so how can I use that equation to predict from 2013-2022 Do I change something in it?
or bring the costs back in?
For 2013 - 2022, you have to plug the x-values into the equation to get predictions. For example: 2013 - 1885 = 128 So plug in 128 into your equation y = 0.009538e^(0.03*128) y = 0.44376 y = $0.44
$0.46
yeah $0.46
awesome! Okay I just want to thank you so much for everything. Like you are a life saver. <333 The best
You're welcome :)

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