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Math Help - Algrbra help

  1. #1
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    Algrbra help

    I am working on a project for school and there is one question I do not understand. Could someone please help me?

    Project:

    1.Find the high temperature for a certain city in the US. Use the internet to find temperature on July 1 of each year from 1987 to 2007. Make a table with the year in the first column and the corresponding temperature in the second column.

    Ok, I gathered all of my data for and made a table.

    2.Let x = 7 for 1987, x=8 for 1988, etc… Using Excel input the x values into column A and the high temperatures in column B. Highlight the cells A1 to B21. Click the insert tab at the top of the screen. Then click on the “Scatter” icon in the middle of the tool bar. Select the first icon in the first row for your graph type. This will make a scatter plot of your data. Select a chart type from the tool bar that allows a title, an x-axis label, and a y-axis label. Label each of these for your specific project. Save and print the scatter plot.

    I created my scatter plot:


    http://i333.photobucket.com/albums/m...g?t=1217452117


    Ok, here is the part I do not understand what to do:

    3. Calculate by hand the line passing through your ordered pairs for the year 1987 and 2007. You must show your work. Now use this line to predict the values at 5 other years from 1987 – 2007. Record the difference from your model with the actual temperature for those years.

    Am I supposed to use the point slope formula y = mx + b for finding 1987 and 2007?
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  2. #2
    Member Jonboy's Avatar
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    Use the coordinates you've found and create a slope intercept equation.

    Then plug in five different years from 1987 to 2007 so you'll get the temperature. Then make your line.
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  3. #3
    MHF Contributor Quick's Avatar
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    Quote Originally Posted by robasc View Post
    Am I supposed to use the point slope formula y = mx + b for finding 1987 and 2007?
    Nice post by the way, you clearly stated your problem

    You are supposed to find the line that passes through the points at x=7 and x=27 (1987 and 2007)

    So yeah, you need to find the line that passes through the two temperatures.

    on a side note: y=mx+b is standard form, y-y_1=a(x-x_1) is the point-slope formula.
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  4. #4
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    So they tell you which two ordered-pairs your will use. For 1987, the x-coordinate is 7 and the y-coordinate looks to be about 75 (let's just say 75 for now even if it is not exact... then you can change it if you need to).

    For 2007, the x-coordinate is 27 and the y-coordinate appears to be around 78 (again, I will just assume it is 78).

    So were have 2 ordered pairs (points) and we want a line in  y = mx + b form. Follow my steps below:

    point1 =  (x_{1}, y_{1}) = (7, 75)

    point2 =  (x_{2}, y_{2}) = (27, 78)

    First we compute the slope of the line connecting these two points. The general equation for the slope of a line, m, is:

     m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}} .

    Note that it does not matter which ordered pair is  (x_{1}, y_{1}) and which is  (x_{2}, y_{2}) , as long as you stay CONSISTENT. Therefore,  m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}} is the same thing as  m = \frac{y_{1} - y_{2}}{x_{1} - x_{2}} .

    So, for your problem, we have:

     x_{1} = 7

     x_{2} = 27

     y_{1} = 75

     y_{2} = 78

    So,  m = \frac{78 - 75}{27 - 7} = \frac{3}{20} .

    Now that we know our slope, m is  \frac{3}{20} , we have:

     y = mx + b

     y = \frac{3}{20}x + b

    Now finally, pick a point we know is on the line. For instance, since we are given two points, I will pick the first point (point1). That is, the point (7, 75).

    Plug the x-coordinate (which is 7) into x and the y-coordinate (which is 75) into y in our y=mx+b equation. Then we solve for b.

     75 = \frac{3}{20}(7) + b

     75 = \frac{21}{20} + b

     b = 75 - \frac{21}{20} = 73.95

    Now we have m and b. That is enough information. The equation of our line is  y = \frac{3}{20}x + 73.95 . Choose either decimals or fractions. I would simply write:

     y = 0.15x + 73.95 .

    -Andy
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  5. #5
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    4. Ok, for the next part below it is asking to use technology to find the regression line.



    And I came up with r = 0.2726285576 is this right?



    Using technology find the regression line for the high temperatures. Give the equation including the regression coefficient, r. Now use this model to predict the same 5 years from part 3. Record the difference from this model with the actual temperature for those years. How does this model compare to the model generated in part 3?
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