Howard-Wolowitz
  • Howard-Wolowitz
If an object is dropped from a height, its downward speed theoretically increases linearly over time because the object is subject to the steady pull of gravity. Here are observational data on the speed of a ball dropped from a certain height at time x = 0: Time (seconds) X 0 0.2 0.4 0.6 0.8 Speed (m/sec) Y 0 1.92 3.58 6.01 7.88 11. Determine the least-squares regression line (LSRL). Write the equation below
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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xMissAlyCatx
  • xMissAlyCatx
Here's your graph first of all..
1 Attachment
xMissAlyCatx
  • xMissAlyCatx
I might need @jigglypuff314 to help me. D:
xMissAlyCatx
  • xMissAlyCatx
or @ParthKohli

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Howard-Wolowitz
  • Howard-Wolowitz
isnt this my graph
Michele_Laino
  • Michele_Laino
please you don't draw the segments between experimental points. In other words, what we get is a scatter plot of @xMissAlyCatx
Michele_Laino
  • Michele_Laino
@Howard-Wolowitz
Michele_Laino
  • Michele_Laino
according to the scatter plot of @xMissAlyCatx we see that the relation between space and time is not a linear relation, since we can not fit such data with a straight line
Howard-Wolowitz
  • Howard-Wolowitz
ok so my plot is correct for a scatterplot
Michele_Laino
  • Michele_Laino
yes! Nevertheless you have to delete the segments between experimental points
Howard-Wolowitz
  • Howard-Wolowitz
how on I dont know what you mean? can you show me
Michele_Laino
  • Michele_Laino
Sorry, linear relation can be applied to your experimental data
Howard-Wolowitz
  • Howard-Wolowitz
ok so you do believe that my scatter plot is 100% correct for the points given?
Michele_Laino
  • Michele_Laino
yes! Here is what I meant:
Michele_Laino
  • Michele_Laino
1 Attachment
Howard-Wolowitz
  • Howard-Wolowitz
ok, but thats what i have? so i dont understand
Michele_Laino
  • Michele_Laino
now, we have to conjecture this relation: \[\huge v = At + B\] where \(v\) is the speed after \(t\) seconds. Therefore we have to apply the usual formulas in order to get the values of both constants \(A,\;B\) starting from the coordinates of the experimental points
Howard-Wolowitz
  • Howard-Wolowitz
are you talking about this question now: 11. Determine the least-squares regression line (LSRL). Write the equation below
Michele_Laino
  • Michele_Laino
yes!
Howard-Wolowitz
  • Howard-Wolowitz
ok so i am with yoy so far with what you put
Howard-Wolowitz
  • Howard-Wolowitz
how do we conjecture that relation
Michele_Laino
  • Michele_Laino
as I said before, we have to evaluate, starting from your experimental data, both constants \(A,\;B\)
Michele_Laino
  • Michele_Laino
here are the corresponding formulas:
Michele_Laino
  • Michele_Laino
please wait a moment I'm writing such formulas...
Michele_Laino
  • Michele_Laino
\[\Large \begin{gathered} \Delta = N\left( {\sum {t_i^2} } \right) - {\left( {\sum {{t_i}} } \right)^2} \hfill \\ \hfill \\ B = \frac{{\left( {\sum {t_i^2} } \right)\left( {\sum {{v_i}} } \right) - \left( {\sum {{t_i}} } \right)\left( {\sum {{t_i}{v_i}} } \right)}}{\Delta }, \hfill \\ \hfill \\ A = \frac{{N\left( {\sum {{t_i}{v_i}} } \right) - \left( {\sum {{t_i}} } \right)\left( {\sum {{v_i}} } \right)}}{\Delta } \hfill \\ \end{gathered} \]
Michele_Laino
  • Michele_Laino
where \(\Large N=5\)
Michele_Laino
  • Michele_Laino
of course, the index \(\Large i\) runs from \(\Large 1\) to \(\Large 5\)
Howard-Wolowitz
  • Howard-Wolowitz
ok so we have to use the foeumla to get the final equation
Michele_Laino
  • Michele_Laino
that's right! Once we know both constants \(A,\;B\) we can write the above formula
Howard-Wolowitz
  • Howard-Wolowitz
ok, the only part i dont get is how do we know them
Michele_Laino
  • Michele_Laino
what is the value for \(\Large \Delta\) ?
Howard-Wolowitz
  • Howard-Wolowitz
thats the variable
Michele_Laino
  • Michele_Laino
Please wait, I'm checking your values...
Michele_Laino
  • Michele_Laino
I got: \(A=9.925\)
Michele_Laino
  • Michele_Laino
and \(B=-0.092\)
Howard-Wolowitz
  • Howard-Wolowitz
show me your work please
Michele_Laino
  • Michele_Laino
ok! Please wait...
Michele_Laino
  • Michele_Laino
\[\Large \begin{gathered} \Delta = N\left( {\sum {t_i^2} } \right) - {\left( {\sum {{t_i}} } \right)^2} = 5 \cdot \frac{6}{5} - {2^2} = 2 \hfill \\ \hfill \\ B = \frac{{\left( {\sum {t_i^2} } \right)\left( {\sum {{v_i}} } \right) - \left( {\sum {{t_i}} } \right)\left( {\sum {{t_i}{v_i}} } \right)}}{\Delta } = \hfill \\ \hfill \\ = \frac{{\left( {6/5} \right) \cdot \left( {19.39} \right) - \left( 2 \right) \cdot \left( {11.726} \right)}}{2} = - 0.092, \hfill \\ \hfill \\ A = \frac{{N\left( {\sum {{t_i}{v_i}} } \right) - \left( {\sum {{t_i}} } \right)\left( {\sum {{v_i}} } \right)}}{\Delta } = \hfill \\ \hfill \\ = \frac{{5 \cdot \left( {11.726} \right) - \left( 2 \right) \cdot \left( {19.39} \right)}}{2} = 9.925 \hfill \\ \end{gathered} \]
Howard-Wolowitz
  • Howard-Wolowitz
ok i have a question, I dont get what the question means by the equation... do we set up a new equation since the we have the values? thats confusing me sorry
Michele_Laino
  • Michele_Laino
we can say that the straight line of best fit, has the subsequent equation: \[\huge v\left( t \right) = - 0.092 + 9.925t\]
Michele_Laino
  • Michele_Laino
please try to draw such line, inside your scatter plot
Howard-Wolowitz
  • Howard-Wolowitz
ok give me a second plz
Howard-Wolowitz
  • Howard-Wolowitz
|dw:1449512098957:dw|
Michele_Laino
  • Michele_Laino
please, I don't see the straight line
Michele_Laino
  • Michele_Laino
for example, at \(t=0\) we have: \(v=-0.092\) so the first point is: \(P=(0,-0.092)\) whereas at \(t=1\), we have \(v=9.833)\), so the second point is: \(Q=(1,9.833)\) So, please try to draw a line which passes at point P and at point Q
Howard-Wolowitz
  • Howard-Wolowitz
its n this correct?
Michele_Laino
  • Michele_Laino
no, I'm sorry, it is not correct
Howard-Wolowitz
  • Howard-Wolowitz
no i mean i add the points on this correct?
Michele_Laino
  • Michele_Laino
please add your experimental points, to this graph:
Michele_Laino
  • Michele_Laino
second step: the second inequality, can be rewritten as below: \[\huge y \geqslant \frac{2}{3}x - 4\] am I right?
Michele_Laino
  • Michele_Laino
oops.. sorry I was answering to another student
Howard-Wolowitz
  • Howard-Wolowitz
lol i wsa like im so confused
Howard-Wolowitz
  • Howard-Wolowitz
ok so why would i put the points on your graph, your points arent accurate
Michele_Laino
  • Michele_Laino
so you can see that the best fit line, really fits to your experimental points
Michele_Laino
  • Michele_Laino
nevertheless, my equation above, solves completely your exercise
Howard-Wolowitz
  • Howard-Wolowitz
i would like to see your final line plz

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