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

- Howard-Wolowitz

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- xMissAlyCatx

Here's your graph first of all..

##### 1 Attachment

- xMissAlyCatx

I might need @jigglypuff314 to help me. D:

- xMissAlyCatx

or @ParthKohli

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## More answers

- Howard-Wolowitz

isnt this my graph

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- 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

@Howard-Wolowitz

- 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

ok so my plot is correct for a scatterplot

- Michele_Laino

yes! Nevertheless you have to delete the segments between experimental points

- Howard-Wolowitz

how on I dont know what you mean? can you show me

- Michele_Laino

Sorry, linear relation can be applied to your experimental data

- Howard-Wolowitz

ok so you do believe that my scatter plot is 100% correct for the points given?

- Michele_Laino

yes! Here is what I meant:

- Michele_Laino

##### 1 Attachment

- Howard-Wolowitz

ok, but thats what i have? so i dont understand

- 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

are you talking about this question now: 11. Determine the least-squares regression line (LSRL). Write the equation
below

- Michele_Laino

yes!

- Howard-Wolowitz

ok so i am with yoy so far with what you put

- Howard-Wolowitz

how do we conjecture that relation

- Michele_Laino

as I said before, we have to evaluate, starting from your experimental data, both constants \(A,\;B\)

- Michele_Laino

here are the corresponding formulas:

- Michele_Laino

please wait a moment I'm writing such formulas...

- 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

where \(\Large N=5\)

- Michele_Laino

of course, the index \(\Large i\) runs from \(\Large 1\) to \(\Large 5\)

- Howard-Wolowitz

ok so we have to use the foeumla to get the final equation

- Michele_Laino

that's right! Once we know both constants \(A,\;B\) we can write the above formula

- Howard-Wolowitz

ok, the only part i dont get is how do we know them

- Michele_Laino

what is the value for \(\Large \Delta\) ?

- Howard-Wolowitz

thats the variable

- Michele_Laino

Please wait, I'm checking your values...

- Michele_Laino

I got: \(A=9.925\)

- Michele_Laino

and \(B=-0.092\)

- Howard-Wolowitz

show me your work please

- Michele_Laino

ok! Please wait...

- 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

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

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

please try to draw such line, inside your scatter plot

- Howard-Wolowitz

ok give me a second plz

- Howard-Wolowitz

|dw:1449512098957:dw|

- Michele_Laino

please, I don't see the straight line

- 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

its n this correct?

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- Michele_Laino

no, I'm sorry, it is not correct

- Howard-Wolowitz

no i mean i add the points on this correct?

- Michele_Laino

please add your experimental points, to this graph:

##### 1 Attachment

- 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

oops.. sorry I was answering to another student

- Howard-Wolowitz

lol i wsa like im so confused

- Howard-Wolowitz

ok so why would i put the points on your graph, your points arent accurate

- Michele_Laino

so you can see that the best fit line, really fits to your experimental points

- Michele_Laino

nevertheless, my equation above, solves completely your exercise

- Howard-Wolowitz

i would like to see your final line plz

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