Question

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

Answer 1

Here's your graph first of all..

Answer 2

I might need @jigglypuff314 to help me. D:

Answer 3

or @ParthKohli

Answer 4

isnt this my graph

Answer 5

please you don't draw the segments between experimental points. In other words, what we get is a scatter plot of @xMissAlyCatx

Answer 6

@Howard-Wolowitz

Answer 7

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

Answer 8

ok so my plot is correct for a scatterplot

Answer 9

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

Answer 10

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

Answer 11

Sorry, linear relation can be applied to your experimental data

Answer 12

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

Answer 13

yes! Here is what I meant:

Answer 14

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

Answer 15

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

Answer 16

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

Answer 17

yes!

Answer 18

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

Answer 19

how do we conjecture that relation

Answer 20

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

Answer 21

here are the corresponding formulas:

Answer 22

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

Answer 23

\[\Large \begin{gathered}
\Delta = N\left( {\sum {t_i^2} } \right) - {\left( {\sum {{t_i}} } \right)^2} \hfill \\
\hfill \\
B = \frac{{\left( {\sum {t_i^2} } \right)\left( {\sum {{v_i}} } \right) - \left( {\sum {{t_i}} } \right)\left( {\sum {{t_i}{v_i}} } \right)}}{\Delta }, \hfill \\
\hfill \\
A = \frac{{N\left( {\sum {{t_i}{v_i}} } \right) - \left( {\sum {{t_i}} } \right)\left( {\sum {{v_i}} } \right)}}{\Delta } \hfill \\
\end{gathered} \]

Answer 24

where \(\Large N=5\)

Answer 25

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

Answer 26

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

Answer 27

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

Answer 28

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

Answer 29

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

Answer 30

thats the variable

Answer 31

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

Answer 32

I got: \(A=9.925\)

Answer 33

and \(B=-0.092\)

Answer 34

show me your work please

Answer 35

ok! Please wait...

Answer 36

\[\Large \begin{gathered}
\Delta = N\left( {\sum {t_i^2} } \right) - {\left( {\sum {{t_i}} } \right)^2} = 5 \cdot \frac{6}{5} - {2^2} = 2 \hfill \\
\hfill \\
B = \frac{{\left( {\sum {t_i^2} } \right)\left( {\sum {{v_i}} } \right) - \left( {\sum {{t_i}} } \right)\left( {\sum {{t_i}{v_i}} } \right)}}{\Delta } = \hfill \\
\hfill \\
= \frac{{\left( {6/5} \right) \cdot \left( {19.39} \right) - \left( 2 \right) \cdot \left( {11.726} \right)}}{2} = - 0.092, \hfill \\
\hfill \\
A = \frac{{N\left( {\sum {{t_i}{v_i}} } \right) - \left( {\sum {{t_i}} } \right)\left( {\sum {{v_i}} } \right)}}{\Delta } = \hfill \\
\hfill \\
= \frac{{5 \cdot \left( {11.726} \right) - \left( 2 \right) \cdot \left( {19.39} \right)}}{2} = 9.925 \hfill \\
\end{gathered} \]

Answer 37

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

Answer 38

we can say that the straight line of best fit, has the subsequent equation:
\[\huge v\left( t \right) = - 0.092 + 9.925t\]

Answer 39

please try to draw such line, inside your scatter plot

Answer 40

ok give me a second plz

Answer 41

please, I don't see the straight line

Answer 42

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

Answer 43

its n this correct?

Answer 44

no, I'm sorry, it is not correct

Answer 45

no i mean i add the points on this correct?

Answer 46

please add your experimental points, to this graph:

Answer 47

second step:
the second inequality, can be rewritten as below:
\[\huge y \geqslant \frac{2}{3}x - 4\]
am I right?

Answer 48

oops.. sorry I was answering to another student

Answer 49

lol i wsa like im so confused

Answer 50

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

Answer 51

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

Answer 52

nevertheless, my equation above, solves completely your exercise

Answer 53

i would like to see your final line plz