Find the position function of a particle moving along a coordinate line that satisfies the given conditions. 2sint-cost

Mathematics

1 answer:

dimulka [17.4K]3 years ago

5 0

Answer:

the position of a particle moving at a coordinate say(y) will satisfy the given conditions t=0 (say) if y=2sint-cost

Step-by-step explanation:

Clearly by the above we can see that if y=2sint-cost at t=0, then y=-1 because at t=0 sint vanishes and leaves us with only cost and at t=0 cos0=1

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We are asked to determine the correlation factor "r" of the given table. To do that we will first label the column for "Quality" as "x" and the column for "Easiness" as "y". Like this:

Now, we create another column with the product of "x" and "y". Like this:

Now, we will add another column with the squares of the values of "x". Like this:

Now, we add another column with the squares of the values of "y":

Now, we sum the values on each of the columns:

Now, to get the correlation factor we use the following formula:

$r=\frac{n\Sigma xy-\Sigma x\Sigma y}{\sqrt{(n\Sigma x^2-(\Sigma x)^2)(n\Sigma y^2-(\Sigma y)^2)}}$

Where:

$\begin{gathered} \Sigma xy=\text{ sum of the column of xy} \\ \Sigma x=\text{ sum of the column x} \\ \Sigma y=\text{ sum of the column y} \\ \Sigma x^2=\text{ sum of the column x\textasciicircum2} \\ \Sigma y^2=\text{ sum of the column y\textasciicircum2} \\ n=\text{ number of rows} \end{gathered}$

Now we substitute the values, we get:

$r=\frac{\left(6)(70.56)-(25.2)(16.4\right)}{\sqrt{((6)(107.12)-(25.2)^2)((6)(47.82)-(16.4)^2)}}$

Solving the operations:

$r=0.858$

Therefore, the correlation factor is 0.858. If the correlation factor approaches the values of +1, this means that there is a strong linear correlation between the variables "x" and "y" and this correlation tends to be with a positive slope.