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mafiozo [28]
3 years ago
8

Q(–1, a) is 4 5 units away from P(3, –2). Find all possible values of a

Mathematics
1 answer:
Natali [406]3 years ago
5 0

Answer:

Step-by-step explanation:

Using the formula for calculating the distance between 2 points.

D = √(x2-x1)²+(y2-y1)²

Given

D = 45 units

The coordinates Q(–1, a) and P(3,2)

45 = √(3-(-1))²+(-2-a)²

45 = √4²+(-2-a)²

45 = √16+(-2-a)²

45² = 16+(-2-a)²

45² = 16+4+4a+a²

45² = 20+4a+a²

a²-4a+20-2025 = 0

a²+4a-2005 = 0

a = -4±√16+4(2005)/2

a = -4±√16+8020/2

a = -4±√8036/2

a =  -4±90/2

a = -4+90/2

a = 86/2

a = 43

The possible values of a is 43

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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":

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

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\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:

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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.

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