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erastovalidia
9 days ago
14

The table below represents the closing prices of stock ABC for the last 5 days. What is the r-value of the linear regression tha

t fits these data:
Day Value
1 472.08
2 454.26
3 444.95
4 439.49
5 436.55

Mathematics
2 answers:
Zina [12.3K]9 days ago
4 0
For this query, you can utilize a scientific calculator that can compute the r value in regression analysis. Assign the days as x values and the corresponding figures as y values. Upon executing the calculations with the scientific calculator, the r value was determined to be -0.947.
Zina [12.3K]9 days ago
3 0

The r value of the linear regression for this data is -0.947110707.

Here's the detailed explanation:

The equation for the linear regression follows the format of

y=bx+a

Where

b=\frac{n(\sum xy)-(\sum x)(\sum y)}{n(\sum x^2)-(\sum x)^2}

a=\frac{(\sum y)(\sum x^2)-(\sum x)(\sum xy)}{n(\sum x^2)-(\sum x)^2}

The r value is computed using

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

The corresponding values are

\sum x=15,\sum y=2247.35,\sum x^2=55,\sum y^2=1010938.423,\sum xy=6656.18, n=5

By applying the formula, it leads to

r=-0.947110707

Consequently, the r value for the linear regression that matches this data is -0.947110707.

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Response: The outcome is (x+7)(3x+1)=33x^2+22x+7.

Detailed explanation: Suppose that

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We have the polynomial division algorithm as follows

\textup{If }a(x)\times b(x)=c(x),\\\textup{then, we have }\\\\\dfrac{c(x)}{b(x)}=a(x)~~~~~\textup{or}~~~~~~c(x)\div b(x)=a(x).

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\textup{since }(3x2 + 22x + 7) \div(x + 7) = 3x + 1,\\\\\textup{so, }\\\\(x+7)(3x+1)=3x^2+22x+7=0.

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