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konstantin123 [22]
3 years ago
14

Suppose the line of best fit is being found for some points that have an r-value of 0.769. If the standard deviation of the x-co

ordinates is 5.508, what is the slope of the line to three decimal places?
Mathematics
1 answer:
jeyben [28]3 years ago
7 0

r value i.e Correlation between x value and y value = 0.769

A_{x}=Standard deviation of the x-coordinate = 5.508

 A_{y}=Standard deviation of the y-coordinate = ?=k

Slope of line is given by formula

If slope of line is m, then

m = r\times \frac{ A_{x}}{A _{y}}

m= 0.769 \times \frac{k}{5.508}

Substitute the value of k, i.e Standard deviation of the y-coordinate and then you can get the slope of line to three decimal places.

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(−t 4 −5t 3 −10t 2 )+(9t 3 +3t 2 −1)
klemol [59]

Answer:

\left(-t^4-5t^3-10t^2\right)+\left(9t^3+3t^2-1\right)=-t^4+4t^3-7t^2-1

Step-by-step explanation:

Given the expression

\left(-t^4\:-5t^3\:-10t^2\:\right)+\left(9t^3\:+3t^2\:-1\right)

Remove parentheses:  (a)=a

=-t^4-5t^3-10t^2+9t^3+3t^2-1

Group like terms

=-t^4-5t^3+9t^3-10t^2+3t^2-1

Add similar elements        

=-t^4-5t^3+9t^3-7t^2-1         ∵ -10t^2+3t^2=-7t^2

Add similar elements        

=-t^4+4t^3-7t^2-1                  ∵  -5t^3+9t^3=4t^3

Thus, the equivalent expression in simplified form:

\left(-t^4-5t^3-10t^2\right)+\left(9t^3+3t^2-1\right)=-t^4+4t^3-7t^2-1

4 0
3 years ago
To increase or decrease an amount by 30% what single multiplier would you use?
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to decrese by 30%, multiply by 70%
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simplify each term

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remove unnecessary parentheses

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