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NeTakaya
3 years ago
13

You use a line of best fit for a set of data to make a prediction about an unknown value. the correlation coeffecient is -0.833

can you be confident that your predicted value will be reasonably to the actual value? why or why not?
please help.
Mathematics
1 answer:
alina1380 [7]3 years ago
5 0

Answer: The square root of π has attracted attention for almost as long as π itself. When you’re an ancient Greek mathematician studying circles and squares and playing with straightedges and compasses, it’s natural to try to find a circle and a square that have the same area. If you start with the circle and try to find the square, that’s called squaring the circle. If your circle has radius r=1, then its area is πr2 = π, so a square with side-length s has the same area as your circle if s2  = π, that is, if s = sqrt(π). It’s well-known that squaring the circle is impossible in the sense that, if you use the classic Greek tools in the classic Greek manner, you can’t construct a square whose side-length is sqrt(π) (even though you can approximate it as closely as you like); see David Richeson’s new book listed in the References for lots more details about this. But what’s less well-known is that there are (at least!) two other places in mathematics where the square root of π crops up: an infinite product that on its surface makes no sense, and a calculus problem that you can use a surface to solve.

Step-by-step explanation: this is the same paragraph The square root of π has attracted attention for almost as long as π itself. When you’re an ancient Greek mathematician studying circles and squares and playing with straightedges and compasses, it’s natural to try to find a circle and a square that have the same area. If you start with the circle and try to find the square, that’s called squaring the circle. If your circle has radius r=1, then its area is πr2 = π, so a square with side-length s has the same area as your circle if s2  = π, that is, if s = sqrt(π). It’s well-known that squaring the circle is impossible in the sense that, if you use the classic Greek tools in the classic Greek manner, you can’t construct a square whose side-length is sqrt(π) (even though you can approximate it as closely as you like); see David Richeson’s new book listed in the References for lots more details about this. But what’s less well-known is that there are (at least!) two other places in mathematics where the square root of π crops up: an infinite product that on its surface makes no sense, and a calculus problem that you can use a surface to solve.

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Explanation:

The easiest way to do this is to make use of the 2-point form of the equation for a line. For points (x₁, y₁) and (x₂, y₂), the equation is ...

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Filling in your given points, the equation becomes ...

  y=\dfrac{7-1}{\frac{-7}{2}-\frac{9}{2}}(x-\frac{9}{2})+1

After you fill in the values, it is a matter of simplifying the resulting equation.

y=\dfrac{6}{\frac{-16}{2}}(x-\frac{9}{2})+1=\dfrac{-12}{16}(x-\frac{9}{2})+1=-\dfrac{3}{4}x+\left(\dfrac{-3}{4}\cdot\dfrac{-9}{2}\right)+1\\\\y=-\dfrac{3}{4}x+\dfrac{27+8}{8}\\\\y=-\dfrac{3}{4}x+\dfrac{35}{8}

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Step-by-step explanation:

4 0
4 years ago
Read 2 more answers
Pleaeeeeeeeeeeeeeeeeeeeeeeeeeeee
LekaFEV [45]
4.) 2(x-1)+3= 2x+1

Distribute the 2, multiply the 2 by x and -1.

2*x= 2x (all you need to do is put 2 in front of x)

2*-1= -2

2x-2+3= 2x+1

Combine like terms. 

2x+1=2x+1

Subtract one on both sides. 

2x=2x

Divide by 2 into both sides. 

x=1

We can check this by plugging it in. 

2(1-1)+3= 2(1)+1
2(0)+3= 2+1
0+3=3
3=3 <== this works

I hope this helps!
~kaikers

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3 years ago
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lilavasa [31]

Answer:

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If your goal was to write it in slope intercept form, heres your answer.

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