Lines of best fit

Phobosdeimos

New member
Joined
Apr 4, 2015
Messages
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The average annual exchange rate in Canada for the US Dollar from 1998-2007 is shown in the following table. Draw a scatter plot, without using graphing technology

Year Exchange Rate
1998 .67
1999 .67
2000 .70
2001 .74
2002 .80
2003 .81
2004 .86
2005 .87
2006 .90
2007 .99

To determine the Slope I did the following
1998 - 2006 = -8
.67 - .90 = -.23

y= .23
Divide
x = -8

The slope of the line is - 0.02875

I then tried the y intercept
y = mx +b

.90 = - 0.02875 (2006) + b

.90 = -57.6725 + b

.90
- 57.6725 = b

b = 58.5725

This is what I came up for the y Intercept (58.5725)

Doesn't seem right to me

Please Help

Phobos
 
The average annual exchange rate in Canada for the US Dollar from 1998-2007 is shown in the following table. Draw a scatter plot, without using graphing technology

Year Exchange Rate
1998 .67
1999 .67
2000 .70
2001 .74
2002 .80
2003 .81
2004 .86
2005 .87
2006 .90
2007 .99

To determine the Slope I did the following
1998 - 2006 = -8
.67 - .90 = -.23

y= .23
Divide
x = -8

The slope of the line is - 0.02875

I then tried the y intercept
y = mx +b

.90 = - 0.02875 (2006) + b

.90 = -57.6725 + b

.90
- 57.6725 = b

b = 58.5725

This is what I came up for the y Intercept (58.5725)

Doesn't seem right to me
Why does this not seem right to you? ;)
 
I guess I just wanted to know if I'm right or wrong. This is new material for me and I'm not very confident at the moment in my answers.

Thanks

Phobos
 
...
To determine the Slope I did the following
1998 - 2006 = -8
.67 - .90 = -.23

y= .23
Divide
x = -8

The slope of the line is - 0.02875

...
You have y as -0.23 in one place and as y=0.23 in another. Which is correct and will that change the slope?
 
To determine the Slope I did the following
1998 - 2006 = -8
.67 - .90 = -.23

y= - .23 " This should have been negative"
Divide
x = -8

The slope of the line is 0.02875 "This should be positive"
 
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