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March 4, 2015

March 4, 2015

Posted by **Anna** on Sunday, October 25, 2009 at 8:33pm.

The linear equation

y=0.15x + 0.79

represents an estimate of the average cost of gas for year x starting in 1997. The year 1997 would be represented by x = 1, for example, as it is the first year in the study. Similarly, 2005 would be year 9, or x = 9.

a)What year would be represented by x = 4?

My answer is = year 2000

b)What x-value represents the year 2018?

x= 22

c)What is the slope (or rate of change) of this equation?

How or where do I start on this one??

d)What is the y-intercept?

e)What does the y-intercept represent?

f)Assuming this growth trend continues, what will the price of gasoline be in the year 2018? How did you arrive at your answer

- equations -
**DrBob222**, Sunday, October 25, 2009 at 8:44pmrepresents an estimate of the average cost of gas for year x starting in 1997. The year 1997 would be represented by x = 1, for example, as it is the first year in the study. Similarly, 2005 would be year 9, or x = 9.

a)What year would be represented by x = 4?

My answer is = year 2000**2000 is correct. You might make the next part easier by looking at how this was done, instead of just counting up 4.**

1996 + 4 = 2000.

b)What x-value represents the year 2018?**1996 + 18 = ??**

x= 22

c)What is the slope (or rate of change) of this equation?**The equation for a straight line is**

y = mx + b where m is the slope. The equation given is

y=0.15x + 0.79; therefore, 0.15 is the slope.

How or where do I start on this one??

d)What is the y-intercept?**The y intercept is the 0.79 in the original equation. It is the point at which the line crosses the y axis (and x = 0 at that point).**

e)What does the y-intercept represent?**See answer above.**

f)Assuming this growth trend continues, what will the price of gasoline be in the year 2018? How did you arrive at your answer**Use the equation, plug in x for the number of years (check me out but I think that will be 22) and solve for y.**

- equations -
**Anna**, Sunday, October 25, 2009 at 9:00pmthank you dr. bob

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