APlot the given ordered pairs to create a scatter plot. The horizontal axis represents the years since 1900.
BChoose two points from the line that you made in Part A and substitute their coordinates into slope formula.
CCalculate the number of years between 1900 and 2020 and then substitute it into the equation from Part B.
AScatter Plot:
Example Line:
BExample Equation: y=0.29x+50
CExample Conjecture: The life expectancy of a person born in 2020 is 84.8 years.
Practice makes perfect
We are given a table that shows the life expectancy, in years, for people born in certain years.
Years Since 1900
0
10
20
30
40
50
60
70
80
90
100
Life Expectancy
47.3
50.0
54.1
59.7
62.9
68.2
69.7
70.8
73.7
75.4
77.1
We are asked to construct a scatter plot and draw a line of best fit. To construct a scatter plot, we need to sketch each point from the table on a coordinate plane. The horizontal axis will represent the years since 1900 and the vertical axis will represent the life expectancy of the people born.
For example, let's sketch the third point from the table ( 20, 54.1).
In a similar way, we will now graph the other points from the table.
We constructed a scatter plot of the data! Now we will draw a straight line that best represents the data. Remember that we should try for the line to be as close to the data points as possible.
The points are very close to the line and a similar number of points is above and below the line. Therefore, this is a line of best fit. Notice that, if you sketch a different line that is close to the data points, it would also be a correct answer.
In Part A, we drew the following line of best fit.
Let's write an equation in slope-intercept form for this line!
y=mx+bIn this equation, m is the slope and b is the y-intercept. First, we will choose any two points on the line.
The first point has coordinates x_1= 70 and y_1= 70.8. The second point has coordinates x_2= 80 and y_2= 73.7. Let's substitute these values into slope formula.
We found that the slope m is equal to 0.29. This means that the life expectancy of a person born in a certain year is longer by 0.29 years than the life expectancy of a person born in a previous year.
y=mx+b ⇔ y= 0.29+b
The y-intercept b is the y-value when x=0. We can find it on the graph. Let's do it!
We can see that the y-intercept is equal to about 50. This means that in the year 1900, the life expectancy was approximately equal to 50 years. Finally, we can write a complete equation in slope-intercept form for our line of fit.
y=0.29x+b ⇕ y=0.29x+ 50
In Part B, we wrote an equation that estimates the life expectancy y for people born x years after the year 1900.
y=0.29x+50Here, we are asked to use this equation to make a conjecture about the life expectancy for a person born in 2020. To do that, we first need to calculate the number of years x between the year 1900 and 2020.
x=2020-1900 ⇔ x=120
Now we can substitute x= 120 into the equation from Part B to estimate the life expectancy for a person born in 2020.
