Price of a Stamp The table lists the price P of a
first-class postage stamp for selected year’s t.
t 1963 1975 1987 2002 2007
p $.05 $.13 $.25 $.37 $.41
(a) Find the slope of each line segment connecting consecutive data points in the table. Can these data be modeled with a line? Explain.
(b) Does the polynomial 0.835t – 1635 model the
data in the table exactly? Explain
Your going to have to have a graphing calculator its not solvable without it
(a) To find the slope of each line segment connecting consecutive data points in the table, we can use the formula for the slope of a line passing through two points. The formula is:
Slope = (y2 - y1) / (x2 - x1)
Let's apply this formula to find the slopes between each pair of consecutive points in the table:
For 1963-1975:
Slope = ($0.13 - $0.05) / (1975 - 1963) = $0.08 / 12 = $0.0067
For 1975-1987:
Slope = ($0.25 - $0.13) / (1987 - 1975) = $0.12 / 12 = $0.01
For 1987-2002:
Slope = ($0.37 - $0.25) / (2002 - 1987) = $0.12 / 15 = $0.008
For 2002-2007:
Slope = ($0.41 - $0.37) / (2007 - 2002) = $0.04 / 5 = $0.008
Now, let's analyze if these data points can be modeled with a line. If the slopes between each pair of consecutive points are approximately equal, then a linear model could be a good fit for the data.
Comparing the slopes, we see that the slopes for the line segments connecting consecutive data points are not exactly equal. This suggests that the data points cannot be modeled perfectly by a single straight line.
(b) To determine if the polynomial 0.835t - 1635 exactly models the data in the table, we need to check if the values of P calculated using this polynomial expression match the actual values in the table.
Let's calculate the values of P for each year using the given expression P = 0.835t - 1635:
For 1963: P = 0.835 * 1963 - 1635 ≈ $0.12
For 1975: P = 0.835 * 1975 - 1635 ≈ $0.50
For 1987: P = 0.835 * 1987 - 1635 ≈ $0.89
For 2002: P = 0.835 * 2002 - 1635 ≈ $1.61
For 2007: P = 0.835 * 2007 - 1635 ≈ $1.89
As we can see, the values calculated using the given polynomial expression do not exactly match the actual values in the table. Therefore, the polynomial 0.835t - 1635 does not model the data in the table exactly.