Questions: a) Find the value of m/n(580), s (1590). b) Use functional notation to express the minimum wage in 1985. (The actual value was 3.35.) c) Use the value of m for 1980 and 1990 given in the table to estimate the minimum wage in 1985. (The actual value was 3.35.) Year m = Minimum wage 1960 1.00 1970 1.60 1980 3.10 1990 3.80 2000 5.15 2010 7.25

Solution Steps

a) To find the value of m/n(580) and s(1590), we need to understand the context of the functions m and s. Assuming m and s are functions of time or some other variable, we need to evaluate these functions at the given points. However, the problem does not provide explicit functions for m and s, so we cannot proceed without additional information.

b) To express the minimum wage in 1985 using functional notation, we can denote it as m(1985).

c) To estimate the minimum wage in 1985 using the values for 1980 and 1990, we can use linear interpolation. Linear interpolation involves finding the value of a function between two known values by assuming the function changes at a constant rate between those points.

We are provided with the following data for the minimum wage (\(m\)) over various years: \[ \begin{array}{|c|c|} \hline \text{Year} & m \\ \hline 1960 & 1.00 \\ 1970 & 1.60 \\ 1980 & 3.10 \\ 1990 & 3.80 \\ 2000 & 5.15 \\ 2010 & 7.25 \\ \hline \end{array} \]

To express the minimum wage in 1985 using functional notation, we denote it as: \[ m(1985) \]

To estimate \(m(1985)\) using the values for 1980 and 1990, we use linear interpolation. The formula for linear interpolation is: \[ m(x) = m(x_0) + \frac{m(x_1) - m(x_0)}{x_1 - x_0} \cdot (x - x_0) \] where \(x_0 = 1980\), \(x_1 = 1990\), \(m(x_0) = 3.10\), and \(m(x_1) = 3.80\).

Substituting the values, we get: \[ m(1985) = 3.10 + \frac{3.80 - 3.10}{1990 - 1980} \cdot (1985 - 1980) \]

\[ m(1985) = 3.10 + \frac{0.70}{10} \cdot 5 = 3.10 + 0.35 = 3.45 \]

Final Answer