Problem 415

 

Minimum Wage:- The table shows the minimum wage for three different years.

Year 1940 1968 1997
Wage 0.25 1.67 5.15

 

(a)    Make a scatterplot of the data in the viewing rectangle [1930, 2010, 10] by [0, 6, 1].

(b)   Find a quadratic function given by f(x) =a(x-h)^{2} + k that models the data.

(c)    Estimate the minimum wage in 1976 and compare it the actual value of 2.30.

(d)   Estimate when the minimum wage was 1.00.

(e)   If current trends continue, predict the minimum wage in 2009. Compare it to the projected value of 7.25.

 

Solution

A.

a2

B.

f(x) =a(x - h) ^{2} + k

From the given table h = 1940 and k = 0.25

f(x) =a(x - 1940)^{2} + 0.25

from table f(x) = 5.15 when x = 1997

5.15 =a(1997 - 1940)^{2} + 0.25

5.15 – 0.25 = a(57)^{2}

4.90 = 3249α

α = \frac{4.9}{3249}

α = 0.0015

SO f(x) = 0.0015(x - 1940)^{2}+ 0.25

 

C.

Minimum ways in 1976

Plus in x = 1976 in the part B equation

f(x) = 0.0015(1976 - 1940)^{2} + 0.25

f(x) = 0.0015(36)^2 + 0.25

f(x) = 0.0015(1296) + 0.25

f(x) = 1.944 + 0.25

f(x) = 2.19

Estimate value is 2.19 is less than the actual value 2.30

 

D.

When minimum ways is 1.00

Given value is f(x) =1

Now plug in this value in part B equation

1.00 = 0.0015(x - 1940)^{2} + 0.25

1 – 0.25 = 0.0015(x - 1940)^{2} + 0.25

0.75 = 0.0015(x - 1940)^{2}

\frac{0.75}{0.0015} =(x - 1940)^{2}

500 =(x - 1940)^2

x – 1940 =\sqrt{500}

x – 1940 = 22.36

x = 1940 + 22.36

x = 1962.36 \approx 1962

So minimum ways 1.00 will be in 1962

 

E.

Minimum ways in 2009

Plug in x = 2009 in the part B equation

f(x) =0.0015(2009 - 1940)^{2} + 0.25

f(x) = 0.0015(69)^{2}+ 0.25

f(x) = 0.0015(4761) + 0.25

f(x) = 7.1415 + 0.25

f(x) = 7.3915 \approx 7.39

Estimate value is 7.39 is greater than the actual value 7.25.

 

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