anonymous
  • anonymous
Minimum Wage The table shows the minimum wage for three different years. Year/wages: 1940/0.25, 1968/1.60, 1997/5.15 (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 that models the data. f(x)=a(x - h)^2 + k (c) Estimate the minimum wage in 1976 and compare it to 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.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
katieb
  • katieb
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anonymous
  • anonymous
can someone help me solve this so i can understand it
amistre64
  • amistre64
it looks like they give you 3 points of reference, right?
anonymous
  • anonymous
yea

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amistre64
  • amistre64
this turn into a system of 3 equations with 3 unknown variables; the unknowns are the constant coefficients of our quadratic equation that we are trying to find
amistre64
  • amistre64
lets adjust the years so that the computations are smaller, but equivalent 1940/0.25, 1968/1.60, 1997/5.15 -1940 -1940 -1940 ------------------------------- 0, 0.25 28, 1.60 57, 5.15 in essense, all ive done by this is to shift the graph so to the left; same shape, just a different location |dw:1332971979117:dw|
amistre64
  • amistre64
makes sense so far?
anonymous
  • anonymous
yes so far
amistre64
  • amistre64
i could also move the graph up or down if need be by adjusting the "y" values, but they look simple enough that i wont try that .... yet :)
anonymous
  • anonymous
oh boy ok math isn't my strongest subject
amistre64
  • amistre64
0, 0.25 28, 1.60 57, 5.15 when x=0, y = 0.25 ; this is our y intercept by definition, whenever x = 0 we have a y intercept ax^2 +bx + 0.25 = 0.25 ; when x = 0 ax^2 +bx + .25 = 1.60 ; when x = 28 ax^2 +bx + .25 = 5.15 ; when x = 57
amistre64
  • amistre64
if math aint your strongest subject then i take it trying a matrix on this wont make alot of sense to you :)
anonymous
  • anonymous
probably not..i can usally figure it out but for some reason i can't wrap my head around this problem i appreciate your help
amistre64
  • amistre64
lets apply our x values in each case and see if we can solve for a and b 0a +0b + 0.25 = 0.25 ; this ones not going to give us any new information so we can ignore it for now 28^2a +28b = 1.60-.25 57^2a +57b = 5.15-.25 28^2a +28b = 1.35 57^2a +57b = 4.90 now we are left with a system of 2 equations in 2 unknowns to solve for a and b
amistre64
  • amistre64
if we forget about trying to find the smallest value of this and that, we can simply eliminate some prospects by multiplying the top by 57^2 and the bottom by -28^2 28^2 (57^2)a +28 (57^2)b = 1.35(57^2) 57^2(-28^2)a +57(-28^2)b = 4.90(-28^2) --------------------------------------- 0 +(28 (57^2)+57(-28^2))b = 1.35(57^2) + 4.90(-28^2) dividing off that messy stuff gives us: \[b=\frac{1.35(57^2) + 4.90(-28^2)}{28 (57^2)+57(-28^2) }\]
anonymous
  • anonymous
if the problem is ax^2+bx +.25 would the equation be a(28)^2+b(28)+.25? or am i wrong in assuming that?
amistre64
  • amistre64
with the aid of a calculator; b = abt .012 and the same concept can be applied to get rid of the b to solve for a
amistre64
  • amistre64
your right, I just subtracted the .25 from each side to get a and b all alone
anonymous
  • anonymous
ok im confused 28^2 (57^2)a +28 (57^2)b = 1.35(57^2) 57^2(-28^2)a +57(-28^2)b = 4.90(-28^2) --------------------------------------- 0 +(28 (57^2)+57(-28^2))b = 1.35(57^2) + 4.90(-28^2) dividing off that messy stuff gives us: \[b=\frac{1.35(57^2) + 4.90(-28^2)}{28 (57^2)+57(-28^2) }\] this comes from ?
amistre64
  • amistre64
28^2(57)a + 28(57)b = 1.35(57) 57^2(-28)a +57(-28)b = 4.90(-28) ------------------------------- (28^2(57)+57^2(-28))a +0 = 1.35(57) + 4.90(-28) \[a=\frac{1.35(57) + 4.90(-28)}{28^2(57)+57^2(-28)}=-60.25\] ill explain the best i can after i post this bit of it
amistre64
  • amistre64
the numbers are messy, but the concept is still to use elimination to get rid of one of the variables in order to solve whats left for the remaing variable
amistre64
  • amistre64
\[\begin{array}\ a_1x+b_1c=n\\a_2x+b_2c=m\end{array}\] \[\begin{array}\ (a_1x+b_1c=n)*a_2\\(a_2x+b_2c=m)*(-a_1)\end{array}\] \[\begin{array}\ \cancel{a_1a_2}x+a_2b_1c=a_2n\\\bcancel{-a_1a_2}x-a_1b_2c=-a_1m\\------------\\(a_2-a_1)b_2c=a_2n-a_1m\end{array}\]
amistre64
  • amistre64
ugh, forgot a few parts in that; but it should still make sense i hope lol
anonymous
  • anonymous
not really im an epic fail at this point of no sleep
amistre64
  • amistre64
thats ok, these things can get complicated; i just tested out my "solution" and apparently i missed it someplace.
anonymous
  • anonymous
ok well im lost so i cant help you much :/
amistre64
  • amistre64
i hit a wrong button for the "a"; with any luck its about .001 so i want to test out the following to see how close I get to the points given y = .001x^2 +.012x+.25
anonymous
  • anonymous
ok i'll be right here :)
amistre64
  • amistre64
its yelling at me :/ with my approximations the wolf is giving me a value that is .25 cents less. ima gonna try the matrix route and see what it gives me .... but the concepts i presented are sound, its just working the numbers themselves thats a bit helter skelter
anonymous
  • anonymous
oh don't let it get you like it has gotten me I still need to explain this to others I am working with so the better you are the more it helps me :) cuz right now i am a deer in head lights just looking at the equation coming
amistre64
  • amistre64
http://www.wolframalpha.com/input/?i=rref%7B%7B0%2C0%2C1%2C.25%7D%2C%7B28%5E2%2C28%2C1%2C1.60%7D%2C%7B57%5E2%2C57%2C1%2C5.15%7D%7D well, this is what it pops out to
amistre64
  • amistre64
which is what i had to begin with .... so i think the wolf is getting a little mixed up these days
anonymous
  • anonymous
oh my
anonymous
  • anonymous
:/
amistre64
  • amistre64
the "exact" values for our coeefs are monster looking fractions; and my approximation to those is making my quadratic miss the mark the further it moves away from 1940
anonymous
  • anonymous
:( oh no
amistre64
  • amistre64
\[y = \frac{241}{185136}(x-1940)^2 +\frac{10891}{925680}(x-1940)+\frac{1}{4}\] that works lol
anonymous
  • anonymous
omg what is that
amistre64
  • amistre64
thats the quadratic equation that fits your data points; i got no idea how that would look in the a(x-h)^2+k format tho
anonymous
  • anonymous
hmm
amistre64
  • amistre64
for x=1997 i get 2.36; and the actual they presented in 2.30
amistre64
  • amistre64
it estimates about 1960 for a minimum wage of 1.00
amistre64
  • amistre64
and it estimates 7.26 for 2009
anonymous
  • anonymous
ok well thats awesome thank you know i just have to figure out how I am going to show and explain it
amistre64
  • amistre64
i think youd have to use computer generated values in order to "show" it; but the explanation is the basic notion that i started with
amistre64
  • amistre64
we create a system of 3 equations in 3 unknown and solve it using approved methods
amistre64
  • amistre64
either substitution, elimination, or matrix methods
anonymous
  • anonymous
ok thank you at least you got me headed in the right direction :)
amistre64
  • amistre64
lol, good luck with it ;)
anonymous
  • anonymous
thanks

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