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
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
The question is too long. I feel too lazy to read it :(
amistre64
  • amistre64
ive never made a scatter plot tha tI know of...
amistre64
  • amistre64
just three plots aint much of a scatter, but I can make a quad out of it if you want :)

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amistre64
  • amistre64
1 Attachment
amistre64
  • amistre64
0.001301745743669519x^2 -5.0390080805461945x + 4876.675395385017 maybe
anonymous
  • anonymous
I'm going nuts on this. I need to learn how to do this not just answers. Thanks for everyone's help. If anyone else knows how to solve these. Please let me know. Thank you.
anonymous
  • anonymous
I'm going nuts on this. I need to learn how to do this not just answers. Thanks for everyone's help. If anyone else knows how to solve these. Please let me know. Thank you.
anonymous
  • anonymous
I'm going nuts on this. I need to learn how to do this not just answers. Thanks for everyone's help. If anyone else knows how to solve these. Please let me know. Thank you.
anonymous
  • anonymous
I'm going nuts on this. I need to learn how to do this not just answers. Thanks for everyone's help. If anyone else knows how to solve these. Please let me know. Thank you.
anonymous
  • anonymous
I'm going nuts on this. I need to learn how to do this not just answers. Thanks for everyone's help. If anyone else knows how to solve these. Please let me know. Thank you.
anonymous
  • anonymous
I'm going nuts on this. I need to learn how to do this not just answers. Thanks for everyone's help. If anyone else knows how to solve these. Please let me know. Thank you.
amistre64
  • amistre64
lol...you still there?
amistre64
  • amistre64
(1940/0.25), (1968/1.60), (1997/5.15 ) the numbers are huge, but the process is the same: a(1940)^2 + b(1940) +c = .25 a(1968)^2 + b(1968) +c = 1.6 a(1997)^2 + b(1997) +c = 5.15 we can reduce the numbers and still get the same effect by subtracting 1940 from everything...
amistre64
  • amistre64
these numbers are more managable; just dont forget to "+40" to the year when your done. a(0)^2 + b(0) +c = .25 <-- from this we see that c = .25 thats a given; write it down :) a(28)^2 + b(28) +c = 1.6 a(57)^2 + b(57) +c = 5.15
amistre64
  • amistre64
Eq simply means "Equation": a(28)^2 + b(28) +.25 = 1.6 a(28)^2 + b(28) = 1.6 - .25 = 1.35 <-Eq2 .............................................. a(57)^2 + b(57) +.25 = 5.15 a(57)^2 + b(57) = 5.15 - .25 = 4.90 <-Eq3 ....................................... a(28)^2 + b(28) = 1.35 <-Eq2 ; solve for b, just a random pick nothing special about it 1.35 - a(28)^2 b = ------------- <- use this "value" of b in the other Eq. 28 .................................................
amistre64
  • amistre64
Eq3: a(57)^2 + b(57) = 4.90 (1.35 - a(28)^2) (57) a(57)^2 + ------------------ = 4.90 ; solve for "a" :) 28
amistre64
  • amistre64
a(57)^2(28) + (1.35 - a(28)^2) (57) = 4.90(28) a(57)^2(28) + 1.35(57) - a(28)^2(57) = 4.90(28) a(57)^2(28) - a(28)^2(57) = 4.90(28) - 1.35(57) a [(57)^2(28) -(28)^2(57)] = 4.90(28) - 1.35(57) 4.90(28) - 1.35(57) 60.25 a = -------------------- = ------- ; reduce as wanted :) (57)^2(28) -(28)^2(57) 46284 ............................................. a = 60.25/46284 ;c = .25
amistre64
  • amistre64
now lets see what "b" equals with this monstrocity: recall that "b" =: 1.35 - a(28)^2 b = ------------- 28 1.35 - (60.25/46284)(28)^2 b = ------------------------ 28 1.35(46284) - (60.25)(28)^2 b = ----------------------------- ; reduce as wanted :) 48284(28)
amistre64
  • amistre64
now you plug those values into the equation: y = a(x-40)^2 + b(x-40) + c : and you get your quadratic equation.
anonymous
  • anonymous
Amistre64, You are awesome. I'm a little confused but I will look it over better in the morning in hope that I can follow. Once again thank you so much.
amistre64
  • amistre64
Youre welcome :) I just hope its useful lol
anonymous
  • anonymous
amistre64
  • amistre64
that doesnt wanna open on this computer....
amistre64
  • amistre64
your probably gonna have to chisel it on a stela and throw it at me :)
anonymous
  • anonymous
I think I'm just a lost cause. I'm lost. Its crazy. Sorry
anonymous
  • anonymous
So I plug in y=60.25(x-40)^2+28(x-40)+.25
anonymous
  • anonymous
Then if I want to know the wages for 1976 Do I plug that in for x or am I crazy lost.
amistre64
  • amistre64
plug in 1976 yes
amistre64
  • amistre64
did you get 60.25 and 28 from those monster fractions?
anonymous
  • anonymous
Ok when I did that I got y=225876992.25
anonymous
  • anonymous
I just plugged a=60.25, b = 28, and c = .25 Into y=a(x-40)^2+b(x-40)+c
anonymous
  • anonymous
I'm sorry I don't know why I'm having so much trouble with this. I'm just not understanding it.
amistre64
  • amistre64
when I use my "quad" finder I programed I get the:
amistre64
  • amistre64
I programmed that thing myself just for cases like these lol
amistre64
  • amistre64
0.001301745743669519x^2 + -5.0390080805461945x + 4876.675395385017 the quadratic equation is this monster of a mess...
anonymous
  • anonymous
Wow...I don't know. errrr lol
amistre64
  • amistre64
when given 3 sets of data; you can find a quad to match it by doing all that stuff I did up there; so instead, I just told the computer to do it all and give me an answer ;)
anonymous
  • anonymous
I don't know how to explain my answer in a, b,c, d and e in these problems. I have to show how I got the answers.
amistre64
  • amistre64
then you show them what I typed out ;) cause thats the most basic way to derive it.
anonymous
  • anonymous
(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 when 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.
amistre64
  • amistre64
This is the formula I worked out up there :) 0.0013017457436695185(x-40)^2 + 0.011765404891539196(x-40) + 0.25
amistre64
  • amistre64
where you got 60.25 and 28 from .... I dunno. Unless I got typoed; which is a possibility
anonymous
  • anonymous
I so need a tutor like you in person
amistre64
  • amistre64
From 1976 as an input we get: 2.36 which is pretty close to 2.30.
amistre64
  • amistre64
:) if your near tampa....
anonymous
  • anonymous
LOL. Near KC, MO
amistre64
  • amistre64
id have to do a little more programming to determine the "$1.00" part. just have to tell it to get it from the other input box and give me an answer for it...
amistre64
  • amistre64
i wrote an algorithym a while back to zone in on a particular value by seeing when it gets over to back up and add halves....
anonymous
  • anonymous
You are too smart. LOL. I haven't had a math class in over 12 years. This is what I get for having kids and getting married then a divorce makes me want to finishing school
amistre64
  • amistre64
yeah, it was 20 years for me.
amistre64
  • amistre64
tried back in 92 but kids and bills got more important
anonymous
  • anonymous
I hear you
amistre64
  • amistre64
now that im "unemployable" I am going back to get some degrees to teach in colleges
anonymous
  • anonymous
Wow. That is awesome
amistre64
  • amistre64
awesome that im unemployable ? :)
anonymous
  • anonymous
No that you are going back to school now
amistre64
  • amistre64
yeah. Im surrounded by students that are not much older than my kids...
anonymous
  • anonymous
amistre, need your help
anonymous
  • anonymous
Hey...who cares...I think it is great that you are back...trying...Thats amazing
amistre64
  • amistre64
thanx :) BTM, itll be about 30 mins prolly
anonymous
  • anonymous
Ok Thank you for everything
anonymous
  • anonymous
ok
amistre64
  • amistre64
just finishing up some last minute programming and it should be set....
amistre64
  • amistre64
lol..... still got a bug that needs squashed; sqrt(64) is NOT zero lol
amistre64
  • amistre64
if (q=g){...} is meaningless, if(q==g){.......} is proper. I forgot to put 2 "=" lol
amistre64
  • amistre64
im just gonna plug in dates till I get near 1 ;) just gonna be easier
amistre64
  • amistre64
I get 1959 = .9434 cents 1960 = 1.006 :)
amistre64
  • amistre64
2009 prediction is: $7.259 as compared to $7.25
amistre64
  • amistre64
and thats all the answers :)
anonymous
  • anonymous
Ok quick question. I know (from what you gave me) that a=60.25/46284, c =.25, and what you gave me for b, is not working out for me. What would b be again. and can a be broke down smaller. so in my problem: y=a(x-40)^2+b(x-40)+c I plug in y=60.25/46284(x-40)^2+b(x-40)+.25 So for x I can replace it with 1976, right? and b I can replace it with (?)
anonymous
  • anonymous
What are plugging the dates into? Please show me?
amistre64
  • amistre64
b is a constant that the equation requires in order to function correctly. b :=approximately: 0.011765404891539196 round that to .01 if need be
amistre64
  • amistre64
a :=approx" 0.0013017457436695186241465733298764; rounded to .001 if need be
amistre64
  • amistre64
plugging the "dates" into is: x = date. we use the date as the input variable for the value of "X".
amistre64
  • amistre64
c = .25 ;)
amistre64
  • amistre64
a, b and c are constants that do not change in this equation. They allow us to pinpoint with a great deal of accuracy the proper amounts that the inputs require.
amistre64
  • amistre64
output = .001 x^2 +.01 x +.25
amistre64
  • amistre64
that looks more manageable ;)
amistre64
  • amistre64
opps...(x-40)^2 and (x-40) is what it should read
amistre64
  • amistre64
output = .001(x-40)^2 +.01(x-40) +.25
amistre64
  • amistre64
for 1975 we get: .001(35)^2 +.01(35) + .25 = 1.825 with this simplified version of it.
anonymous
  • anonymous
When I try this and put 1976 in for x I get y=3767.706
amistre64
  • amistre64
but they ask for 1976 lol..... i have a memory like a steel seive :)
amistre64
  • amistre64
1976-40=36 right? so: .001(36^2) + .01(36) + .25 = 1.906 is what I get for 1976
amistre64
  • amistre64
to get it more accurate, you will have to include more decimal points from the original setup
amistre64
  • amistre64
This setup right here gives you the most accurate: .0013017457436695185 (x-40)^2 + .011765404891539196 (x-40) + .25
amistre64
  • amistre64
are you making sure you subtract 40 from the year?
anonymous
  • anonymous
That was my problem. Thank You. I get it now. You are amazing. Thank you Thank you Thank you!!!
amistre64
  • amistre64
Google gives me this ;) (.0013017457436695185 * (36^2)) + (.011765404891539196 * 36) + .25 = 2.36061706
amistre64
  • amistre64
you can copy and past an equation into a google search bar and it calculates it for you. its pretty nifty lol I said nifty
anonymous
  • anonymous
I didn't know that. Cool. Thanks again. Yay!!!
amistre64
  • amistre64
yay!! we got new stuff to play with in here...

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