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.
The question is too long. I feel too lazy to read it :(
ive never made a scatter plot tha tI know of...
just three plots aint much of a scatter, but I can make a quad out of it if you want :)
0.001301745743669519x^2 -5.0390080805461945x + 4876.675395385017 maybe
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.
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.
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.
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.
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.
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.
lol...you still there?
(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...
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
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 .................................................
Eq3: a(57)^2 + b(57) = 4.90 (1.35 - a(28)^2) (57) a(57)^2 + ------------------ = 4.90 ; solve for "a" :) 28
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
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)
now you plug those values into the equation: y = a(x-40)^2 + b(x-40) + c : and you get your quadratic equation.
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.
Youre welcome :) I just hope its useful lol
that doesnt wanna open on this computer....
your probably gonna have to chisel it on a stela and throw it at me :)
I think I'm just a lost cause. I'm lost. Its crazy. Sorry
So I plug in y=60.25(x-40)^2+28(x-40)+.25
Then if I want to know the wages for 1976 Do I plug that in for x or am I crazy lost.
plug in 1976 yes
did you get 60.25 and 28 from those monster fractions?
Ok when I did that I got y=225876992.25
I just plugged a=60.25, b = 28, and c = .25 Into y=a(x-40)^2+b(x-40)+c
I'm sorry I don't know why I'm having so much trouble with this. I'm just not understanding it.
when I use my "quad" finder I programed I get the:
I programmed that thing myself just for cases like these lol
0.001301745743669519x^2 + -5.0390080805461945x + 4876.675395385017 the quadratic equation is this monster of a mess...
Wow...I don't know. errrr lol
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 ;)
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.
then you show them what I typed out ;) cause thats the most basic way to derive it.
(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.
This is the formula I worked out up there :) 0.0013017457436695185(x-40)^2 + 0.011765404891539196(x-40) + 0.25
where you got 60.25 and 28 from .... I dunno. Unless I got typoed; which is a possibility
I so need a tutor like you in person
From 1976 as an input we get: 2.36 which is pretty close to 2.30.
:) if your near tampa....
LOL. Near KC, MO
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...
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....
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
yeah, it was 20 years for me.
tried back in 92 but kids and bills got more important
I hear you
now that im "unemployable" I am going back to get some degrees to teach in colleges
Wow. That is awesome
awesome that im unemployable ? :)
No that you are going back to school now
yeah. Im surrounded by students that are not much older than my kids...
amistre, need your help
Hey...who cares...I think it is great that you are back...trying...Thats amazing
thanx :) BTM, itll be about 30 mins prolly
Ok Thank you for everything
ok
just finishing up some last minute programming and it should be set....
lol..... still got a bug that needs squashed; sqrt(64) is NOT zero lol
if (q=g){...} is meaningless, if(q==g){.......} is proper. I forgot to put 2 "=" lol
im just gonna plug in dates till I get near 1 ;) just gonna be easier
I get 1959 = .9434 cents 1960 = 1.006 :)
2009 prediction is: $7.259 as compared to $7.25
and thats all the answers :)
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 (?)
What are plugging the dates into? Please show me?
b is a constant that the equation requires in order to function correctly. b :=approximately: 0.011765404891539196 round that to .01 if need be
a :=approx" 0.0013017457436695186241465733298764; rounded to .001 if need be
plugging the "dates" into is: x = date. we use the date as the input variable for the value of "X".
c = .25 ;)
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.
output = .001 x^2 +.01 x +.25
that looks more manageable ;)
opps...(x-40)^2 and (x-40) is what it should read
output = .001(x-40)^2 +.01(x-40) +.25
for 1975 we get: .001(35)^2 +.01(35) + .25 = 1.825 with this simplified version of it.
When I try this and put 1976 in for x I get y=3767.706
but they ask for 1976 lol..... i have a memory like a steel seive :)
1976-40=36 right? so: .001(36^2) + .01(36) + .25 = 1.906 is what I get for 1976
to get it more accurate, you will have to include more decimal points from the original setup
This setup right here gives you the most accurate: .0013017457436695185 (x-40)^2 + .011765404891539196 (x-40) + .25
are you making sure you subtract 40 from the year?
That was my problem. Thank You. I get it now. You are amazing. Thank you Thank you Thank you!!!
Google gives me this ;) (.0013017457436695185 * (36^2)) + (.011765404891539196 * 36) + .25 = 2.36061706
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
I didn't know that. Cool. Thanks again. Yay!!!
yay!! we got new stuff to play with in here...
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