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.

- anonymous

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- schrodinger

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- anonymous

The question is too long. I feel too lazy to read it :(

- amistre64

ive never made a scatter plot tha tI know of...

- 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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## More answers

- amistre64

##### 1 Attachment

- amistre64

0.001301745743669519x^2
-5.0390080805461945x
+ 4876.675395385017 maybe

- 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

- anonymous

- anonymous

- anonymous

- amistre64

lol...you still there?

- 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

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

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

Eq3:
a(57)^2 + b(57) = 4.90
(1.35 - a(28)^2) (57)
a(57)^2 + ------------------ = 4.90 ; solve for "a" :)
28

- 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

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

now you plug those values into the equation:
y = a(x-40)^2 + b(x-40) + c : and you get your quadratic equation.

- 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

Youre welcome :) I just hope its useful lol

- anonymous

##### 1 Attachment

- amistre64

that doesnt wanna open on this computer....

- amistre64

your probably gonna have to chisel it on a stela and throw it at me :)

- anonymous

I think I'm just a lost cause. I'm lost. Its crazy. Sorry

- anonymous

So I plug in y=60.25(x-40)^2+28(x-40)+.25

- anonymous

Then if I want to know the wages for 1976 Do I plug that in for x or am I crazy lost.

- amistre64

plug in 1976 yes

- amistre64

did you get 60.25 and 28 from those monster fractions?

- anonymous

Ok when I did that I got y=225876992.25

- anonymous

I just plugged a=60.25, b = 28, and c = .25
Into y=a(x-40)^2+b(x-40)+c

- anonymous

I'm sorry I don't know why I'm having so much trouble with this. I'm just not understanding it.

- amistre64

when I use my "quad" finder I programed I get the:

##### 1 Attachment

- amistre64

I programmed that thing myself just for cases like these lol

- amistre64

0.001301745743669519x^2 + -5.0390080805461945x + 4876.675395385017
the quadratic equation is this monster of a mess...

- anonymous

Wow...I don't know. errrr lol

- 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

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

then you show them what I typed out ;) cause thats the most basic way to derive it.

- 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

This is the formula I worked out up there :)
0.0013017457436695185(x-40)^2
+ 0.011765404891539196(x-40)
+ 0.25

- amistre64

where you got 60.25 and 28 from .... I dunno. Unless I got typoed; which is a possibility

- anonymous

I so need a tutor like you in person

- amistre64

From 1976 as an input we get:
2.36 which is pretty close to 2.30.

- amistre64

:) if your near tampa....

- anonymous

LOL. Near KC, MO

- 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

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

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

yeah, it was 20 years for me.

- amistre64

tried back in 92 but kids and bills got more important

- anonymous

I hear you

- amistre64

now that im "unemployable" I am going back to get some degrees to teach in colleges

- anonymous

Wow. That is awesome

- amistre64

awesome that im unemployable ? :)

- anonymous

No that you are going back to school now

- amistre64

yeah. Im surrounded by students that are not much older than my kids...

- anonymous

amistre, need your help

- anonymous

Hey...who cares...I think it is great that you are back...trying...Thats amazing

- amistre64

thanx :) BTM, itll be about 30 mins prolly

- anonymous

Ok
Thank you for everything

- anonymous

ok

- amistre64

just finishing up some last minute programming and it should be set....

- amistre64

lol..... still got a bug that needs squashed; sqrt(64) is NOT zero lol

- amistre64

if (q=g){...} is meaningless,
if(q==g){.......} is proper. I forgot to put 2 "=" lol

- amistre64

im just gonna plug in dates till I get near 1 ;) just gonna be easier

- amistre64

I get 1959 = .9434 cents
1960 = 1.006 :)

- amistre64

2009 prediction is:
$7.259 as compared to $7.25

- amistre64

and thats all the answers :)

- 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

What are plugging the dates into? Please show me?

- 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

a :=approx" 0.0013017457436695186241465733298764; rounded to .001 if need be

- amistre64

plugging the "dates" into is: x = date.
we use the date as the input variable for the value of "X".

- amistre64

c = .25 ;)

- 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

output = .001 x^2 +.01 x +.25

- amistre64

that looks more manageable ;)

- amistre64

opps...(x-40)^2 and (x-40) is what it should read

- amistre64

output = .001(x-40)^2 +.01(x-40) +.25

- amistre64

for 1975 we get:
.001(35)^2 +.01(35) + .25 = 1.825 with this simplified version of it.

- anonymous

When I try this and put 1976 in for x I get y=3767.706

- amistre64

but they ask for 1976 lol..... i have a memory like a steel seive :)

- amistre64

1976-40=36 right? so:
.001(36^2) + .01(36) + .25 = 1.906 is what I get for 1976

- amistre64

to get it more accurate, you will have to include more decimal points from the original setup

- amistre64

This setup right here gives you the most accurate:
.0013017457436695185 (x-40)^2
+ .011765404891539196 (x-40)
+ .25

- amistre64

are you making sure you subtract 40 from the year?

- anonymous

That was my problem. Thank You. I get it now. You are amazing. Thank you Thank you Thank you!!!

- amistre64

Google gives me this ;)
(.0013017457436695185 * (36^2)) + (.011765404891539196 * 36) + .25 = 2.36061706

- 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

I didn't know that. Cool. Thanks again. Yay!!!

- amistre64

yay!! we got new stuff to play with in here...

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