anonymous
  • anonymous
ALGEBRA HELP- Does anyone with patience know to do the “Least Squares Linear Regression Method”? I watched a YouTube video on how to do it but I don’t know how to apply it here. Sorry this one is long…and I really want to understand all of it, but if you can help with even one part- please do. The table lists the distances (in megaparsecs; 1 megaparsec= 3.085 × 10^24 cm, and 1 megaparsec= 3.26 million light-years) and velocities (in kms per second) of four galaxies moving rapidly away from Earth. (I’ll put the table in the comments) (a) Plot the data using distances for the x-values
Mathematics
  • Stacey Warren - Expert brainly.com
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katieb
  • katieb
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anonymous
  • anonymous
(cont.) and velocities for the y-values. What type of relationship seems to hold between the data? (b) Find a linear equation in the form y=mx that models these data using the points (520, 40,000) and (0, 0). Graph your equation with the data on the same coordinate axes. (c) The galaxy Hydra has a velocity of 60,000 km per sec. How far away is it according to the model in part (b)? (d) The value of m is called the Hubble constant. The Hubble constant can be used to estimate the age of the universe A (in years) using the formula A= (9.5)(10^11)/m (e) Astronomers currently place the value of the Hubble constant between 50 and 100. What is the range for the age of the universe A?
anonymous
  • anonymous
|dw:1437111756117:dw|
jtvatsim
  • jtvatsim
No one brave enough to tackle this one? lol

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jtvatsim
  • jtvatsim
Did you want to go through all the parts?
jtvatsim
  • jtvatsim
This actually doesn't look too bad. I'm assuming that your class wants you to do this by hand?
anonymous
  • anonymous
Yes. I know it should be easy which makes me feel stupid for asking...
jtvatsim
  • jtvatsim
No, no worries. It's only easy because I've had to pull my hair out on these in the past.
jtvatsim
  • jtvatsim
Have you already gone ahead and graphed the data for part a)?
anonymous
  • anonymous
Yes. I dont see the relation though
jtvatsim
  • jtvatsim
OK, I'm going to go ahead and use technology on my side to see the graph. Hold on one moment.
anonymous
  • anonymous
Alright, thank you!
jtvatsim
  • jtvatsim
OK, here's what I see:
1 Attachment
anonymous
  • anonymous
Your graph is much more proportional than mine,haha, and I see the curve now
jtvatsim
  • jtvatsim
Well, I cheated with technology. :) Alright, so we see the curve now.
jtvatsim
  • jtvatsim
It's actually very straight, so a line should do a good approximation.
anonymous
  • anonymous
So its linear? is that the relationship?
jtvatsim
  • jtvatsim
Yep, one of the few words in math that actually makes some sense. :)
jtvatsim
  • jtvatsim
Alright, so now for part b).... whoa, hang on... this is a lot simpler than least square regression... you won't need that level of complexity...
anonymous
  • anonymous
So how do we make this into a linear equation? Take a random pair of coordinates?
anonymous
  • anonymous
Yeah I don't know. Our teacher had as watch a video about it before answering this question.
jtvatsim
  • jtvatsim
Yeah, probably not needed for this one at least... your intuition into taking a random pair of coordinates is the best move. Actually the question says to use the points (520,42000) and (0,0)
jtvatsim
  • jtvatsim
So, they've chosen the random points for us.
jtvatsim
  • jtvatsim
So, part b) wants us to write something for y = mx. The main challenge right now is to find m, the slope. Do you have any ideas based on the points they gave us?
anonymous
  • anonymous
So, 420000=m(520) Because you're just subracting the first coordinantes by the second pair which are zeros
jtvatsim
  • jtvatsim
OK, looks good
anonymous
  • anonymous
So, m=807.692
jtvatsim
  • jtvatsim
I think you mean 80.769, I believe it was 42,000 vs. 420,000, but I might be mistaken.
jtvatsim
  • jtvatsim
Actually, wait... The points were (0,0), and (520,40,000) typo on my part. Sorry.
anonymous
  • anonymous
Oops it was 40,000. Guess we dragged the 2 from 520 by mistake
jtvatsim
  • jtvatsim
Yep, that happened. :) So, I'm getting 76.923
anonymous
  • anonymous
yeah haha. So 40000=m(520) So m=76.923
anonymous
  • anonymous
Yay!!
jtvatsim
  • jtvatsim
So our equation is y = 76.923 x, where x is the distance and y is the velocity.
jtvatsim
  • jtvatsim
I'll go ahead and plot this over my previous graph, one second.
jtvatsim
  • jtvatsim
Here we go
1 Attachment
anonymous
  • anonymous
So for part C, I plug in the velocity and y=(76.923)(60,000) ?
jtvatsim
  • jtvatsim
Careful... the velocity goes into the y, not the x.
anonymous
  • anonymous
Almost linear, do I needto change answer A?
jtvatsim
  • jtvatsim
Actually it would be a good idea to write your equation as: v = 76.923 d to keep that clear.
jtvatsim
  • jtvatsim
No need to change your answer. In the real world, there is almost NOTHING that is perfectly described by an equation. It is "close enough" so we would say it is linear.
jtvatsim
  • jtvatsim
Actually, this equation does an amazing job of fitting the data, and any scientist would be very happy with this. :)
anonymous
  • anonymous
Okay, then linear it is. And galaxy Hydra is 780 (megaparsec?) away?
jtvatsim
  • jtvatsim
Let me check.
jtvatsim
  • jtvatsim
Yep, I agree.
jtvatsim
  • jtvatsim
Now to estimate the age of the universe with our slope.
anonymous
  • anonymous
Does the m stand for megaparsec?
jtvatsim
  • jtvatsim
I don't think so, mega is usual a capital M in science. So I'm pretty sure they are referring to m as in the slope of the line. Also, is the formula written like this? \[\frac{9.5 \times 10^{11}}{m}\]
anonymous
  • anonymous
yes
jtvatsim
  • jtvatsim
OK, I think all they want is to just plug in our m = 76..... into the equation.
anonymous
  • anonymous
Do I have to simplify it at all?
jtvatsim
  • jtvatsim
Sure, it will be easier to understand that way... you'll need a powerful calculator though. Your computer calculator can probably handle it or ask google. :)
jtvatsim
  • jtvatsim
I'm getting a really big number...
anonymous
  • anonymous
Me too :(
anonymous
  • anonymous
I'm using mathway, what are you using?
jtvatsim
  • jtvatsim
Just searching directly in Google the following 9.5*10^11 / 76.923 gives me this 12350012350
jtvatsim
  • jtvatsim
Looks like it has a built in calculator, this agrees with my other one, it just dropped a few decimals.
anonymous
  • anonymous
I got 12348888600.026 so... 1.23x10^11 ? Does that sound reasonable?
jtvatsim
  • jtvatsim
Sure, our model isn't perfect anyway, so that's a good estimate.
anonymous
  • anonymous
So I have no lead for what part E wants me to do
jtvatsim
  • jtvatsim
Just real quick I think you want 1.23x10^10
anonymous
  • anonymous
Oh okay. Sounds good. Good to double check, ty
jtvatsim
  • jtvatsim
OK, so E is just a really complicated way of them telling you to do part D again. Except this time use m = 50 and see what you get. Then use m = 100 and see what you get. Then you'll see that this is a range of possible ages for the universe. m = 50 gives the maximum age and m = 100 gives the minimum age.
anonymous
  • anonymous
Ohhh okay!! Thank you so much!!!
jtvatsim
  • jtvatsim
Your welcome! Not so bad hopefully! :)
anonymous
  • anonymous
Not as bad as I thought it would be at all XD guess I'm just afraid of big numbers
jtvatsim
  • jtvatsim
They are scary, but we can still deal with them. Just takes more paper and typing. :) Good work! Take care!
jtvatsim
  • jtvatsim
If you want me to check the other two answers you get, just let me know. In fact, I'll post mine below so you can double check.
jtvatsim
  • jtvatsim
m = 100 -> 9,500,000,000 = 9.5 x 10^9 m = 50 -> 19,000,000,000 = 1.9 x 10^10
anonymous
  • anonymous
Yes! B) Yay team! Have a good *enter time of the day here*
jtvatsim
  • jtvatsim
lol Sounds good, you too!

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