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

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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

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(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?
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No one brave enough to tackle this one? lol

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Other answers:

Did you want to go through all the parts?
This actually doesn't look too bad. I'm assuming that your class wants you to do this by hand?
Yes. I know it should be easy which makes me feel stupid for asking...
No, no worries. It's only easy because I've had to pull my hair out on these in the past.
Have you already gone ahead and graphed the data for part a)?
Yes. I dont see the relation though
OK, I'm going to go ahead and use technology on my side to see the graph. Hold on one moment.
Alright, thank you!
OK, here's what I see:
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Your graph is much more proportional than mine,haha, and I see the curve now
Well, I cheated with technology. :) Alright, so we see the curve now.
It's actually very straight, so a line should do a good approximation.
So its linear? is that the relationship?
Yep, one of the few words in math that actually makes some sense. :)
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...
So how do we make this into a linear equation? Take a random pair of coordinates?
Yeah I don't know. Our teacher had as watch a video about it before answering this question.
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)
So, they've chosen the random points for us.
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?
So, 420000=m(520) Because you're just subracting the first coordinantes by the second pair which are zeros
OK, looks good
So, m=807.692
I think you mean 80.769, I believe it was 42,000 vs. 420,000, but I might be mistaken.
Actually, wait... The points were (0,0), and (520,40,000) typo on my part. Sorry.
Oops it was 40,000. Guess we dragged the 2 from 520 by mistake
Yep, that happened. :) So, I'm getting 76.923
yeah haha. So 40000=m(520) So m=76.923
Yay!!
So our equation is y = 76.923 x, where x is the distance and y is the velocity.
I'll go ahead and plot this over my previous graph, one second.
Here we go
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So for part C, I plug in the velocity and y=(76.923)(60,000) ?
Careful... the velocity goes into the y, not the x.
Almost linear, do I needto change answer A?
Actually it would be a good idea to write your equation as: v = 76.923 d to keep that clear.
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.
Actually, this equation does an amazing job of fitting the data, and any scientist would be very happy with this. :)
Okay, then linear it is. And galaxy Hydra is 780 (megaparsec?) away?
Let me check.
Yep, I agree.
Now to estimate the age of the universe with our slope.
Does the m stand for megaparsec?
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? \[\frac{9.5 \times 10^{11}}{m}\]
yes
OK, I think all they want is to just plug in our m = 76..... into the equation.
Do I have to simplify it at all?
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. :)
I'm getting a really big number...
Me too :(
I'm using mathway, what are you using?
Just searching directly in Google the following 9.5*10^11 / 76.923 gives me this 12350012350
Looks like it has a built in calculator, this agrees with my other one, it just dropped a few decimals.
I got 12348888600.026 so... 1.23x10^11 ? Does that sound reasonable?
Sure, our model isn't perfect anyway, so that's a good estimate.
So I have no lead for what part E wants me to do
Just real quick I think you want 1.23x10^10
Oh okay. Sounds good. Good to double check, ty
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.
Ohhh okay!! Thank you so much!!!
Your welcome! Not so bad hopefully! :)
Not as bad as I thought it would be at all XD guess I'm just afraid of big numbers
They are scary, but we can still deal with them. Just takes more paper and typing. :) Good work! Take care!
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.
m = 100 -> 9,500,000,000 = 9.5 x 10^9 m = 50 -> 19,000,000,000 = 1.9 x 10^10
Yes! B) Yay team! Have a good *enter time of the day here*
lol Sounds good, you too!

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