## A community for students. Sign up today

Here's the question you clicked on:

## anonymous one year ago 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

• This Question is Closed
1. 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?

2. anonymous

|dw:1437111756117:dw|

3. jtvatsim

No one brave enough to tackle this one? lol

4. jtvatsim

Did you want to go through all the parts?

5. jtvatsim

This actually doesn't look too bad. I'm assuming that your class wants you to do this by hand?

6. anonymous

Yes. I know it should be easy which makes me feel stupid for asking...

7. jtvatsim

No, no worries. It's only easy because I've had to pull my hair out on these in the past.

8. jtvatsim

Have you already gone ahead and graphed the data for part a)?

9. anonymous

Yes. I dont see the relation though

10. jtvatsim

OK, I'm going to go ahead and use technology on my side to see the graph. Hold on one moment.

11. anonymous

Alright, thank you!

12. jtvatsim

OK, here's what I see:

13. anonymous

Your graph is much more proportional than mine,haha, and I see the curve now

14. jtvatsim

Well, I cheated with technology. :) Alright, so we see the curve now.

15. jtvatsim

It's actually very straight, so a line should do a good approximation.

16. anonymous

So its linear? is that the relationship?

17. jtvatsim

Yep, one of the few words in math that actually makes some sense. :)

18. 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...

19. anonymous

So how do we make this into a linear equation? Take a random pair of coordinates?

20. anonymous

Yeah I don't know. Our teacher had as watch a video about it before answering this question.

21. 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)

22. jtvatsim

So, they've chosen the random points for us.

23. 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?

24. anonymous

So, 420000=m(520) Because you're just subracting the first coordinantes by the second pair which are zeros

25. jtvatsim

OK, looks good

26. anonymous

So, m=807.692

27. jtvatsim

I think you mean 80.769, I believe it was 42,000 vs. 420,000, but I might be mistaken.

28. jtvatsim

Actually, wait... The points were (0,0), and (520,40,000) typo on my part. Sorry.

29. anonymous

Oops it was 40,000. Guess we dragged the 2 from 520 by mistake

30. jtvatsim

Yep, that happened. :) So, I'm getting 76.923

31. anonymous

yeah haha. So 40000=m(520) So m=76.923

32. anonymous

Yay!!

33. jtvatsim

So our equation is y = 76.923 x, where x is the distance and y is the velocity.

34. jtvatsim

I'll go ahead and plot this over my previous graph, one second.

35. jtvatsim

Here we go

36. anonymous

So for part C, I plug in the velocity and y=(76.923)(60,000) ?

37. jtvatsim

Careful... the velocity goes into the y, not the x.

38. anonymous

Almost linear, do I needto change answer A?

39. jtvatsim

Actually it would be a good idea to write your equation as: v = 76.923 d to keep that clear.

40. 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.

41. jtvatsim

Actually, this equation does an amazing job of fitting the data, and any scientist would be very happy with this. :)

42. anonymous

Okay, then linear it is. And galaxy Hydra is 780 (megaparsec?) away?

43. jtvatsim

Let me check.

44. jtvatsim

Yep, I agree.

45. jtvatsim

Now to estimate the age of the universe with our slope.

46. anonymous

Does the m stand for megaparsec?

47. 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? $\frac{9.5 \times 10^{11}}{m}$

48. anonymous

yes

49. jtvatsim

OK, I think all they want is to just plug in our m = 76..... into the equation.

50. anonymous

Do I have to simplify it at all?

51. 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. :)

52. jtvatsim

I'm getting a really big number...

53. anonymous

Me too :(

54. anonymous

I'm using mathway, what are you using?

55. jtvatsim

Just searching directly in Google the following 9.5*10^11 / 76.923 gives me this 12350012350

56. jtvatsim

Looks like it has a built in calculator, this agrees with my other one, it just dropped a few decimals.

57. anonymous

I got 12348888600.026 so... 1.23x10^11 ? Does that sound reasonable?

58. jtvatsim

Sure, our model isn't perfect anyway, so that's a good estimate.

59. anonymous

So I have no lead for what part E wants me to do

60. jtvatsim

Just real quick I think you want 1.23x10^10

61. anonymous

Oh okay. Sounds good. Good to double check, ty

62. 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.

63. anonymous

Ohhh okay!! Thank you so much!!!

64. jtvatsim

Your welcome! Not so bad hopefully! :)

65. anonymous

Not as bad as I thought it would be at all XD guess I'm just afraid of big numbers

66. jtvatsim

They are scary, but we can still deal with them. Just takes more paper and typing. :) Good work! Take care!

67. 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.

68. jtvatsim

m = 100 -> 9,500,000,000 = 9.5 x 10^9 m = 50 -> 19,000,000,000 = 1.9 x 10^10

69. anonymous

Yes! B) Yay team! Have a good *enter time of the day here*

70. jtvatsim

lol Sounds good, you too!

#### Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy