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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 lightyears) 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 xvalues
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 lightyears) 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 xvalues

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0(cont.) and velocities for the yvalues. 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
 one year ago
Best ResponseYou've already chosen the best response.0dw:1437111756117:dw

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.1No one brave enough to tackle this one? lol

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.1Did you want to go through all the parts?

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.1This actually doesn't look too bad. I'm assuming that your class wants you to do this by hand?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yes. I know it should be easy which makes me feel stupid for asking...

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.1No, no worries. It's only easy because I've had to pull my hair out on these in the past.

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.1Have you already gone ahead and graphed the data for part a)?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yes. I dont see the relation though

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.1OK, I'm going to go ahead and use technology on my side to see the graph. Hold on one moment.

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.1OK, here's what I see:

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Your graph is much more proportional than mine,haha, and I see the curve now

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.1Well, I cheated with technology. :) Alright, so we see the curve now.

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.1It's actually very straight, so a line should do a good approximation.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So its linear? is that the relationship?

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.1Yep, one of the few words in math that actually makes some sense. :)

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.1Alright, 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
 one year ago
Best ResponseYou've already chosen the best response.0So how do we make this into a linear equation? Take a random pair of coordinates?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yeah I don't know. Our teacher had as watch a video about it before answering this question.

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.1Yeah, 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
 one year ago
Best ResponseYou've already chosen the best response.1So, they've chosen the random points for us.

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.1So, 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
 one year ago
Best ResponseYou've already chosen the best response.0So, 420000=m(520) Because you're just subracting the first coordinantes by the second pair which are zeros

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.1I think you mean 80.769, I believe it was 42,000 vs. 420,000, but I might be mistaken.

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.1Actually, wait... The points were (0,0), and (520,40,000) typo on my part. Sorry.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Oops it was 40,000. Guess we dragged the 2 from 520 by mistake

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.1Yep, that happened. :) So, I'm getting 76.923

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yeah haha. So 40000=m(520) So m=76.923

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.1So our equation is y = 76.923 x, where x is the distance and y is the velocity.

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.1I'll go ahead and plot this over my previous graph, one second.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So for part C, I plug in the velocity and y=(76.923)(60,000) ?

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.1Careful... the velocity goes into the y, not the x.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Almost linear, do I needto change answer A?

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.1Actually it would be a good idea to write your equation as: v = 76.923 d to keep that clear.

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.1No 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
 one year ago
Best ResponseYou've already chosen the best response.1Actually, this equation does an amazing job of fitting the data, and any scientist would be very happy with this. :)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Okay, then linear it is. And galaxy Hydra is 780 (megaparsec?) away?

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.1Now to estimate the age of the universe with our slope.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Does the m stand for megaparsec?

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.1I 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? \[\frac{9.5 \times 10^{11}}{m}\]

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.1OK, I think all they want is to just plug in our m = 76..... into the equation.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Do I have to simplify it at all?

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.1Sure, 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
 one year ago
Best ResponseYou've already chosen the best response.1I'm getting a really big number...

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I'm using mathway, what are you using?

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.1Just searching directly in Google the following 9.5*10^11 / 76.923 gives me this 12350012350

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.1Looks like it has a built in calculator, this agrees with my other one, it just dropped a few decimals.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I got 12348888600.026 so... 1.23x10^11 ? Does that sound reasonable?

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.1Sure, our model isn't perfect anyway, so that's a good estimate.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So I have no lead for what part E wants me to do

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.1Just real quick I think you want 1.23x10^10

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Oh okay. Sounds good. Good to double check, ty

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.1OK, 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
 one year ago
Best ResponseYou've already chosen the best response.0Ohhh okay!! Thank you so much!!!

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.1Your welcome! Not so bad hopefully! :)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Not as bad as I thought it would be at all XD guess I'm just afraid of big numbers

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.1They are scary, but we can still deal with them. Just takes more paper and typing. :) Good work! Take care!

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.1If 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
 one year ago
Best ResponseYou've already chosen the best response.1m = 100 > 9,500,000,000 = 9.5 x 10^9 m = 50 > 19,000,000,000 = 1.9 x 10^10

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yes! B) Yay team! Have a good *enter time of the day here*

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.1lol Sounds good, you too!
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