core help...please :) URGENT

[chemkid_23]

can someone explain where the 141 came from for mc qs 8 core exam1

that isnt the answer is it? Because the y-int is clearly 167

no report clearly says B

[Keo]

can anyone help with 2010 VCAA CORE questions 7, 8 and 12?

i would love you forever   

Q7) It's positive, and the data is spread out so it can't be 0.98 as it isnt close to the regression line
it can't be 0.23 as it isn't spread out enough. So it is B) 0.78

Q8) to find the gradient you choose two easy points on the regression line so lets say (20,167) and (34,185)
we do the formula m= y2-y1/x2-x1
185-167/34-20 = approx 1.286

we can see from the graph that on the x axis 20 is the first value so the y-int can't be 167, so the answer must be B

Q12) The value that is in the middle of the 5 numbers of 359,375,385,378 is 375
thereofre answer is D

[Keo]

Is anyone reading this doing Graphs and Relations? I may need to ask help for a Q if I cannot work it out - am about to attempt it. Stay tuned.

Yes, I do graphs and relations. Will try my best to help

Ohh, thanks! Yea I have no idea. It is question 9 of GR, 2010 exam 1 ^_^ I never got around to asking about this Q and just found the exam in my folder.

oh dear

let h= height of the roof in metres
so we assume at length=b that,

h=1/2*b+3  (1)
h-1=-1/6*b+5  (2)

now we use simultaneous equation to solve this (this is where i use my cas)
so i solved it on my cas and it states b=4.5 and h=5.25
we only need to find the value b so therefore the answer is B
hope it helped guys

haha are you "oh dear"ing at my stupidity? This question just threw me haha, esp. being in G&R. why are those constraints to the right there? you didn't seem to use them in working it out. and why is the 10m there if we didnt need it? or am I just really retarded at this question

nonoon not at that no way!! 'oh dear' because i did this before and its a pain in the ass
i didnt use the constraints as i was just finding b, if that makes sense sorry im not the greatest at explaining online D:

ohk. i hate when they give info that you don't really need and so you think you're wrong cos you haven't used it to find the answer... haha

yes mid year chem exam is a classical example..

[tea.squaredd]

can someone explain where the 141 came from for mc qs 8 core exam1

that isnt the answer is it? Because the y-int is clearly 167

It's 141.

The yint is not 167, look at the axis carefully, its (20,167)

To work out the answer you must work out the equation of the line.

(:

[Deank]

can anyone help with 2010 VCAA CORE questions 7, 8 and 12?

i would love you forever   

Q7) It's positive, and the data is spread out so it can't be 0.98 as it isnt close to the regression line
it can't be 0.23 as it isn't spread out enough. So it is B) 0.78

Q8) to find the gradient you choose two easy points on the regression line so lets say (20,167) and (34,185)
we do the formula m= y2-y1/x2-x1
185-167/34-20 = approx 1.286

we can see from the graph that on the x axis 20 is the first value so the y-int can't be 167, so the answer must be B

Q12) The value that is in the middle of the 5 numbers of 359,375,385,378 is 375
thereofre answer is D

Basically what he said, for Q7 said .78 because even though the line is a strong positive linear line, the data is still quite spread out.

Q8 rise/run, i picked the points (34,185) and (20,167), doing y2-y1/x2-x1, you get 185-167/34-20 = 1.286 = 1.3, thinking logically, you cannot have C as your answer, as 167 is at x=20, subbing in x=20 into the equation, you would get a value greater than 167 for x =20, also, the x-axis starts at 20, not 0, therefore 167 is not the y-int.

[harlequinphoenix]

can someone explain where the 141 came from for mc qs 8 core exam1

that isnt the answer is it? Because the y-int is clearly 167

It's 141.

The yint is not 167, look at the axis carefully, its (20,167)

To work out the answer you must work out the equation of the line.

(:

omg! I am retarded. see people this is what you're up against, me the idiot haha. be confident. I have a feeling ill be making similar mistakes on the exam D:

can someone explain where the 141 came from for mc qs 8 core exam1

that isnt the answer is it? Because the y-int is clearly 167

no report clearly says B

Ohk yea keo explained it haha thanks for that

[spinaway]

can anyone help with 2010 VCAA CORE questions 7, 8 and 12?

i would love you forever   

Q7) It's positive, and the data is spread out so it can't be 0.98 as it isnt close to the regression line
it can't be 0.23 as it isn't spread out enough. So it is B) 0.78

Q8) to find the gradient you choose two easy points on the regression line so lets say (20,167) and (34,185)
we do the formula m= y2-y1/x2-x1
185-167/34-20 = approx 1.286

we can see from the graph that on the x axis 20 is the first value so the y-int can't be 167, so the answer must be B

Q12) The value that is in the middle of the 5 numbers of 359,375,385,378 is 375
thereofre answer is D

Basically what he said, for Q7 said .78 because even though the line is a strong positive linear line, the data is still quite spread out.

Q8 rise/run, i picked the points (34,185) and (20,167), doing y2-y1/x2-x1, you get 185-167/34-20 = 1.286 = 1.3, thinking logically, you cannot have C as your answer, as 167 is at x=20, subbing in x=20 into the equation, you would get a value greater than 167 for x =20, also, the x-axis starts at 20, not 0, therefore 167 is not the y-int.

thankyou both question 8 was so silly i just didn't see that x stared at 20...no mistakes like that tomorrow

[harlequinphoenix]

oh, and for that q you helped me with keo, i don't have a CAS haha.... i have an older graphics calc. would you be able to help me work those equations out manually

[Keo]

oh, and for that q you helped me with keo, i don't have a CAS haha.... i have an older graphics calc. would you be able to help me work those equations out manually

borrow one harlequin!! (just noticed your name is related to a disease)
ill get back to you, some nice atar notes person will do it most liekly for you

[DannyN]

Keo is busy writing up a further trial exam 2, so i'll attend to you with your question regarding module 3 question 9.

As you can see there are two equations
h=L/2 +3
h=-L/6+5
they are asking for value for b, since we know the difference in height for the first and second equation at point b is 1. You substitute b into the two equations

h=b/2+3    and h=-b/6+5

top equation minus second equation=1
(b/2+3)-(-b/6+5)=1  expand
4b/6=3
solve for b and therefore b=4.5 hope that makes sense

[harlequinphoenix]

Keo is busy writing up a further trial exam 2, so i'll attend to you with your question regarding module 3 question 9.

As you can see there are two equations
h=L/2 +3
h=-L/6+5
they are asking for value for b, since we know the difference in height for the first and second equation at point b is 1. You substitute b into the two equations

h=b/2+3    and h=-b/6+5

top equation minus second equation=1
(b/2+3)-(-b/6+5)=1  expand
4b/6=3
solve for b and therefore b=4.5 hope that makes sense

oh, sorry keo! i'll leave you to it. thanks heaps danny I would borrow a cas but i dont actually know how to use them. I can do everything else on my graphics calc just not simultaneous equations. though I guess I could use matrices to do them. thanks!

[DannyN]

Keep asking questions guys! it's the ones you dont know that wil stump you on the exam! =)

[Keo]

Keep asking questions guys! it's the ones you dont know that wil stump you on the exam! =)

f

This guy is a genius, he only loses 1 mark on a spesh exam.. i will kiss your feet dude just to be in your presence

[harlequinphoenix]

Keep asking questions guys! it's the ones you dont know that wil stump you on the exam! =)

hang on. im working through that bloody question still... gah. so what do each of the equations mean exactly? like what exactly are we finding the height of for each?

[chemkid_23]

 qs 9 geo/trig exam 1 2009
how do we do it?