VCE General & Further Maths Question Thread!

[Patches]

You can't reduce the duration of an activity to 0, only to one hour/day/whatever - I think that's your problem.

[plato]

i dont understand how this question is 4 hours, i got it wrong when i did it in timed  conditions, then i went and did it after i finished and got 5, i took the maximum time which is 19, minus by the minimum which is 14 = 5''

(Image removed from quote.)

heres pic, apparently its D-4 hours

See  the following  Further Question Thread!

***EDIT - Corrected the URL

[Green]

how do you suppose solve part c and part d on calculator 2006 vcaa exam 2 squence and series question 3 please

[plato]

The easiest way to solve these is with the Sequence function on the calculator.

***EDIT - For question d, solve the equation where S(n+1) = S(n)

[BLACKCATT]

If we are given a sequence, tn= -3n+44, and the first term is not given.
Would 44 be the first term or 41?

[plato]

There is obviously room for ambiguity here. In a sequence that does not refer to t(n+1) or t(n-1), then assume n will start at one rather than zero.
Your first term would be t(1) = a = 41

[BLACKCATT]

There is obviously room for ambiguity here. In a sequence that does not refer to t(n+1) or t(n-1), then assume n will start at one rather than zero.
Your first term would be t(1) = a = 41

Yeah thanks, i think the initial term would be 44.

[plato]

If t_n=-3n+44 and the first term = t₁=a, then

   t₁=-3x1 + 44 = 41

And, since a = 41 and also t_n = a +(n-1)d, then
  -3n+44 = 41 + (n-1)d
  -3n + 3 = (n-1)d
  (n - 1) x -3 = (n-1)d   and so d = -3

The sequence described by t_n=-3n+44 is an arithmetic sequence where a = 41 and d = -3

[itsalmostover]

If we are told to find the proportion of x after n days, and we end up with a matrix such as:
   _               _
  |   590.18...  |
  |   410.43...  |
  |_ 499.39..._|

Say we started off with 1500 in total at the beginning, rounding to the nearest whole number would only give 590+410+499=1499.

In this case, should we round up the value with the highest decimal so that 410 becomes 411, and we once again have a total of 1500?

[Yacoubb]

If we are told to find the proportion of x after n days, and we end up with a matrix such as:
   _               _
  |   590.18...  |
  |   410.43...  |
  |_ 499.39..._|

Say we started off with 1500 in total at the beginning, rounding to the nearest whole number would only give 590+410+499=1499.

In this case, should we round up the value with the highest decimal so that 410 becomes 411, and we once again have a total of 1500?

Because this is a closed system matrix problem, you must ALWAYS have 1500. Thus, you can choose to round up any of the elements (I'd go with the one closest to being rounded up (e.g. 410.43 over 499.39 because .43 is closer to .5 than .39 is). But, on the exam, its highly unlikely that there will be something like this (or so I've heard from Further Maths lectures I've attended).

[duhherro]

when asked to draw a regression line, do you need to only draw at where the points start and end of the original data ? Or are you meant to extend it so it hits the Y-axis and goes further  than the most furthest x-value on the plot?

And If anyone has MAV 2012 exam 2, how do you get Q3)a) of core? I just did a statistics table and used the given linear regression to get my y-values then i just applied reciprocal on those y-values and then sketched the new regression. Or am I meant to simply look at the given graph and its data and just insert in calc with plain eye?

[Zealous]

And If anyone has MAV 2012 exam 2, how do you get Q3)a) of core? I just did a statistics table and used the given linear regression to get my y-values then i just applied reciprocal on those y-values and then sketched the new regression. Or am I meant to simply look at the given graph and its data and just insert in calc with plain eye?

The data was given in Question 2. So take all the BMI values and find the reciprocal of each one, then fit a linear regression line to the sleep time and 1/BMI.

when asked to draw a regression line, do you need to only draw at where the points start and end of the original data ? Or are you meant to extend it so it hits the Y-axis and goes further  than the most furthest x-value on the plot?

Plot it like you're drawing a linear line. So you can extend (extrapolate) the line past the data given.

[tcstudent]

Hi tcstudent!
The graph is showing the distance Mike travels. During the hours where the graph is in a slope he's travelling at a constant speed, and where the graph is just a horizontal line he is stationary (as his distance is not changing.

a) Average speed is given by the total distance he travels/total time taken.
He travels 16km in 7 hours so, 16/7=2.3 km/h

b) The equation with the variables they want you to find is the equation of the second slope (between 3 to 7 hours).
You can use simultaneous equations to solve this.
The variables from the graph are t (for the x axis) and d for the y axis, (as the equation is in the form y=mx+c). So pick two points on the line, say (5,13) and (3,10) and sub them in to the equation. (Pick clear points so you don't have to estimate).
The two equations I got from these points are:
13=a5+b  and 10=a3+b. Plug them in to your CAS and you get a=3/2 and b=11/2 which in decimals are 1.5 and 5.5.

c) I made a mistake with this one first time I did it.
Sketching the line for the distance traveled by Katie first is a good thing to do. This also helps with the next question!
The line intersects Michael's line at t=2. Therefore they meet at 2 hours.
However just to be super careful incase it's not like 2.1 or something, double check in the exam if you have time!
Use the equation for the first slope in Michael travels (d=5t, get the five from the rise/run of the line) and make it equal the the equation for Katies distance travelled.
Plug in 5t=-3t+16 into the cas to get t=2 hrs.

d) As you can see from the lines on the graph, the distance between Michael and Katie becomes 3km somewhere between 1-2 hours until they finally meet at 2.
So since the distance is the y value here, makes Katies distance-Michael's=3 to find t when their distance apart is 3
-3t+16-5t=3
solve on CAS, t=1.625
After two when they're moving away from each other, you can see just from the graph that at t=3, Katie is at 7km and Michael is at 10km from his destination, difference of 3km.
So they are only 3km near each other from t=1.625 and t=3
3-1.625=1.38 km

Hope that helps!

thank you, however i still don't understand question D about the radio transmitters to talk to each other.

[tcstudent]

Hey guys 2012 Exam 2 , Question 2B about some stupid weather LOl here is the question



Question B

[duhherro]

The data was given in Question 2. So take all the BMI values and find the reciprocal of each one, then fit a linear regression line to the sleep time and 1/BMI.
Plot it like you're drawing a linear line. So you can extend (extrapolate) the line past the data given.

Thanks! And there will always be a range of answers given due to reading variables right?