VCE Methods Question Thread!

[RKTR]

This seems like quite an easy question, but my mind is dead

The probability of Vanessa's car starting on a cold morning is 0.6, while on a normal morning the chance of it starting is 0.9. The probability of any morning being a cold is 0.3. If Vanessa's car starts tomorrow morning, find the probability that the morning is cold.

probability of car starting= 0.3 x 0.6 + 0.7x0.9 = 0.81
probability ( cold n car starting ) =0.18
0.18 / 0.81 = 0.22

[Daenerys Targaryen]

probability of car starting= 0.3 x 0.6 + 0.7x0.9 = 0.81
probability ( cold n car starting ) =0.18
0.18 / 0.81 = 0.22

Can you explain how you got the numbers that aren't in the question? Thanks
#backtoschoolhurstmyhead

[RKTR]

Can you explain how you got the numbers that aren't in the question? Thanks
#backtoschoolhurstmyhead

Let P(cold morning)= P(C) , normal morning = P(C'), car starting =P(S),not starting = P(S')
P(C)=0.3, P (C')=0.7 
the 0.6 is P(S / C)  P(S / C) = P(S n C )  /  P (C)
                               0.6 = P(S n C ) / 0.3
                                  P(S n C ) =0.18

P( S / C')=0.9     0.9= P(S n C') /  0.7
                              P(S n C' ) =0.63

P(S)=0.18 + 0.63 =0.81

the question asking for P ( C / S ) , so P ( C /  S)= P ( C n S ) /  P (S)
                                                                           =0.18/ 0.81
                                                                            =0.22


                               

[b^3]

Let be the event that the car starts.
Let be the event that the morning is cold.
We then know that

Spoiler

Now using conditional probability we have

From the above you can get the result that RKTR has.

Which is also know as the Law of Total Probability.

Hope that helps

EDIT: 22 seconds too late

[RKTR]

i think i really need to learn how to type in latex haha

[Jaswinder]

The graph of a cubic function f has a local maximum at (a, –3) and a local minimum at (b, –8).
The values of c,such that the equation f(x) + c = 0 has exactly one solution, are
A. 3 < c < 8
B. c > –3 or c < –8
C. –8 < c < –3
D. c < 3 or c > 8
E. c < –8

I keep getting b, but its not right :s

[Alwin]

The graph of a cubic function f has a local maximum at (a, –3) and a local minimum at (b, –8).
The values of c,such that the equation f(x) + c = 0 has exactly one solution, are
A. 3 < c < 8
B. c > –3 or c < –8
C. –8 < c < –3
D. c < 3 or c > 8
E. c < –8

I keep getting b, but its not right :s

It looks like you missed a negative somewhere!
f(x)+c=0
f(x)=-c  <-- negative

Full explanation here:

Full explanation

Basically, its asking for values of c when f(x)=-c has only one solution. This can be restated as:
There is only the one intersection between f(x) and the line y=-c

we know that f(x) is a function, so it passes the vertical line test.
for y=f(x) and y=-c to have only one intersection, it must pass the horizontal line test (for that value of c)
Hence, we are looking for one-to-one relation sections of f(x)

looking at a quick sketch of f(x):


we can see this occurs above and below of the turning points
Hence, -c<-8 or -c>-3
Remember that the sign changes when we times through by -1,
  which is D

Note, we could also use vertical translations of the cubic function (graphical approach) but harder to explain and it sounds like you got it! just not the last negative bit

hope it helps!

[Jaswinder]

could someonoe please explain to me how they got the answer to the last question on exam 2 extended response (2root2)?

kinda confused with the formula they write in the assessors report. Thanks

[SocialRhubarb]

So we have two functions, and

If we equate the two:









Using the quadratic formula, the x-intercepts are:



This gives us the points where the two graphs intersect. The midpoint of the two intercepts is then . We also know that the x-coordinate of the midpoint of the two points of intersection is .

Equating the two we get:

[Jaswinder]

alright thanks mate.

how would you write the equation for the area of the graph?  the answers say integrate f(x) - g(x) between m and 0.

but wouldn't you need to consider that some of g(x) is under the x-axis and some isnt and write seperate integrals? :s

not sure. explanation would be appreciated thanks 

[Professor Polonsky]

g(x) is entirely above the x-axis.

While f(x) is not, its signed area would be negative - and subtracting a negative leads to a positive.

What you are doing is finding the area of g(x), and then subtracting the area of f(x) which is a negative, so you get the positive of that area. Essentially, you are adding the area of g(x) and the non-signed area of f(x), the sum of which is the area between g(x) and f(x).

I hope this makes sense

[Jaswinder]

but wouldn't you need to consider that some of g(x) is under the x-axis and some isnt and write seperate integrals? :s

whoops i ment f(x) with that line

and im not quite sure what you mean

what im trying to say is, wouldn't you need two seperate integrals? one from 0 to the x-intercept between 0 and m. and the other integral from the x-intercept to m for the area of g(x)

how can they just  integrate f(x) - g(x) between m and 0

[lzxnl]

If you want, add a constant c to both functions so that both f(x) and g(x) are wholly above the x axis. Now draw yourself a diagram and you'll see clearly that the area between the curves is found by integrating the difference between these new functions, which is f(x) + c - (g(x) + c) which is still f(x) - g(x). Therefore you can just stick to f(x) - g(x)

[Sanguinne]

Checkpoint 131
Vcaa 2011

In an orchard of 2000 apple trees it is found that 1735 trees have a height greater than 2.8 metrers. The heights are distributed normally with a mean u and standard deviation o . The value of u is closest to

[Professor Polonsky]

The standard deviation is 0.2 according to the VCAA question, might be a typo?