VCE Methods Question Thread!

[chemdeath]

Ok, sorry, my bad.

Question 3:
I forgot that the probability of getting a 6 for the biased coin is 0.1*1/3
So the answer is (0.1*1/3)/(13/90)=3/13

That's alright better than i did! Thanks

Question 4:
Solve: -(10+x)*13/90+(1-13/90)*x=0<=>x=$2.03

That's what i got as well but apparently the answers $1.69? 

[09Ti08]

They must have solved this: -10*13/90+(1-13/90)*x=0
Actually, I find the wording of this question very confusing. From what they say, initially I think that Sarah is not the player, but then the bit in the braket confuses me, why don't they just use the word "player" instead of "she"?

[chemdeath]

They must have solved this: -10*13/90+(1-13/90)*x=0
Actually, I find the wording of this question very confusing. From what they say, initially I think that Sarah is not the player, but then the bit in the braket confuses me, why don't they just use the word "player" instead of "she"?

Yeah i know!! Thanks though

[chemdeath]

d: Conditional probability. Find the interesection between the two events. If the first is the sum of rolls is six, the second event being one of the rolls is a three, the intersection will be when the sum of a roll is six and one of the rolls is a three (which in this case can only happen once, if both dies are three), then set up conditional probability like usual.
Let me know if i made any errors.. it's kinda late and I'm tired =p

I can't get this written up right... 

[Zealous]

I can't get this written up right... 

=

=

more detail.

Spoiler

The probability of event A given B, is equal to the probability of both event A and B occuring (the intersection between the events), divided by the probability of event B occuring.

In your question, event A is rolling a 3, and event B is the sum of two rolls is 6.
The probability of both event A and B occuring will only happen when on both rolls, the die shows 3 (the sum is 6 when there are two 3's and it satisfies the criteria that one roll is a 3).

[zhe0001]

prease help,

The height (in cm) that a clock’s pendulum swings above its base can be approximated by the function,
at any time t seconds after being released.

f. Find the number of times the pendulum swings in 1 minute.

The pendulum is found to be losing time and needs its swing adjusted to 75 swings per minute.

g. Find the new function H(t) that approximates the height of the pendulum.

thanks!

[Zealous]

The height (in cm) that a clock’s pendulum swings above its base can be approximated by the function,
at any time t seconds after being released.

f. Find the number of times the pendulum swings in 1 minute.

The pendulum is found to be losing time and needs its swing adjusted to 75 swings per minute.

g. Find the new function H(t) that approximates the height of the pendulum.

f) Find the period of the function f. As t is in seconds, if you divide 60 seconds by the period (also in seconds), you'll get the amount of times the pendulum swings back and forth in 1 minute.

Spoiler

Period =



The pendulum swings 52.5 times per minute or 52 if rounded down to how many swings within a minute.

g) Find the period of the function required in which 75 swings will be made in one minute. So 60 seconds divided by 75 swings. Then substitute it into the period of a sin/cos graph formula and replace the coefficient of t.

Spoiler

So the period will now have to be 0.8 seconds in order for the pendulum to swing 75 times in one minute.

Sub into the formula for the period:





New function:

[Alwin]

prease help,

The height (in cm) that a clock’s pendulum swings above its base can be approximated by the function,
at any time t seconds after being released.

you're missing a 't' variable in that question

f. Find the number of times the pendulum swings in 1 minute.

The pendulum is found to be losing time and needs its swing adjusted to 75 swings per minute.

g. Find the new function H(t) that approximates the height of the pendulum.

thanks!

Assuming the number of times it swings is left - bottom - right - bottom - left, then the number of swings is:


Now, for 75 swings per minute:


If the hight and stuff are the same, then:





EDIT: beaten

[darklight]

Hi guys,

Just a quick question from the ATARnotes Methods book; Probability Tech-Free Test 2.

For q4, it states that Mr.Li will win if he jumps over 13 m in both the heat and the final on any given day. He has learnt that if he jumps over 78 cm in the vertical jump, he will always jump over 13 m, but he has no chance of winning if he does not jump this height ...

Therefore, shouldn't the probability of him winning (ignoring the bits about driving/riding) be Pr (X>78) * Pr (X> 78) because he has to jump over 13 m for both heat and final? In the solutions, it said it has to be X>78 * 1 because there is no consideration of what happens when he doesn't jump over 78 cm ..... [therefore] he always wins when he jumps over 78 cm. I don't understand the connection here. Sure he won't win if he doesn't jump over 13 for the heat but just because he jumps over 13 for the heat, for example doesn't means he will jump over 13 for the final and therefore win.

Can someone please explain this to me?

[BubbleWrapMan]

Hi guys,

Just a quick question from the ATARnotes Methods book; Probability Tech-Free Test 2.

For q4, it states that Mr.Li will win if he jumps over 13 m in both the heat and the final on any given day. He has learnt that if he jumps over 78 cm in the vertical jump, he will always jump over 13 m, but he has no chance of winning if he does not jump this height ...

Therefore, shouldn't the probability of him winning (ignoring the bits about driving/riding) be Pr (X>78) * Pr (X> 78) because he has to jump over 13 m for both heat and final? In the solutions, it said it has to be X>78 * 1 because there is no consideration of what happens when he doesn't jump over 78 cm ..... [therefore] he always wins when he jumps over 78 cm. I don't understand the connection here. Sure he won't win if he doesn't jump over 13 for the heat but just because he jumps over 13 for the heat, for example doesn't means he will jump over 13 for the final and therefore win.

Can someone please explain this to me?

I believe the solution is botched. The part about not considering when he loses is confusing since the question wasn't worded in a completely logical way. In any case, I think you're right; the probabilities should be squared.

[Frozone]

I'm sorry that this isn't  unit 3/4 but 1/2 doesn't have a question thread.

The curve with the equation y=ax^2+bx has a gradient of 3 at the point (2,-2).
A) find the values of a and b
B) find the coordinates of the turning point.


Thanks in advance.

[clıppy]

I'm sorry that this isn't  unit 3/4 but 1/2 doesn't have a question thread.

The curve with the equation y=ax^2+bx has a gradient of 3 at the point (2,-2).
A) find the values of a and b
B) find the coordinates of the turning point.


Thanks in advance.

The point (2,-2) implies that


Or if we divide by 2
is our first equation

It has a gradient of 3 at x = 2 so,

Sub in our point of (2,3) because now it's a gradient function

is our second equation

Simultaneously solve for a and b.
You should get and

For part b)

Spoiler

If we know that we can do a bunch of things to find the turning point. We could complete the square, find the two x-ints and take the middle or use calculus.
Since we already found the derivative in terms of a and b we can just use that.


Solve for when


Sub into your original equation to find the y-value

Hopefully that clears some things up (and that my math is correct)

[shadows]

A. Let dy/dx equal to 3 (As dy/dx is the gradient of the function at a given x value (at x=2)

It also passes through (2,-2)
Sub x=2 into the function, and let that equal to -2.

You will get two equations in terms of a and b.
Solve them simultaneously to find a and b

B.
T.P (and stationary points) occur at the x value when dy/dx=0.

Just sub that x value into the original parabola to find the y value. (these x and y values are your co-ordinates of the t.p )


Haha BEATEN xD.
Well this is the approach you would take towards these type of questions.  hope i still helped

[JieSun92]

Hey everyone

I need help with a binomial distribution question.

Rex is shooting at a target. His probability of hitting the target is 0.6. What is the minimum number of shots needed for the probability of Rex hitting the target exactly five times to be more than 25%?

Pls help.

[lzxnl]

Hey everyone

I need help with a binomial distribution question.

Rex is shooting at a target. His probability of hitting the target is 0.6. What is the minimum number of shots needed for the probability of Rex hitting the target exactly five times to be more than 25%?

Pls help.

Probability that he hits the target exactly five times is nC5 * (0.6)^5*(0.4)^(n-5).
You want this to be more than 0.25...so keep guessing values of n until this probability is greater than 0.25. There's not much you really can do.