VCE Methods Question Thread!

[keltingmeith]

I think you've got a typo in P(-dZ>0), but you'd do it through the inverse normal function on your calculator.

[MNM101]

How would u calculate the average speed for this  question :
A jogger runs 10km in 45 mins, rests for 10mins, and then runs for another 8km in 1 hour , calculate the average speed

[IndefatigableLover]

How would u calculate the average speed for this  question :
A jogger runs 10km in 45 mins, rests for 10mins, and then runs for another 8km in 1 hour , calculate the average speed

To calculate the average speed, we simply use the formula: Distance/Time.

However, in this case we have our units mixed up so we'll make sure that everything is done in the right units. Generally because you're working in kilometres, I'll make the end result to be in km/h. To do this, we need to convert the times into 'hours' and not minutes.
From collecting our times, we can see that we get 115 minutes. Divide this amount by 60 and that value will be in hours (that is time).

From there we just sub it into the formula:







Therefore Speed is roughly equal to 9.391 km/h (3dp)

[~V]

Help!
Find the value of a such that the area of the region enclosed by the graph of and the x axis, the y axis and the line x=-0.04 equals 2 units squared.

[MNM101]

How would I do this question , a painter used 4.5 litres of paint to cover a wall with an area of 50m^2, calculate the average rate of paint coverage in m^2/L. I got the answer here 50/4.5=11.11, but part b says that the painter works for 1.5 hours and then rests for half an hour before completing the painting in another hour. It is asking to find the average rate at which the painter paints the wall in m^2/h , I'm stuck on part b , the answer is suppose to be 16.67 but I've been getting different numbers maybe it's cos I worked it out wrong 😖

[Zealous]

How would I do this question , a painter used 4.5 litres of paint to cover a wall with an area of 50m^2, calculate the average rate of paint coverage in m^2/L. I got the answer here 50/4.5=11.11, but part b says that the painter works for 1.5 hours and then rests for half an hour before completing the painting in another hour. It is asking to find the average rate at which the painter paints the wall in m^2/h , I'm stuck on part b , the answer is suppose to be 16.67 but I've been getting different numbers maybe it's cos I worked it out wrong 😖

A troll user did respond to your post with the solution before, but couldn't do so without letting out a personal attack at you (which was completely uncalled for and has been removed). Here's the solution for you:

1.5 hrs + 0.5 hrs + 1 hr = 3 hrs.
He paints 50m^2 in 3 hrs.
Question wants m^2 (total) / hours.
50/3 = 16.667

So, the painter painted the wall at an average rate of 16.67m² per hour.

Hopefully that helps!

[Jason12]

1. do we ever use normal pdf in methods 3/4?

2. I'm confused about the inverse normal and what it measures.
E.g. if a question asks to find the value of c with pr (z > or = to 0. what exactly does it mean?
on Ti nspire CAS when you use inverse normal and it asks for 'area' do you put in the value of z or 1- value of z to find c?

[LiquidPaperz]

can you tell me why its C and E?

thanks

[brightsky]

1. no. useless function. if you input an x-value, the normalpdf function will give you the corresponding y-value using the equation of pdf. but since the y-axis of a pdf represents absolutely nothing, the function is more or less redundant. 
2. the inverse normal function finds you the x-value, the area to the left of which is a specified probability. for example, if you were given Pr(X < k) = 0.8 and asked to find k, then the inverse normal function will help you to do so. you should type the area to the left of x = k (i.e. 0.8 ) into the cell labelled 'area'. bear in mind that area always means area to the left as far as the inverse normal function on the calculator is concerned. so if you were given Pr(X > k) = 0.8, in order to find k, you would have to first deduce from the given equation that Pr(X<k) = 0.2, and then type in 0.2 as your area.

hope this makes sense!

[brightsky]

can you tell me why its C and E?

thanks

the limit is wrong in C. we want to make h as small as possible (i.e. we want to let h --> 0). in C, h --> infinity, which is not what we want. E does not give the gradient at a particular point. it gives you the gradient of the line connecting (x,f(x)) and (x+h, f(x+h)). our objective is to find the gradient of the tangent at (x,f(x)). certainly, if we let h be arbitrarily small, then the gradient of the line connecting (x,f(x)) and (x+h, f(x+h)) would approach the gradient of the tangent at (x,f(x)) (which is the definition of the derivative), but in E there is no 'limit when h --> 0' so it is wrong.

[LiquidPaperz]

Alright thanks! i was a bit unsure, what makes A, B and D right then?

Thanks

[kinslayer]

Alright thanks! i was a bit unsure, what makes A, B and D right then?

Thanks

A and D are the same -- the only difference is that in A, the graph's dependent variable is called f(x) and in D, it is called y.

B is just the definition of the derivative- it will be in your textbook.

[lzxnl]

Z<5/3? There should be a normal cdf function on your calculator. Use that, not inverse norm

Remember that if you use inverse norm, the first number you put in is the area. 5/3 can't be an area; it's greater than 1

[psyxwar]

ok potentially stupid question, but how can a continuous pdf have a mode if the probability of any given x is zero? i mean i know how to find it, it just doesnt make sense

[keltingmeith]

ok potentially stupid question, but how can a continuous pdf have a mode if the probability of any given x is zero? i mean i know how to find it, it just doesnt make sense

Consider the PDF of . Now, if you were to draw infinitely small areas (lengths, really), you would see that the "area" under the peak is larger than all the other "area"s. So, even though realistically the probability of getting that single point is 0, that still has a larger infinitely small area than the others.

Also consider this - if you pick any range a, then the probability of being in that range around is going to be larger than the probability of that range around any other point.

Hence, while not intuitive sense, it does make sense that a mode exists.