Pr(A' intersection B') is equal to 1-Pr(A union B), not 1-Pr(A intersection B). If you consider a Venn diagram, (A' intersection B') is the region outside of the two circles, and (A union B) comprises the three regions inside the circles, so (A union B) is the complement of (A' intersection B'). (A intersection B) and (A' intersection B') are only two of the four regions in the Venn diagram, hence they don't add to one. Having established that Pr(A' intersection B')=1-Pr(A union B), you then need to calculate Pr(A union B) so that you can answer the question. Given that you know Pr(A), can use the same process for Pr(B) and also know Pr(A intersection B), the addition rule is an easy way to find Pr(A union B). I hope that makes sense, it's hard to explain without being able to draw or use maths symbols! Good luck for tomorrow.
No. The domain of a composite function is the domain of the "inside" function. In the case of f(g(x)), g is the "inside" function. So, the domain of the composite function is the domain of g. In part i., we just worked out what the domain of g needed to be in order to ensure that the composite function is defined. The domain of g is (-inf, -3]. This ensures that the range of g is [0, inf), which is the domain of f and hence the composition exists only if the original x-values (ie the domain of the function) are (-inf, -3]. This is hard to explain without visual references. Perhaps you need to go back to my video on composite functions: th-cam.com/video/aNQW8UlzpbU/w-d-xo.htmlsi=hUGOWUvHwyjwAAGj
for question 7bii, why are you able to relate the range of g to the domain of f? why would the range of f give you the range of fog? i just thought they were separate things
The range of f doesn't equal the range of fog, but if you think through how the composition works, it can be a useful stepping stone. The range of fog requires you to work out the range of y=sqrt(x^2+4x+1) when the domain is (-infinity,-3], but if you don't know what the graph of y=sqrt(x^2+4x+1) looks like (which I don't off the top of my head!), then you will need other ways to think it through. When you consider the composition f(g(x)), you start with the domain of g, those x-values are then substituted into g, and g spits out a set of numbers which is called the range of g. Those numbers then become the domain of f (because g is being substituted into f), and are substituted into f which then turns them into the range of f(g(x)). The original input values (domain of g) work their way through both functions (g and then f) to produce the range of fog. This is how a composite function works. So, in this instance, you start with domain g = (-infinity, -3], those values are substituted into g(x)=x^2+4x+3 and are turned into range g = [0, infinity), those values then become domain f and are substituted into f(x)=sqrt(x+1), which will turn them into the values [1, infinity). So, we start with (-infinity, -3] and work those values through the composition f(g(x)) (ie. through g and then through f) to end up with the values [1, infinity). This means that if domain fog = (-infinity, -3], then range fog = [1, infinity). I hope that makes sense. It's hard to explain clearly without access to diagrams, graphs, mathematical notation and general handwaving!
Hi Mustaffa. Yes, Q5 is still relevant in the 2020 adjusted study design. Basic probability (tree diagrams, Venn diagrams, intersections, unions, conditional prob, independence, etc.) are all still part of the course. Q5 can be solved by setting up a tree diagram.
You'll have to take that gripe up with VCAA! 😊 It's the last question, so will always be tough, but this did at least assess core parts of the course content. Sometimes that can go a little off-course when they try to be tricky. To that end, I think it's a fair question for the last question of Exam 1.
Thanks a lot, helped me understand the last question!
For Q8(b), why isn't it correct to do 1-Pr(A intersection B)? Why do you have to use the addition rule?
Pr(A' intersection B') is equal to 1-Pr(A union B), not 1-Pr(A intersection B). If you consider a Venn diagram, (A' intersection B') is the region outside of the two circles, and (A union B) comprises the three regions inside the circles, so (A union B) is the complement of (A' intersection B'). (A intersection B) and (A' intersection B') are only two of the four regions in the Venn diagram, hence they don't add to one. Having established that Pr(A' intersection B')=1-Pr(A union B), you then need to calculate Pr(A union B) so that you can answer the question. Given that you know Pr(A), can use the same process for Pr(B) and also know Pr(A intersection B), the addition rule is an easy way to find Pr(A union B).
I hope that makes sense, it's hard to explain without being able to draw or use maths symbols! Good luck for tomorrow.
@@LMKMaths It does make sense ! Think I just needed to visualize it first...
thankyou so much :)
For Question 7bii, Does the range of fog not change from f(x) because the range of g(x)fi into f(x)?
No. The domain of a composite function is the domain of the "inside" function. In the case of f(g(x)), g is the "inside" function. So, the domain of the composite function is the domain of g. In part i., we just worked out what the domain of g needed to be in order to ensure that the composite function is defined. The domain of g is (-inf, -3]. This ensures that the range of g is [0, inf), which is the domain of f and hence the composition exists only if the original x-values (ie the domain of the function) are (-inf, -3]. This is hard to explain without visual references. Perhaps you need to go back to my video on composite functions: th-cam.com/video/aNQW8UlzpbU/w-d-xo.htmlsi=hUGOWUvHwyjwAAGj
@@LMKMaths Thanks for replying just watched the video. Thanks for all the help really appreciate all the help with the videos during this period
thank you, this was very useful, especially Q7
Glad it helped!
Thank you so so much, highly appreciate your help!
You're welcome! Good luck in the exams over the next 2 days.
for question 7bii, why are you able to relate the range of g to the domain of f? why would the range of f give you the range of fog? i just thought they were separate things
The range of f doesn't equal the range of fog, but if you think through how the composition works, it can be a useful stepping stone. The range of fog requires you to work out the range of y=sqrt(x^2+4x+1) when the domain is (-infinity,-3], but if you don't know what the graph of y=sqrt(x^2+4x+1) looks like (which I don't off the top of my head!), then you will need other ways to think it through.
When you consider the composition f(g(x)), you start with the domain of g, those x-values are then substituted into g, and g spits out a set of numbers which is called the range of g. Those numbers then become the domain of f (because g is being substituted into f), and are substituted into f which then turns them into the range of f(g(x)). The original input values (domain of g) work their way through both functions (g and then f) to produce the range of fog. This is how a composite function works.
So, in this instance, you start with domain g = (-infinity, -3], those values are substituted into g(x)=x^2+4x+3 and are turned into range g = [0, infinity), those values then become domain f and are substituted into f(x)=sqrt(x+1), which will turn them into the values [1, infinity). So, we start with (-infinity, -3] and work those values through the composition f(g(x)) (ie. through g and then through f) to end up with the values [1, infinity). This means that if domain fog = (-infinity, -3], then range fog = [1, infinity).
I hope that makes sense. It's hard to explain clearly without access to diagrams, graphs, mathematical notation and general handwaving!
@@LMKMaths That makes sense, thank you for the reply and also for the helpful videos!
@@101BFFs4eva101 You're welcome! I'm glad you've found the videos helpful. Good luck for the exams tomorrow and Weds!
Is Q5 in the current study design?
Hi Mustaffa. Yes, Q5 is still relevant in the 2020 adjusted study design. Basic probability (tree diagrams, Venn diagrams, intersections, unions, conditional prob, independence, etc.) are all still part of the course. Q5 can be solved by setting up a tree diagram.
Appreciated Thank you
You are very welcome
bro the last questionis so dumb
You'll have to take that gripe up with VCAA! 😊
It's the last question, so will always be tough, but this did at least assess core parts of the course content. Sometimes that can go a little off-course when they try to be tricky. To that end, I think it's a fair question for the last question of Exam 1.