Solved CAIE A Level Math Paper 51 October/November 2023 (9709/51/O/N/2023)
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- Опубліковано 18 вер 2024
- This video shows step by step solution of CAIE A level & AS Mathematics Paper 51 for October/November 2023 with Code: 9709/51/O/N/2023.
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THAT IS ACTUALLY SO HELPFUL MA'AM.
having my cie on 13th
Best of luck!
@@MathWorld-yp9od ma'am can you please link the trickiest s1 papers you solved on videos? would mean alot. thanks
me too bro
Thank you so much for your help !!
Glad it helped!
Thank you for your video, it is really helpful to me.
Glad it was helpful! Thank you for your support
I love your videos thanks a lot
Thank you for your support! Please share this channel with everyone. 🙏
Your help during this dire hours are a Godly blessings.
Thank you for compliment! Please continue supporting this channel by sharing it with your friends.
@@MathWorld-yp9od i did already.
I really wanted to ask u in advance that, it would be a hugeeee help if u uploaded the upcoming 2024 FM papers right after their exams.
I am counting on my maths solely on ur videos. So please do. Will be forever glad.
And personally i got Country highest in Law A's in 2023. I will legally represent u if u ever need help in future.😃
@@GalibHasan-ps5dy Thank you on your request and your offer. I will upload the videos when the papers are published
@@GalibHasan-ps5dy yes miss please consider this gentleman's suggestion we want your help
@@GalibHasan-ps5dy u got country distinction? Which country btw?😄
Miss your videos are very helpful but sometimes I am not able to understand what you are saying so it makes things difficult. Can you please speak a bit clearer that will be very appreciated. Thank you Ma'am.
yes this lady is speaking very wisely, please consider her brilliant suggestion miss
Noted with thanks
@@MathWorld-yp9od that's very kind of you🙃
Hello, can you please solve Paper 52 for October/November 2023? Thank you.
Hi, the video will be available on the day after tomorrow. Thanks
@22:00 for these kind of questions can i just use my calculator to give me the answer or should i solve it manually using the table provided
You need to check your calculator whether you are getting the same answer as in the ms
Hi so in todays exam i did the distribution table wrong, which means the variance is wrong, so would I lose all the marks or will I gain some marks cuz of the method?
It depends on the remark on the mark scheme
Hi, have a question. When do we apply 0.5(n)th value and 0.5(n+1)th value for Median? Same for Quartiles. My doubt is how do we differentiate when to do (n)th value or (n+1)th value.
For grouped data, median=0.5(n)th value, Q1= 0.25(n)th value, Q3=0.75(n)th value.
For ungrouped data, median= 0.5(n+1)th value, then to get the Q1 and Q3, just divide equally from the first half and second half of the data after determined the median.
@@MathWorld-yp9od Alright, understood! Your channel has really been helping me with my preparation. Keep pumping out those videos! 💪
Thanks for your support! Please share this channel for everyone you know. Stay tuned for more upcoming videos 😊
Hi for q3b how do u decide that its ( x> a) and not (x
Least possible weight means minimum weight. I let a be the minimum weight, then the rest of the weights must be greater than a. Since weight is continuous variable, hence x>a
@MathWorld-yp9od oh! Thank you so much
@@MathWorld-yp9od TY this came in the exam today i really appreciate ur reply
@@MathWorld-yp9od but why there is no equal sign? as at least a does not mean a is also included?
@@Iris_is_me Are you referring to q3b? If yes, supposedly at least a means greater than and equals to a, but the weight is normally distributed means it is a continuous random variable, hence with or without equal makes no difference
ma'am at 1:04:01 arent we supposed to use 2P1 and 4P1 instead of Combination 'C', since the question says arrangements. We know the answer is same but isnt that technically write to use P instead of C
In general 4P1 means choose 1 letter from 4 different letters followed by arranging the letter, but since only 1 letter is chosen, basically no arrangement involved. Therefore instead of 4P1, it is same as 4C1 even though we are solving for arrangements.