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Mathematics: Post your doubts here!

PlanetMaster

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can you figure out why the gradient of a curve is never negative ?
Can you please provide the question?

Generally, if a question is asking why the gradient of a curve is never negative, you need to find the derivative of y (gradient) and you'll get an equation which can never be negative for e.g.:
\((3x-2)^2,\text{ which is always}>0\)
 
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qr3bdg7t6fd.png

STATISTICS HELP NEEDED
I got the first two parts but I am so confused about solving (iii). Can anyone help me out by providing a good explanation so that I can, too, develop that way of thinking.
 
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PlanetMaster

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STATISTICS HELP NEEDED
I got the first two parts but I am so confused about solving (iii). Can anyone help me out by providing a good explanation so that I can, too, develop that way of thinking.
There are several ways to tackle this kind of question.
I prefer the plain probability way (as opposed to permutations).

So since Mrs Brown is a front row hogger, she could sit in any of the 3 seats from 14 total i.e. 3/14

Now for Mrs Lin to sit directly behind a student, she can sit anywhere except the front row so that's any of the seats highlighted in red.
qr3y5eruer67.png
So that's 11 possible seats but since Mrs Brown has occupied 1 of the 3 seats in front, Mrs Lin can't sit behind her so we have 10 possibilities here i.e. 10/13. (Its 13 because 1 seat is no longer an option and we have 13 to choose from)

Finally, there are 5 students and 12 passengers so the probably of a student sitting in the seat in front of Mrs Lin is 5/12.

So,
\(p=\frac{3}{14}\times\frac{10}{13}\times\frac{5}{12}\)

Basically, we staged the entire scenario step by step here. First, we put Mrs Brown in one of the front seats. Then we put Mrs Lin in one of the possible seats where a student could be in front of her. Then, finally, we forced a student in that seat against their will!

Hope this helps!
 
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There are several ways to tackle this kind of question.
I prefer the plain probability way (as opposed to permutations).

So since Mrs Brown is a front row hogger, she could sit in any of the 3 seats from 14 total i.e. 3/14

Now for Mrs Lin to sit directly behind a student, she can sit anywhere except the front row so that's any of the seats highlighted in red.
View attachment 64998
So that's 11 possible seats but since Mrs Brown has occupied 1 of the 3 seats in front, Mrs Lin can't sit behind her so we have 10 possibilities here i.e. 10/13. (Its 13 because 1 seat is no longer an option and we have 13 to choose from)

Finally, there are 5 students and 12 passengers so the probably of a student sitting in the seat in front of Mrs Lin is 5/12.

So,
\(p=\frac{3}{14}\times\frac{10}{13}\times\frac{5}{12}\)

Basically, we staged the entire scenario step by step here. First, we put Mrs Brown in one of the front seats. Then we put Mrs Lin in one of the possible seats where a student could be in front of her. Then, finally, we forced a student in that seat against their will!

Hope this helps!

It does, Thank You so much!
 

PlanetMaster

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can someone quickly explain convergent and divergent series to me and how to verify? as math
Keeping it as simple as possible, a sequence has a limit that it is converging to if it gets arbitrary close to some value if you go out far enough.

A series is a sum of terms, each of which is from a sequence.
The sum of the first terms of a series up to a point is another sequence called the partial sum.

A series is convergent if its partial sum has a limit.
You can say that a convergent series is equal to some finite value.

A series is divergent if its partial sum has no limit.

Hope this helps but if still unclear, lemme know and I'll be happy to explain in more detail.
 
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Hey guys so it's been close to 22 hours after the AS Maths exam 9709 12, I'm writing down the answers that I remember, please do recheck and let me know if something doesn't match:
1) a= -4
2) 3y+2x=14
3)y=5root(X) - 6


[ I don't remember the order here onwards so I'm just categorizing the questions according to the chapter]

4) the vector question
*PB= 5i+8j-5k
*PQ=4i+8j+5k
* closer to Q
than B
* angle was something like 57.4

5) the circle question
*perimeter= 2r(@+tan@)
(assuming that @ is theta for simplicity lol)
*shaded area was around 34.3 ish

6) calculus-volume question
*proved
that question
*h= root(75)=8.66cm; volume is 1360 (this part might be unnecessary but I calculated it to be on the safe side); Maximum value

7) series question
*37 km
on day 21
*525 km overall
*(geometric series) x=9
*
r=2/3 and therefore t4= 8/3
*
sum to infinity is 27

8) function(s) questions
*k is 7/2
for the line to be a tangent
*-1<x<4 for f(x)>G(x) (I don't remember if it was greater than or less than)
*G{inv}f(x) = 2x^2+8x and so solving gives x=0 and x= -4
*5 (cos^2x)- 2
*a= -2 and b= -7
(this was the question where they asked us to do completing squares) least value as f(x)<= -7
*
I think there was another range question- something like [-2,3] I'm sure abt upper limit being 3 but I dont remember the lower limit

9) calculus question
*dy/dx= 16/ (2x+1)^3
*int. y dx = x+(2/(2x+1))+c
*B was (0,.25) i think and A was (0.5,0) equation of normal was something like y= -0.5x+0.25
* area was (1/16)+(1/2)= 9/16 (i got this wrong, i subtracted instead of adding ugh)

10)trigonometry question
*3tan(2x+1)=1, we had to find the smallest values-- x= 1.23 rad and 2.80 rad

Hope the paper went well!
 
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Hello
I am taking AS further mathematics this year (9231), and I was wondering if anybody can help me by providing resources/books if possible.
Thank You
 

PlanetMaster

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Hello
I am taking AS further mathematics this year (9231), and I was wondering if anybody can help me by providing resources/books if possible.
Thank You
For Further Mathematics, I'd rather watch lectures/videos on YouTube as its mostly Undergraduate Year 1 and 2 Math (from a 4 year degree) and most of the printed resources are just not on par!
If you have any questions or need help with anything for Further Mathematics, feel free to ask here and we'll be happy to help.

Edit: If you still want published resources, here is a list of a few recommended ones: https://bit.ly/2OVfYMT
 
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Please help me to solve the questions

In the holidays Martin spends 25% of the day playing computer games. Martin’s friend phones him once a day at a randomly chosen time.
  1. Find the probability that, in one holiday period of 8 days, there are exactly 2 days on which Martin is playing computer games when his friend phones.
  2. Another holiday period lasts for 12 days. State with a reason whether it is appropriate to use a normal approximation to find the probability that there are fewer than 7 days on which Martin is playing computer games when his friend phones.
  3. Find the probability that there are at least 13 days of a 40-day holiday period on which Martin is playing computer games when his friend phones.
 
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PlanetMaster

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Please help me to solve the questions

In the holidays Martin spends 25% of the day playing computer games. Martin’s friend phones him once a day at a randomly chosen time.
  1. Find the probability that, in one holiday period of 8 days, there are exactly 2 days on which Martin is playing computer games when his friend phones.
  2. Another holiday period lasts for 12 days. State with a reason whether it is appropriate to use a normal approximation to find the probability that there are fewer than 7 days on which Martin is playing computer games when his friend phones.
  3. Find the probability that there are at least 13 days of a 40-day holiday period on which Martin is playing computer games when his friend phones.
1. I like to take the step-by-step approach here. We need to find probability that there are exactly 2 of 8 days \(\text{(}^8C_{2}\text{)}\) on which martin is playing computer games when his friend phones. So there are 2 days when he calls in his gaming time and 6 days when he might call outside this window i.e \(\text{(}0.25^{2}\times0.75^{6}\text{)}\).
So \(p=^8C_{2}\times0.25^{2}\times0.75^{6}\)
\(p=0.3115\)

2. Recall that normal distribution can be used as an approximation to the binomial distribution if X ~ B(n, p) and if n is large and/or p is close to ½, then X is approximately N(np, npq) (where q = 1 - p).
Since \(12\times0.25=3\) is quite small, using normal approximation isn't a great idea here.

3. Recall that \(\text{Z-score}=\frac{X-\mu}{\sigma}\).
Since we need P(X ≥ 13), we can use continuity correction P(X ≥ n) = P(X > n – 0.5) so
\(\text{Z-score}>\frac{12.5-10}{\sqrt{7.5}}=0.91241\)
\(P(X<12.5) = 0.81922\)
\(P(X>12.5) = 1 - P(x<12.5) = 0.18078\)

Hope this helps!
 
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