Mathematics: Post your doubts here!

[lara dalal]

Hello guys. I have a confusion when it comes to drawing argand diagrams, do we draw them as approximations or do we use a compass and protactor. Also do we sketch them on graph paper or on normal paper.
Thank you

 

[awesomaholic101]

Hello guys. I have a confusion when it comes to drawing argand diagrams, do we draw them as approximations or do we use a compass and protactor. Also do we sketch them on graph paper or on normal paper.
Thank you

Just sketches. No graph paper needed, no protractor, no compass (unless you can't draw a circle that doesn't look like an egg )

 

[The Sarcastic Retard]

Need help on this question ! I couldn't get the answer after many times of trials,

2010 O/N paper 33 question 8 ii) b)

b)
Let theta = @
@ = 41.41, -41.41
@/2 - 52.24 = -41.41
@ = 21.7

 

[Saad the Paki]

Part (ii). The ms says graph has line of symmetry at x=3 therefore A=6. What?
The completed square form is 2(x-3)^2-5

 

[sj0007]

Part (ii). The ms says graph has line of symmetry at x=3 therefore A=6. What?View attachment 59662
The completed square form is 2(x-3)^2-5

U know how the turning point obtained from this form is the line of symmetry?
So imagine this: The curve passes through zero, reaches its minimum point at x=3, now when we say that its the line of symmetry, we mean that its the exact half and so there will be an exact reflection on the other side........... So the point at which it will cross the x axis again will be 6.........
Something like this:

 

[Saad the Paki]

U know how the turning point obtained from this form is the line of symmetry?
So imagine this: The curve passes through zero, reaches its minimum point at x=3, now when we say that its the line of symmetry, we mean that its the exact half and so there will be an exact reflection on the other side........... So the point at which it will cross the x axis again will be 6.........
Something like this:

View attachment 59663

Ooh kay. Thank you!

 

[Saad the Paki]

Q5 (a) both parts

 

[awesomaholic101]

View attachment 59665
Q5 (a) both parts
Why can't it simply be 5! for (i) ?

You have four digits to fill. The first one HAS to be the number 5, otherwise you won't get a number between 5000 and 6000.
i) 3 digits left, 5 numbers to choose from. And since none can be repeated, you can choose from 5 numbers for the 2nd digit, from 4 for the 4th and from 3 for the 5th.
So the number of possibilities = 5*4*3 (aka 5P3) = 60

ii) If digits can be repeated, it means you can choose from all 6 numbers for the 3 digits left.
So, no. of possibilities: 6^3 = 216

 

[Saad the Paki]

You have four digits to fill. The first one HAS to be the number 5, otherwise you won't get a number between 5000 and 6000.
i) 3 digits left, 5 numbers to choose from. And since none can be repeated, you can choose from 5 numbers for the 2nd digit, from 4 for the 4th and from 3 for the 5th.
So the number of possibilities = 5*4*3 (aka 5P3) = 60

ii) If digits can be repeated, it means you can choose from all 6 numbers for the 3 digits left.
So, no. of possibilities: 6^3 = 216

Thanks

 

[Saad the Paki]

Q3 part (ii) also please

 

[Anum96]

μ = 100 min
σ = 7 min


X -> journey times of a certain car in the given journey.
Let the lower limit of times to be termed as standard to be μ - a and upper limit to be μ + a
P(μ - a < X < μ + a)
standardizing X, using Z = X - μ /σ

P( -a / 7 < Z < a / 7) = 0.34
P(Z < a / 7) - P(Z < -a / 7) = 0.34
ϕ (a/7) - ϕ(-a/7) = 0.34
2 ϕ(a/7) - 1 = 0.34
ϕ(a/7) = 1.34/2

Use normal distribution tables,

a = 3.08

So the upper limit of time will be 100+3.08 = 103.08 ~ 103.1 min
lowe limit of time wll be 100-3.08 =96.92 ~ 96.9 min

Credit goes to : Rizwan Javed

 

[Saad the Paki]

μ = 100 min
σ = 7 min


X -> journey times of a certain car in the given journey.
Let the lower limit of times to be termed as standard to be μ - a and upper limit to be μ + a
P(μ - a < X < μ + a)
standardizing X, using Z = X - μ /σ

P( -a / 7 < Z < a / 7) = 0.34
P(Z < a / 7) - P(Z < -a / 7) = 0.34
ϕ (a/7) - ϕ(-a/7) = 0.34
2 ϕ(a/7) - 1 = 0.34
ϕ(a/7) = 1.34/2

Use normal distribution tables,

a = 3.08

So the upper limit of time will be 100+3.08 = 103.08 ~ 103.1 min
lowe limit of time wll be 100-3.08 =96.92 ~ 96.9 min

Credit goes to : Rizwan Javed

Wow this is great! Thank you (both of you)

 

[sj0007]

Ooh kay. Thank you!

 

[leenz98]

Someone please explain Q5 part iii. Since X=0 why won't it be 12C0*(0.20)^0 * (0.8)^12
Also any trips and tricks on how to solve the permutations and combination questions like question 6

 

[leenz98]

In the first part why isn't 4C2*9C2*2C2 the right way to get to the answer since we have to choose from 4 9 and 2 types of tress?
And part iiii as well please explain.

 

[leenz98]

For this question p= 5/12 and q= 7/12. Can anyone explain what's wrong with how I am doing it.
1-(12C0*(5/12)^0 * (7/12)^12 + 12C1* (5/12)^1 * (7/12)^11)

 

[Saad the Paki]

For this question p= 5/12 and q= 7/12. Can anyone explain what's wrong with how I am doing it.
1-(12C0*(5/12)^0 * (7/12)^12 + 12C1* (5/12)^1 * (7/12)^11)

It won't be 12C0 or 12C1 it will be 3C0 and 3C1 as the newspaper is delivered to 3 houses not 12

 

[Saad the Paki]

In the first part why isn't 4C2*9C2*2C2 the right way to get to the answer since we have to choose from 4 9 and 2 types of tress?
And part iiii as well please explain.

I suck at permutations and combinations as well so sorry can't help u there

 

[leenz98]

I suck at permutations and combinations as well so sorry can't help u there

Still appreciate the fact that you replied.
We literally studied the whole s1 in like 15 days and I can't make heads or toes of it now.

 

[Rizwan Javed]

In the first part why isn't 4C2*9C2*2C2 the right way to get to the answer since we have to choose from 4 9 and 2 types of tress?
And part iiii as well please explain.

What are the answers? I'll post the solutions if my answers will be correct.