Mathematics HL/SL/Studies Help

[dessskris]

what do you not understand?

P(S|R) is the probability of S happening given that R happens.

P(S|R)=P(SnR)/P®

[Hinuku]

what do you not understand?

P(S|R) is the probability of S happening given that R happens.

P(S|R)=P(SnR)/P®

is P(RnS) = 4/5?

If so, than i simply dont know hot to do ii) (talking about the first problem)

[dessskris]

right, for problem #7 I hope you can complete the tree diagram yourself. coincidentally I'll have a probability quiz tomorrow so this is a good practice

part b i)

P(S|R) = P(RnS) / P( R )

P(RnS) = P(S|R) * P( R )

P(RnS) = 4/5 * 1/3

P(RnS) = 4/15

part b ii)

look at the diagram.

P(S) = 1/3 * 4/5 + 2/3 * 1/4

P(S) = you count ok

part b iii)

P(R|S) = answer to question i / answer to question ii

so you count

and what is the other question? can't seem to find it. could you please take a screenshot of it and put it here so it's easier to see the problem? thank you.

[genepeer]

Let the proposition P(n) be true when 8ⁿ≥n³.

For n=1, 8¹≥1³, therefore P(1) is true.

If P(n) is true for some n≥1, i.e. 8ⁿ≥n³ (Hypothesis)

Then

8ⁿ⁺¹ = 8*8ⁿ

≥ 8*n³ (by hypothesis)

≥ (2n)³

≥ (n+1)³

Then P(n+1) is true.

Since we know that P(1) is true, then by Principle of Mathematical Induction, P(n) is true for all n≥1.

Things to note:

-It is important to show where you used the hypothesis to reach the conclusion P(n+1) is true.

-I assumed "true for some n≥1" but concluded "true for all n≥1".

Edited September 15, 2011 by genepeer

[jmw]

Yeah, have you done radians yet? as that may cause some confusion.

But basically, the question is just using fractions of pi becuase it is a sine function, so increments of pi make it much easier to handle.

But as MR.AHM said, pi/4 = 0.785 so just covert the fraction to a decimal using your calculator and go from there!

[Procrastination]

Hey people, I need some help.

-Find the nth term of an arithmetical progression in which a₄= 13 and a₂+a₁₁=41

-In an arithmetical progression of eight terms, the first term and the last sum up 21. The third term is 6. Write down the progression.

IM Seriously STUCKED

Edited September 18, 2011 by Procrastination

[Lero]

Ok so for the a4 = 13, a4 consists of the initial term (a1), and three arithmetic differences (3d).

For example the sequence 1, 3, 5, 7. The arithmetic difference is 2, and a1 = 1, also a4 = 7. We observe that a4 = a1 + 3d.

Applying this to the two statements you have given, therefore a1 + 3d = 13, and (a1 + d) + (a1 + 10d) = 41.

Now just combine like terms and solve simultaneously .

Edit:

For the second question the same principle applies, I'll leave that for you to solve.

Edited September 18, 2011 by Hus

[Procrastination]

Ok so for the a4 = 13, a4 consists of the initial term (a1), and three arithmetic differences (3d).

For example the sequence 1, 3, 5, 7. The arithmetic difference is 2, and a1 = 1, also a4 = 7. We observe that a4 = a1 + 3d.

Applying this to the two statements you have given, therefore a1 + 3d = 13, and (a1 + d) + (a1 + 10d) = 41.

Now just combine like terms and solve simultaneously .

Edit:

For the second question the same principle applies, I'll leave that for you to solve.

Hey thanks! But what do you mean by "combine like terms and solve simunltaneously"?

This makes perfect sense.

[Emmi]

Ok so for the a4 = 13, a4 consists of the initial term (a1), and three arithmetic differences (3d).

For example the sequence 1, 3, 5, 7. The arithmetic difference is 2, and a1 = 1, also a4 = 7. We observe that a4 = a1 + 3d.

Applying this to the two statements you have given, therefore a1 + 3d = 13, and (a1 + d) + (a1 + 10d) = 41.

Now just combine like terms and solve simultaneously .

Edit:

For the second question the same principle applies, I'll leave that for you to solve.

Hey thanks! But what do you mean by "combine like terms and solve simunltaneously"?

This makes perfect sense.

What he means is to combine similar terms, like if you had 2x and 3x after multiplying/distributing you'd combine them to make 5x.

And solve simultaneously just means to solve after combining like terms.

[Procrastination]

Hey guys, i need some help. What's the nth term of a progression if the first five terms (S5) sum up 2,5 and the first eight terms (S8) sum up 5,2?

[Lero]

Procrastination, what do you mean by 2,5? Is that meant to be 25?

[Procrastination]

No Hus, it means that the total sum of the first five terms in 2.5 .

It is an arithmetical progression.

[Lero]

If S5 = 2.5 and S8 = 5.2, I got the nth term to be (1/10)n + (1/5) or 0.1n + 0.2.

The sequence looks something like: 0.3, 0.4, 0.5 .....

If this is correct let me know and I can show step by step. Not 100% sure.

[Procrastination]

Yes huss you're correct. Could you please show me the step and step that gave you this result?

[Peanut Butter Jelly]

S_n = n/2 [2a₁ +(n-1)d]

S₅ = 5/2 [2a₁ +(5-1)d]

2.5 = 2.5 [2a₁ +4d]

1 = 2a₁ +4d ______equation 1

S₈ = 8/2 [2a₁ +(8-1)d]

5.2 = 4 [2a₁ +7d]

1.3 = 2a₁ +7d _________equation 2

equation 1-equation 2

-0.3 = -3d

d=0.1

sub d=0.1 into equation 1

1 = 2a₁ +4(0.1)

1 = 2a₁ +0.4

0.6 = 2a₁

a₁ = 0.3

Therfore general term for nth term

= a₁+ nd

= 0.3+ n(0.1)

[eelnedross]

Hello, i have a question:

ABCD is a square-based pyramid. E, the apex of the pyramid is vertically above M, the point of intersection of AC and BD. If one of the edges is 200m, find the height of the pyramid.

[Drake Glau]

I'm assuming all edges are 200m...

So from the edge of the square bottom to m would be 200m/2=100m and you know that the hypotenuse edge is 200m so then you just have an a^2+b^2=c^2 problem.

100^2+b^2=200^2

10,000+b^2=40,000

b^2=30,000

sqrt(30,000)=b

Apparently windows vista calculator doesn't have a sqrt button so I leave you with that

[Procrastination]

Hey guys, can you help me by expressing this fraction in the form a (Square root of c) divided b?

4 x(Square root of 45) divided 5 x( Square root of eight).

And by simplyfing the following:

[2 + (square root of 2)] x [3+(square root of five)] x[(square root of five)-2]

-----------------------------------------------------------------------------

[(Square root of five) - 1]x[1+(square root of two)]

[dessskris]

you want it to be (a√c)/b

(4√45)/(5√8)

=(4*3√5)/(5*2√2)

=(2*3√5)/(5√2)

=(√2*3√5)/5

=(3√10)/5

---

(2+√2)(3+√5)(√5-2) / (√5-1)(1+√2)

this is so complicated! I guess you have to expand them...

=(2+√2)(5+√5-6) / (√5-1)(1+√2)

=(2+√2)(√5-1) / (√5-1)(1+√2)

=(2+√2)/(1+√2)

=(2+√2)/(1+√2) * (1-√2)/(1-√2)

=(2+√2)(1-√2)/(1-2)

=(2+√2)(1-√2)/(-1)

=-(2+√2)(1-√2)

=(2+√2)(√2-1)

=(2+√2-2)

=√2

do you have the answer key btw? just to make sure I got it right, because I sometimes make careless mistakes. don't merely copy my work, try to understand why I did it the way I did. if you don't understand, please ask.

[Procrastination]

Yeah Desy you're right, the answear key says that's the correct answear. 1. I'm shocked, how do you type the square root? 2. In the second exercise, why did you erased the /(-1)?