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Not sure you typed those equations correctly.......please put the question in again.....

2/7 - 1/7 =  (2-1)/7 = 1/7  = 1 divided by 7 = .142857

Tangent = opposite / adjacent      tangent 75 = 28/x     x =  28 / tan75 = 7.50 = ~ 8 feet

Two sides   18 x 13 x 2 =468

Two Ends    1/2 x (12 x 5)  x 4= 120

Floor             10 x 18 =180

= 768 sq  ft   total

OK guys and gals. Here is the EXACT probability!:

55,950,584,378,441,149,993,810,452,680 / 2^100=0.04413722864042168774067859............etc.

I'm done.

Thanks very much Alan that is about what I would have expected since I knew I had double counted.

I have done questions like this recently and they were correct but I do it by logic not by formula.

I think this one might be too long to do that way.

Nauseated probably has a formula that will work for this but I don't know it. :/

Nauseated's formulas are excellent especially if you want to program them for different values.

I like Alan's method.  Simulations are great.  They can be used to check logic and formula outputs. Always a good thing when dealing with probablility.

(1+3+7+15+31+..........+n-th term), it is a G.P series. Now find the sumtotal upto n-th term.

This is NOT a GP series.

Guest #2 seems to know whats what.   (I'm not talking about U.S. politics either)

I've just done a Monte-Carlo simulation using 10^6 trials (i.e. tossing 100 coins, one million times and counting in how many trials at least ten consecutive heads appeared).   The result is a probability of approximately 0.044  (i.e. 4.4% of the time you get at least ten consecutive heads).

If \(7 ∣ r\), then \(π = \frac{22}{7}\).

As \(r = 6\), \(6 ∤ 7\). 

\(\therefore π = 3.14\)

\(C = 2πr \\ = 2 \times 3.14 \times 6 \\ = 6.28 \times 6 \\ = 37.68\)

2-(-1)=2+1

2+1=3

3.

TheSmartOne2.0's answers:

a) 30cm

b)90

c)1363

d)2033

e)I'm just that good

x + 2 = y + 3              | - 2

x = y + 1                    | - 1

y = x - 1

If I could edit my answer I would.

I am quite sure that I have double counted however I still say the answer has to be greater then    0.5^10

MWizzard has asked me to comment on this answer.  At least I think that is what I was expected to do ???

Anyway.

what is the probability of at least 10 consecutive heads if a fair coin is tossed 100 times?

I am not really sure...  maybe

The chance of getting all heads in any 10 consecutive rolls is     0.5^10

0.5^10 = 0.0009765625

There are a total of 90 groups of 10 consecutive rolls so maybe the prob is

90*0.5^10

90*0.5^10 = 0.087890625

I maybe be double counting here.    ://

BUT

The probability has to be greater than 0.0009765625  so I do not think yours is correct.

Maybe another mathematician could comment here.  

Alan, could you run this through a simulator maybe, I don't know how to do that but maybe your do????

Hi guest,

We do not mind a few social posts but please

If there are other social posts on the fist page then add yours underneath.

If there are no social posts on the page then you can make one for yourself.

Also the moderators in general tend to be more tolerant of social posts if they are made by members.  It is easy to become a member, you just need a username, password and valid email address.  (At least that is all that was needed last time I looked)  Plus I think your username is only ever used to confirm your membership and as a backup if you forget your password.

A randomiser of n bits (base 2 unsigned)

When n ≥ 2, we have 2ⁿ possible ways to store numbers.

Number of consecutive Heads:

10 heads + 90 random (1st ~ 10th try, 2nd ~ 11th try ... 91st ~ 100th try)

91 tries × 2⁹⁰ = 112 652 543 574 969 605 015 820 304 384

11 heads + 89 random (1st ~ 11th, 2nd ~ 12th ... 90th ~ 100th try)

90 tries × 2⁸⁹ = 55 707 301 767 842 112 370 460 590 080

12 heads + 88 random

89 tries × 2⁸⁸ = 27 544 165 874 099 711 116 505 513 984

...

The information above is an assumption; we do it by working backwards and finding a pattern.

97 heads + 3 random (97H, T, T, T; 97H, T, T, H; 97H, T, H, T; 97H, T, H, H  )

4 tries

98 heads + 2 random (98 heads, 1 tail, 1 head or 98 heads, 2 tails)

2 tries

99 heads + 1 tails

1 try

100 heads

1 try

                       100H   99H         98H            97H        96H            95H                  11H                  10H

Thus, we have 1 + (2¹ - 2⁰) + (2² - 2¹) + (2³ - 2²) + (2⁴ - 2³) + (2⁵ - 2⁴) + ... + (2⁸⁹ - 2⁸⁸) + (2⁹⁰ - 2⁸⁹) = 2⁹⁰ = 

                        ||

                        2⁰

1 237 940 039 285 380 274 899 124 224 ways.

Do some cancellation. Remove the brackets as there is only + outside of brackets, and a + (b - c) = a + b - c.

Thus, the probability of getting at least 10 consecutive heads is:

Percentage : \(8.0779356694631608 \times 10^{-28} %\)

Fraction : \(1 \over 1 237 940 039 285 380 274 899 124 224\)         

My Calculator

Brand: Casio

Model: fx-96SG PLUS

Type: Natural-V.P.A.M.

Digit Display: 10 digits

Calculation Input: 15 digits

Calculation Range: \(1 \times {10}^{-99} ≤ |x| < 1 \times {10}^{100} \, or \, 0\)

Hi Gyrst,   it is nice to meet you :)

G = y 1-(1+r)^-n/r

\(G = y_1-\frac{(1+r)^{-n}}{r}\\ G - y_1=-\frac{(1+r)^{-n}}{r}\\ -r(G - y_1)=(1+r)^{-n}\\ (ry_1-Gr)=(1+r)^{-n}\\ ln(ry_1-Gr)=ln(1+r)^{-n}\\ log(ry_1-Gr)=log(1+r)^{-n}\\ log(ry_1-Gr)=-nlog(1+r)\\ \frac{log(ry_1-Gr)}{log(1+r)}=-n\\ -n=\frac{log(ry_1-Gr)}{log(1+r)}\\ \)

84=(Pi*6.7) + (2*6.7) + (2*a)

Solve for a:
84 = 2 a+34.4487

84 = 2 a+34.4487 is equivalent to 2 a+34.4487 = 84:
2 a+34.4487 = 84

Subtract 34.4487 from both sides:
2 a+(34.4487-34.4487) = 84-34.4487

34.4487-34.4487 = 0:
2 a = 84-34.4487

84-34.4487 = 49.5513:
2 a = 49.5513

Divide both sides of 2 a = 49.5513 by 2:
(2 a)/2 = 49.5513/2

2/2 = 1:
a = 49.5513/2

49.5513/2  =  24.7757:
Answer: |  a = 24.7757

Hi AaratikRoy :)

If a,b,c,d are in G.P ,then prove that-

(i) (a-b)^2 , (b-c)^2 , (c-d)^2 are in G.P 

If a, b, c, d are in GP then there there is a real number r such that

\(a\\ b=ar\\ c=ar^2\\ d=ar^3\\\)

also

\(\frac{b}{a}=\frac{c}{b}=\frac{d}{c}\\ so\\ b^2=ac\qquad c^2=bd\)

Now I will look at the sequence in question (i)

(i) (a-b)^2 , (b-c)^2 , (c-d)^2 are in G.P 

This sequence is a GP if it can be shown that

\(\frac{(b-c)^2}{(a-b)^2}=\frac{(c-d)^2}{(b-c)^2}\\ LHS=\frac{b^2+c^2-2bc}{a^2+b^2-2ab}\\ LHS=\frac{ac+c^2-2bc}{a^2+ac-2ab}\\ LHS=\frac{c(a+c-2b)}{a(a+c-2b)}\\ LHS=\frac{c}{a}\\ LHS=\frac{ar^2}{a}\\ LHS=r^2\\\)

\(RHS=\frac{(c-d)^2}{(b-c)^2}\\ RHS=\frac{(c^2+d^2-2cd)}{(b^2+c^2-2bc)}\\ RHS=\frac{(bd+d^2-2cd)}{(b^2+db-2bc)}\\ RHS=\frac{d(b+d-2c)}{b(b+d-2c)}\\ RHS=\frac{d}{b}\\ RHS=\frac{ar^3}{ar}\\ RHS=r^2\\ RHS=LHS\\ \therefore\\ \mbox{The terms in question (i) make a GP}\)

Yo can't sum up a GP to the "nth term". You have to specify what the "nth term" is. Is it the 10nth term, is it the 20th term, the 30th term........ and so on.