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Thanks, i noticed that you are now over 100,000 now!

Evaluate   c + 1/a

a + 1/b  =  1             b + 1/c  =  1   

a = 1 - 1/b                   1/c   =   1 - b

a = (b - 1)                    c  =  1 

      _____                          ____

          b                               1 - b

1/a  =    b

          ____

            b - 1

So

c +  1/a  =

   1         +         b

____                ____      =         [factor a negative out of the second denominator  ]

(1 - b)               b - 1

  1        -         b

____          _______     =

1 - b             1 - b

( 1 - b)

______   =

(1  - b)

1

I will only solve (a), so then you can apply what I did in (a) to (b) and learn.

tan(angle)=\(\frac{opposite}{adjacent}\)

Lets say a is MP, and b is MN

So tan is a/b

we know that tan(36) and b=28

So \(tan(36)=\frac{a}{28}\)

Solving for A we will get 20.3

Very nice, heureka.....I Iike this problem  !!!

P(6)   =  (2 on first roll) (4 on second roll) + (4 on first roll)(2 on second roll)   =

               (1/4) (2/3)     + (1/2) (1/3)  =

                   2/12  + 1/6  =

                    1/6  + 1/6  =    

                         1/3

P(7)  = (2 on first roll) (5 on second roll) + (5 on first roll)(2 on second roll)  =

              (1/4) (1/3)    + (1/4) (1/3)  =

                    1/12   + 1/12

                        1/6

P(8)   =  (4 on first roll) (4 on secod roll)  =

                   (1/2) (1/3)  =

                          1/6

P(9)  =  ( 4 on first roll )(5 on second roll) + (5 on first roll)(4 on second roll) =

                (1/2) (1/3)    + (1/4) (2/3)  =

                    1/6   +  2/12  =

                         1/3

So

P(6)  = 1/3

P(7 or 8)  =  1/6 + 1/6  = 1/3

P(9)  = 1/3

"A"   provides a fair division

At (1/x62) ... Shoulda held down the shift key to type a ^ instead of a 6. 

1/x^2

Yep...sorry....I missed a sign!!!!

CPhill ... Shouldn't that –32 be a +32 instead?

+

In a particular game, a spinner with four equally-sized sectors labeled 1, 4, 6, and 8 is spun twice. One turn is considered 2 spins of the spinner.

If the sum of the spins is even, you move forward 6 spaces. Otherwise, you move back 2 spaces.

What is the mathematical expectation for the number of spaces moved in one turn?

A) 1 space forward

B) 1 space backward

C) 3 spaces forward

D) 3 spaces backward

Possible outcomes

(1,1)      (4,4)     (6, 6)  (8,8)

(1,4)      (4, 6)    (6, 8)

(4, 1)     (6,4)     (8,6)

(1, 6)     (4,8)

(6.1)      (8,4)

(1,8)

(8,1)

P(even )  = 10/16  =  5/8

P(odd)  =  (3/8)

So.....the expectation is

(5/8)(6)  - (3/8)(2)   =  30/8 - 6/8  =   24/8  =  3 spaces forward

1) Forty eight slips of paper are numbered 1 through 48 and are distributed among a group of people. A random number generator is used to select a single number between 1 and 48, inclusively.

A fair decision is made using this process.

How many people could be in this group?

A) 5

B) 8

C) 9

D) 10

The number of people must be a divisor of 48.......so   "B"  is the answer.....

14-(-5x)+2z-(-32)+4z-2x  =

14 + 5x+2z +32 + 4z  - 2x     [ combine like terms ]

[14 + 32]  + [ 5x -2x ] + [ 2z + 4z ]  =

46 + 3x + 6z

EDIT....

\(\sum_{n=1}^{\infty} \frac{3n-1}{2^n} = 5\)

1) Considering Option A: prime numbers up to 12 are 2, 3, 5, 7, and 11. The probability of obtaining these is \(\frac{15}{36}\) (sum probabilities: P(2) + P(3) + P(5) + P(7) + P(11) = \(\frac{1}{36} + \frac{2}{36} + \frac{4}{36} + \frac{6}{36} + \frac{2}{36}\)). Multiplying the probability of obtaining a prime number by the given point value of 12 gives us 5 as the expected value of Option A.

Considering Option B: multiples of 4 up to 12 are 4, 8, and 12. The probability of obtaining these is \(\frac{9}{36}\) (sum probabilities: P(4) + P(8) + P(12) = \(\frac{3}{36} + \frac{5}{36} +\frac{1}{36}\)). Multiplying the probability of obtaining a multiple of 4 by the given point value of 24 gives us 6 as the expected value of Option B.

Therefore, the answer is C with Option B and the expectation being 6.

Here's a good table for two-dice sum probabilities: http://lh4.ggpht.com/-RSwGhQhT9JI/UblGzz7HNvI/AAAAAAAABRA/8mTpypdmFZA/TwoDiceTable9.jpg?imgmax=800

2) Considering Option A: Sum chart: 

So, \(P(2) = \frac{1}{16}, P(5) = \frac{4}{16}, P(6) = \frac{2}{16}, P(8) = \frac{4}{16}, P(9) = \frac{4}{16}, P(10) = \frac{1}{16}\). Then, \(P(sum > 6) = \frac{4}{16} + \frac{4}{16} + \frac{1}{16} = \frac{9}{16}\) and \(P(sum \leq 6) = 1 - \frac{9}{16} = \frac{7}{16}\). Multiplying the winning probability of \(\frac{9}{16}\) by $8, we get an expected value of $4.50. Multiplying the losing probability of \(\frac{7}{16}\) by $2, we get an expected value of $0.875. Subtracting these two expected values, our overall expectation is $3.625.

Considering Option B: the probability of flipping a coin three times and heads appearing exactly once is \(\frac{3}{8}\). The probability of losing is then \(1 - \frac{3}{8} = \frac{5}{8}\). Multiplying the probability of winning \(\frac{3}{8}\) by $6 gives us an expected value of $2.25. Multplying the probability of losing \(\frac{5}{8}\) by $1 gives us an expected value of $0.625. Subtracting these two expected values, our overall expectation is $1.625.

\(x^2+ax+5a = (x-i_1)(x-i_2)=x^2 - (i_1+i_2)x+i_1i_2 \\ 5a=i_1 i_2\\ i_1+i_2=a\\ 5a = i_1(a-i_1)\\ i_1^2-i_1a+5a=0\\ i_1 = \dfrac{-a\pm\sqrt{a^2-20a}}{2}\\ \sqrt{a^2-20a} \in \mathbb{N}+\{0\}\)

\(a=20\text{ and }a=36 \text{ seem to be the only solutions}\)

CPhill got this answer. I am confused which is correct.

Option A.....possible outcomes

(1, 1)  (2, 2)

(1, 2)  ( 2,5)

(2, 1)   (5,2)

(1, 5)   (5, 5)

(5, 1)

Expected vlalue  = (3/9)(8)  - (6/9)(2)  =   8/3 - 4/3  ≈ $1.33

Option B

Expected value (50/((1/8) - (1)(7/8)  ≈ $5.38

Ptiob B is better by  $ [ 5.38 - 1.33]  =  $4.05

watts to kilowatts

We can also use modular arithmetic here.

Sum up the top and bottom separately and divide:
T = sumfor(n, 1, 2017,(2017 - n) / n) = 14495.83626
B = sumfor(n, 2, 2017,(1 / n) = 7.186830075
T / B = 14495.83626 / 7.186830075 =2017

What the fuck doorknoob?!    Why do you feel the need to inject your opinion on this post or any other?

The asshole has already deleted his question. This post is now useless for anyone wanting to do research on this question. 

“nevermind i got it” means he got his rocks off, but he’s a selfish dick and doesn’t want to return the favor.    

why did you do 23/1000?

\(\text{It hopefully is pretty obvious that it uses}\\ \dfrac{23}{1000}\cdot 336~kw\cdot hr\)

\(1~kw\cdot hr = 3.6 \cdot 10^6~J\\ E=\dfrac{23}{1000}\cdot 336 \cdot 3.6 \cdot 10^6 ~J\)

you can do all the calculator work

This one

We only need to look at the top  two triangles.....they are under the same height.....and the areas of triangles under the same height are to each other as their bases

So.......

x / (6 - x)  = .40      cross-multiply

x =  .40 (6 - x)

x = 2.4 - .4x       

1.4x  = 2.4       

x = 2.4/1.4  =   12/7

Proof :   [ 2*  area of triangle on the left ] /  [ 2 * area of triangle on the right ]  =   

[ 2 * triangle height * base length of left triangle ]  / [ 2 * triangle height * base length of right triangle ] =  

[ base length of triangle on the left ]  / [ base length of triangle on the right ]

[12/ 7 ] / [ 6 - 12/7 ]   =

[ 12/7 ] / [ 42/7 - 12/7 ]  =

[12/7 ] / [ 30/7]   =

[ 12/7] * [ 7/30 ]  =

12/30  =

4/10   =

.40  = 40%

d) is the correct choice.

By the Central Limit theorem the sample average of i.i.d samples from any underlying distribution

will converge in distribution to that of a a normal random variable with mean equal to the mean of the underlying distribution

and variance equal to the variance of the underlying distribution divided by the number of samples taken.