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Thank you very much!!!!!!!

A = 3W2B

B = 4W3B

transfer W from A to B gives: A = 2W2B, B = 5W3B, Probability 3/5

transfer B from A to B gives : A = 3W1B, B = 4W4B, Probability 2/5

Let event X be the event that the moved marble was white

Let event Y be the event that the marble chosen from bag B was white

We want P[X | Y]

\(P[X | Y] = \dfrac{P[Y|X]P[X]}{P[Y]}\)

\(P[Y|X] = \dfrac 5 8\\ \\ P[X] = \dfrac 3 5 \\ \\ P[Y] = \dfrac 5 8 \dfrac 3 5 + \dfrac 4 8 \dfrac 2 5 = \dfrac{23}{40}\)

\(P[X|Y] = \dfrac{\frac 5 8 \cdot \frac 3 5}{\frac{23}{40}} = \dfrac{15}{23}\)

to clarify when you say 1 - x^3 / x^1/2, do you mean 1 - (x^3 / x^1/2) or (1 - x^3) / (x^1/2)?

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If you mean 1 - (x^3 / x^1/2), then it would be (x^0)-(x^(5/2)), assuming x\(\ne\)0.

If you mean (1 - x^3) / (x^1/2), then it would be \(\frac{\sqrt{x}}{x}-x^{\frac{5}{2}}\). However, i cannot find a way to have this answer in the form of x^a-x^b.

the answer is -76

Rom: Have you seen this question? 

https://web2.0calc.com/questions/thanking-in-advance-really-important

Is there a way of solving it, using permutations, combinations, factorials, partitions....etc. other than brute-force computer tabulation of each combination, as is done here in the solution appended to the question? I have tried and failed miserably!.

Thanks for looking at it.

There are 4C2 = 6 ways to select 2 schools.

Given those 2 schools there are 4² = 16 ways to pair up players 1 from each school

There are 3 games played by each of those 3 pairs.

This is a total of 6 x 16 x 3 = 288 games played against other schools.

There are 4C2 = 6 ways to select a pair from the 4 people at a given school.

There are 4 schools

They play 1 game each

This is a total of 6 x 4 x 1 = 24 games played against schoolmates

A total of 288 + 24 = 312 games

Simplify the following:
-23 - -5 + 3 - -20

Multiply -1 and -5 together.
-(-5) = 5:
-23 + 5 + 3 - -20

Multiply -1 and -20 together.
-(-20) = 20:
-23 + 5 + 3 + 20 

Evaluate 5 + 3 + 20 using long addition=28

Subtract 23 from 28.
  =5

Simplify the following:
(-20)/(-5) - (57 + 6)/(-9)

Reduce (-20)/(-5) to lowest terms. Start by finding the GCD of -20 and -5.
The gcd of -20 and -5 is -5, so (-20)/(-5) = (-5×4)/(-5×1) = (-5)/(-5)×4 = 4:
4 - (57 + 6)/(-9)

Evaluate 57 + 6 using long addition.

Reduce 63/(-9) to lowest terms. Start by finding the GCD of 63 and -9.
The gcd of 63 and -9 is 9, so 63/(-9) = (9×7)/(9 (-1)) = 9/9×7/(-1) = 7/(-1):
4 - 7/(-1)

Simplify the sign of 7/(-1).
Multiply the numerator and denominator of 7/(-1) by -1:
4 - -7

Multiply all instances of -1 in -(-7).
(-1)^2 = 1:
4 + 7

=11

simplify  x^(2/3) * x^(1/2) / x^(1/6)
x^(2/3)* x^(1/2)/ x^(1/6)
x^(2/3+1/2)/x^(1/6)
x * x^(1/6)/x^(1/6)
=x

Solve for x:
x^2 + 24 = 0

Isolate terms with x to the left-hand side.
Subtract 24 from both sides:
x^2 = -24

Eliminate the exponent on the left-hand side.
Take the square root of both sides:

x = 2 i sqrt(6)     or         x = -2 i sqrt(6)

Solve for x:
x^2 + 4 x + 13 = 0

Solve the quadratic equation by completing the square.
Subtract 13 from both sides:
x^2 + 4 x = -13

Take one half of the coefficient of x and square it, then add it to both sides.
Add 4 to both sides:
x^2 + 4 x + 4 = -9

Factor the left hand side.
Write the left-hand side as a square:
(x + 2)^2 = -9

Eliminate the exponent on the left-hand side.
Take the square root of both sides:
x + 2 = 3 i or x + 2 = -3 i

Look at the first equation: Solve for x.
Subtract 2 from both sides:
x = -2 + 3 i or x + 2 = -3 i

Look at the second equation: Solve for x.
Subtract 2 from both sides:
 
 x = -2 + 3i     or       x = -2 - 3i

6+4=10

►Y that is the answer

if you didn't know its " X"

Solve for x:
27^x = 3^(2 x + 9/2)

Take the natural logarithm of both sides and use the identity log(a^b) = b log(a):
3 log(3) x = log(3) (2 x + 9/2)

Divide both sides by log(3):
3 x = 2 x + 9/2

Subtract 2 x from both sides:
x = 9/2

Its 2

Use Quadratic Formula

\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\)

a =1   b = -5  c = -8    to find   x =   (5-sqrt57)/2      x = (5+sqrt57)/2

3    collect like terms and simplify

1    given

4    simplify

5    factor ----> equality

2    expand

Take LOG of both sides to obtain:  (I  assume this is  :   2^(x-1) = 4^(x/8)      )

(x-1) LOG 2  = x/8 (LOG 4)    and solve for x

(x-1)(8/x) = LOG4/LOG2

8-8/x =2

-8/x = -6

x= 8/6

Thank you so much... I've given it a go and think I understand it now!!

\(f(x)=(2x^{3/2} -3x^{-3/2} )^2  +5\\ Find \;\;x\;\;when\;\;f(x)=5\\ (2x^{3/2} -3x^{-3/2} )^2  +5=5\\ (2x^{3/2} -3x^{-3/2} )^2  =0\\ 4x^3+9x^{-3}-12x^{0}=0\\ 4x^3+9x^{-3}-12=0\\ 4x^6+9-12x^3=0\\ 4x^6-12x^3+9=0\\ let\;\;m=x^3\\ 4m^2-12m+9=0\\ (2m-3)^2=0\\ m=\pm\left[{\frac{3}{2}}\right]^{0.5}\\ x^3=\pm\left[{\frac{3}{2}}\right]\\ \text{But x is positive so}\\ x^3=\left[{\frac{3}{2}}\right]\\ x=\left[{\frac{3}{2}}\right]^{1/3}\\ \)

1.5^(1/3) = 1.1447142425533319

x= 1.14   (to 3 significant figures)

That it the first bit, the second bit does not look too difficult, give it a try yourself.  :)

Sorry there was a +5

Are you sure that wasn't a +5 on the end? 

\(f(x)=(2x^{3/2} -3x^{-3/2} )^2  -5\\ Find \;\;x\;\;when\;\;f(x)=5\\ (2x^{3/2} -3x^{-3/2} )^2  -5=5\\ (2x^{3/2} -3x^{-3/2} )^2  =10\\ 4x^3+9x^{-3}-12x^{0}=100\\ 4x^3+9x^{-3}-12x=100\\ 4x^6+9-12x^4=100x^3\\ 4x^6-12x^4-100x^3+9=0\\ \)

Thanks guest :)