factorial1: Gud evening here...it been a while, hope you all are doing great...?

This's 1 or 2 qt for the gurus in the house.

1. All possible arrangements of the letters of the word REARRANGE are made. If one of these arrangements is chosen at random, what's the probability that it will end in (a) G, (b) R?

2. A man fires at a target, let a = P (he hits the target) = 1/5. Then b = P (he misses the target) = 1 - 1/5 = 4/5. If he fires four times at the target, what is the probability of his obtaining (a) exactly 2 hits, (b) at least 2 hits?


Do have a wonderful night rest.
#Shalom

personal59: Pls help
an equilateral triangle of side 20cm
is inscribed in a circle.calculate the
distance of a side of the triangle from
th centre of the circle. Pls its urgent need d ans tomorrow morning

personal59: Pls help
an equilateral triangle of side 20cm
is inscribed in a circle.calculate the
distance of a side of the triangle from
th centre of the circle. Pls its urgent need d ans tomorrow morning

smurfy:

Here goes...

The number of ways of arranging the letters of the word REARRANGE is 9!/(3!2!2!) = 15120

(a) If the letter G ends a particular arrangement, then the other letters REARRANE can be arranged in 8!/(3!2!2!) = 1680.

So the probability of choosing an arrangement that ends with G is 1/9.

(b)...

factorial1: Here goes the solution...

From the first statement...it can be inferred that each angle in the triangle will be equal since it's an equilateral triangle... So for the angles to be equal...each angle must be 60¤ each. NB: ¤ means degree, now, the triangle is inscribed in a circle. Lemme just try snap how d picture looks like for better understanding... I've solved it and the ans. is 10/root3cm.
Brief explanation now, u draw a line from two adjacent point of the triangle to meet @ the centre of the circle, dat means...the angle @the centr of the circle will be twice the opposite angle which is 60¤ to form 120¤...hope u getting this. Now another small triangle has been formed and u bisect the triangle to two equal part which will make it base length to be 10cm(i.e 20cm/2), pictured below. Since we are told to find the distance of a side of the triangle from the centre of the circle...we make the distance to be x, so using SOHCAHTOA...u get the value of x which is 10/root3(i.e 5.77cm in decimal). Thanks

benbuks: @smurfy..10x.....drop ur 2go i.d.

smurfy:

Sorry but I'm not on 2go.

smurfy:

Here goes...

The number of ways of arranging the letters of the word REARRANGE is 9!/(3!2!2!) = 15120

(a) If the letter G ends a particular arrangement, then the other letters REARRANE can be arranged in 8!/(3!2!2!) = 1680.

So the probability of choosing an arrangement that ends with G is 1680/15120 = 1/9.

(b) There are three R's. Let the R's be replaced with R1, R2 and R3. We now need to see the number of ways each R could end an arrangement.

The number of ways R1 could end an arrangement is 8!/(2!2!) = 10080.

Obviously, I'd get the same value for R1 and R2. Hence, the total number of ways of the arrangement ending with R1, R2 or R3 is 3*10080 = 30240.

But the R's are all the same so R can end an arrangement in 10080 ways.

So probability of the arrangement ending with R is 10080/15120 = 2/3.

QUESTION 2

P = 1/5, Q = 4/5

(a) All possible combinations:

PPPP, PPPQ, PPQP, PPQQ, PQPP, PQPQ, PQQP, PQQQ, QPPP, QPPQ, QPQP, QPQQ, QQPP, QQPQ, QQQP, QQQQ

Only six of these give exactly two hits, i.e. PPQQ, PQPQ, PQQP, QPPQ, QPQP, QQPP.

Clearly, each of the combinations above gives 1/5 * 1/5 * 4/5 * 4/5 = 16/625.

Therefore, the six give 6 * 16/625 = 96/625.

The probability of having exactly two hits is 96/625.

(b) At least two hits imply two hits or more.

PPPP, PPPQ, PPQP, PQPP, QPPP, including the six in (a) above.

This gives 1/625 + 4(4/625) + 96/625 = 113/625.

So, probability of at least two hits is 113/625.

benbuks: U guys shold try ma questions in d previous page na..(p.107)..

Sm1 should try dis..
Factorise x^4 + x^2b^2 +b^4

factorial1: @smurfy....Correct...except qt(1)b. The ans. is'nt 2/3. Here goes the solution to qt(1)b

If one of d' Rs is put @d end, the remaining eight letters can be arranged in 8!/2!2!2! Ways, as there are now two Rs, two Es and two As.

So P (R at the end) = 8!/2!2!2! divided by 9!/3!2!2! = 8!3!/9!2! = 1/3.

smurfy:

Thanks for the correction but are you sure that the answer is 1/3? I'm not doubting your mathematical giganticality, but I just want to be sure. Are you sure?

benbuks: Compute the derivative of..
*] sqrt [1+sqrt(1+ sqrt(x)) ]

**] sqrt[ (x^2 +1)^2 +sqrt (1 + (x^2 +1)^2 ) ]
nb:"sqrt." Means square root

benbuks: Chairman,@laplacian..
Your boy dey hail o.

benbuks: Compute the derivative of..
*] sqrt [1+sqrt(1+ sqrt(x)) ]

**] sqrt[ (x^2 +1)^2 +sqrt (1 + (x^2 +1)^2 ) ]
nb:"sqrt." Means square root

Laplacian:
*) let y=sqrt [1+sqrt(1+ sqrt(x)) ]
now
let u=1+ sqrt(x),
then
du/dx=1/[2sqt(x)]
let v=1+ sqrt(u)
then
dv/du=1/[2sqt(u)]
now
y=sqr(v)
then
dy/dv=1/[2sqt(v)]
from chain rule,
dy/dx=dy/dv*dv/du*du/dx
substitute and d result follows

**) let
y=sqrt[ (x^2+1)^2+sqrt(1+(x^2+1)^2)]

then
y^2=(x^2 +1)^2+sqrt(1+(x^2+1)^2)
or
differentiatin implicitly;
2yy'=(x^2 +1)*2x+

(x^2 +1)*x/sqrt(1+(x^2+1)^2)

benbuks: ......M.O.G...try again o..incorrect.....9ce attempt sha..

rhydex 247: @ benbuks.
ur question d/dx( sqrt[1+sqrt(1+sqrt(x)].
here is my solution.
d/dx( sqrt[1+sqrt(1+sqrt(x)]= d/du[rut(u)]*du/dx... where u=sqrt[1+sqrt(x)]+1... d/du(sqrt(u)= 1/2sqrt(u).
d/dx(1+sqrt[1+sqrt(x)]/2sqrt[1+sqrt(1+sqrt(x)]
differentiate the sum term by term.
d/dx(1)+d/dx(sqrt[1+sqrt(x)]/2sqrt[1+sqrt(1+sqrt(x)].
the derivative of 1 is zero we av
d/dx(sqrt[1+sqrt(x)]/2sqrt[1+sqrt(1+sqrt(x)].
Using the chain rule, d/dx(sqrt[1+sqrt(x)])=d/du(sqrt(u)*du/dx, where u=sqrt(x)+1 and d/du(sqrt(u))=1/2sqrt(u)
d/dx(1+sqrt(x))/2sqrt[1+sqrt(x)]/2sqrt[1+sqrt(1+sqrt(x)].
simplifying the expression gives
d/dx(1)+d/dx(sqrt(x))/4sqrt[1+sqrt(1+sqrt(x)][sqrt[1+sqrt(x)].
which gives
1/2sqrt(x)/4sqrt[1+sqrt(1+sqrt(x)][sqrt[1+sqrt(x)].
simplifying the expression gives
1/8sqrt[1+sqrt(1+sqrt(x)][sqrt[1+sqrt(x)][sqrt(x)].
thats all. all is well

Mr Calculus: plz guys help me out wit dis......
if 3sin2@=5sin2¥....
find @ & ¥ if(tan@=2tan¥)