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Hi,
Last one - you never know! -
Show that:
can be expressed as:
Hi,
Could anyone help with this:
Show that:
can be expressed as:
Thanks,
Mitch.
You can use induction.
Is the statement true when n has its least value (i.e. when n = 4)? Yes, as 2^4 = 16 > 3 * 4 = 12
Now assume that the statement is true when n = k where k is any positive integer. So we assume that:
2^k > 3k Call this Equation 1
Now what would this assumption imply for 2^(k+1)?
2^(k+1) = 2 * 2^k > 3k (this last inequality is by equation 1)
So we have shown that IF 2^k > 3k then 2^(k+1) > 3k
Well, 2^k IS greater than 3k if k = 4 and so it must also be when k = 5,6,7.............
Mitch.
At the moment when you pick the first card there are 52 cards in the deck of which 12 are face cards. So the chances of the first card being a face card is 12/52 (12 chances out of 52) which cancels down to 3/13.
Now the chances of picking another face card from the remaining pack compound with the previous answer to give a chance of drawing two face cards of:
3/13 * 11/51 (11 because there are now 11 face cards left and 51 because there are now only 51 cards left)
And so for three face cards = 3/13 * 11/51 * 10/50 = 330/33150 = 11/1105 =0.0099
For the second part, the chances are modified as you would expect to produce:
3/13 * 3/13 * 3/13 = 27/2197 = 0.0122
Binomial Expansion is:
(1 + x) ^ n = 1 + n x + n (n - 1) x^2 / 2 + n (n - 1) (n - 2) x^3 / 6 + ...
Here we must substitue x with ax so we have:
a)
na = -6 => a = -6/n
n(n-1)a^2/2 = 27 => 36n(n-1)/n^2 = 54 => 36(n-1) = 54n =>
35n - 36 - 54n = 0 => -18n = 36 => n = -2
Therefore a = -6/-2 = 3
b)
Coefficient of x^3 = n(n-1)(n-2)a^3 / 6 = -2*-3*-4*a^3 / 6 = -4*a^3 = -4 * 9 = -36
c)
General Binomial is valid when |x| < 1 so here we need |3x| < 1 => |x| < 1/3
For a very "natural" way of handling Infinitesimals (and Transfinites) see:
http://www.tondering.dk/claus/sur15.pdf
This equation has no real solutions. The quadratic formula gives T = [-b +- sqrt(b^2 - 4ac)] / 2a here we have a = 1, b = -16 and c = 65 putting these values in we get:
T = [16 +- sqrt(-4)] / 2 = [16 +- 2*sqrt(-1)] / 2 = 8 +- sqrt(-1) = 8 +- i
So the solutions are:
T = 8 + i
and
T = 8 - i
If you plug these into the initial equation you will see that they satisfy it. Let's just do the first one:
T = 8 + i =>
T^2 = (8 + i)*(8 + i) = 64 - 16i -1 = 63 + 16i
Substituting into the LHS of the original equation yields:
63 + 16i - 128 - 16i + 65 which is clearly zero. Similarly for T = 8 - i
Check out http://www.sosmath.com/algebra/factor/fac01/fac01.html it explains what you need to know.
You have to evaluate the function at the upper limit and subtract its value at the lower limit.
So, we have -(x + B)e^(-x/B)
At the lower limit of 0 this is equal to -B (As e^0 = 1)
At the upper limit this is sequal to 0 as e^(-infinity) = 1/e^(infinity) = 0 (one over big number is small number)
So value of this integral between limits is 0 - -B = B (this is the magnitude of the area of the curve between x = 0 and x = infinity) this region is below the x-axis.
Well,
It's a useful result because it means that if you have such new data and you wish to recalculate the variance, you don't have to do all of the larborious summations again. You simply multiply your old result for variance by ac. Imagine you had to do this many many times on many sets of data. You'd use a computer program, but you'd be foolish to make that program recalculate all of the summations each time. You'd simply multiply each time by the new value of ac and thus save many processor ticks of work.
Further, it shows that the offset of the data by b and d has no affect on the variance. Only the multiplying factors affect the variance. Imagine the bell shaped curve, it has a certain "width" or variance. If you translate it along the x-axis left or right, this variance will not change. When you multiply each datum though, you will squash or stretch the curve depending on whether the multiplying factor is less or greater than one respectively.
I'll try:
Our original integral is:
∫(x/B * e^(-x/B)) dx
Let's take the 1/B that is premultiplying the x outside the integral as it is a constant factor, so we now have:
(1/B) ∫(x * e^(-x/B)) dx Call this Eqn (1). Now we use the formula and so we have 'uv - ∫v * (du/dx)' where:
u = x
v = -B*e^(-x/B)
du/dx = 1.
This gives (for our integral):
-Bxe^(-x/B) - ∫-B*e^(-x/B)
Let's take constant B out of integral giving:
-Bxe^(-x/B) + B∫*e^(-x/B)
Let's do the integration by sight, this leave us with:
-Bxe^(-x/B) - B^2 * e^(-x/B)
Finally -Be^(-x/B) is a common factor of the two terms so lets take it out as a common factor:
-Be^(-x/B) * [x + B]
Remember we have a factor of 1/B multiplying out integral from Eqn (1) so our answer is:
-Be^(-x/B)*[x + B] =
-(x + B)e^(-x/B)
Function 2:
(1/B)*e^(tx)*e^(-x/b) = (1/b)*e^(x(t - 1/b)) So our integral is simply:
(1/B)*S[e^x(t - 1/b)] dx =
(1/B) * (1/(t - 1/b)) * e^(x(t - 1/b)) = (1/(bt - 1)) * e^ ((t - 1/b)x)
Mitch.
Function 1:
Use integration by parts formula:
S(u * dv/dx)dx = uv - S(v * du/dx) dx
Let u = x
Let dv/dx = e^(-x/B)
Then du/dx = 1
and v = -B*e^(-x/B) [By sight]
So S(x/B * e^(-x/B)) dx = 1/B * S(x * e^(-x/B)) dx = (1/B) * [-Bx*e^(-x/B) - B^2 * e^(-x/B)] =
-(x + B)e^(-x/B)
Mitch
TWO
y = (1 - bt)^a
Take logs to base e of both sides =>
ln(y) = a*ln(1 - bt)
Differentiate both side wrt t =>
(1/y)(dy/dt) = a*(-b/(1 - bt)) =>
dy/dt = a*(-b/(1 - bt))*(1 - bt)^a = -ab(1-bt)^(a - 1)
ONE
y = e^(ut + s^2t^2/2)
This is a function of a function - has form y = e^(f(t)) so differential is dy/dx = f'(t)*e^(f(t)) - so we have:
dy/dt = (u + s^2t)e^(ut + s^2t^2/2)
THREE
y = e^(u(e^t - 1)) - same thing again -
dy/dt = (u*e^t - 1)e^(u(e^t - 1))
1. Let's evaluate the determinant:
Remember DET|a1 b1 c1|
|a2 b2 c2|
|a3 b3 c3| = a1(b2c3 - c2b3) - b1(a2c3 - c2a3) + c1(a2b3 - b2a3)
Using this we can calucluate our determinant as:
(x+1)(x-2)(x+3) -0 + 0 = (x+1)(x-2)(x+3) so we straight away see that the only value of x listed for which this is NOT zero is 1. Therefore the answer is a. The link provided by George confirms this.
2. Lets just expand the determinant again using the same formula.
(x-1)[(x-2)(x+3) + 6] + 2(-x - 3 + 6) - 1(-3 -3x + 6) =
x^3 + x^2 - 6x + 6x - x^2 - x + 6 - 6 - 2x - 6 + 12 + 3 + 3x - 6 =
x^3 + 3
So the answer is a or d
Mitch.
You need to use two formula:
v = u + at
and
s = ut + at^2/2
Rearrange the first one to give:
t = (v - u) / a
Now convert your value of 64 ft/sec into metres per second and use this as the value of u in the formula.
a = -10 (gravity acting downwards gives -'ve sign, a is not exactly 10 but I'll use 10 here). Finally v = 0 (the final velocity of the ball at the top of the motion is zero).
Put these values into the formula to get a value for t (the time to get to maximium height) now use the second foumla to find a value for s (the distance travelled). Finally add 6ft to this.
That's it.
Area (FGE) = Area (CDE) - AREA(CDFG)
Now length FG = 12cm which is 9 + half the difference between 9 and 15 (As FG is halfway down AD and BC).
So area of trapezoid CDFG is half sum of parallel sides times height = [(15 + 12) / 2] * 2 = 27 sq cm
Now we need area of CDE. We can easilty see that a formula for the width of this triangle as a function of the height from the bottom of the triangle must be linear, in fact it must be W = 15 - (3/2)*H.
**
This is constructed from realising that the foumula must give 15 when height from bottom (H) = 0, and 9 when H = 4 (also 12 when H = 2). So we can imagine a graph of H (x-axis) versus W (y-axis) and we must have a straight line passing through (4,1) and (2,0.5) and we can therefore find its equation easily as y = x/4 so our formula is W = 15 - (H/4)*(15 - 9) = 15 - (3/2)*H (as above)
**
So rearrange this to make H the subject and insert W = 0 (as it does at the top of the triangle) and we get H = 10. So the height of CDE is 10 and the base is 15 therefore area CDE is 0.5 * 10 * 15 = 75 sq cm.
Therefore area of FGE = 75 - 27 = 48 sq cm
1 cubic metre = 1 000 000 cubic cm
1 litre = 0.001 cubic metres
Therefore 1 litre = 0.001 * 1 000 000 = 1000 cubic cm
Now 50 cubic cm of fluid is 80g therefore
20 * 50 = 1000 cubic cm of fluid is 20 * 80 = 1600 g = 1.6 kg
Mitch.
The fallacy is that you are differentiating wrt x, but your function is not definded until x itself is defined, but then x will be a constant. Moreover: in d/dX (X+X+X ..Xtimes) the "function" in the brackets is NOT defined at the time of differentiation and so cannot be differentiated.
For a different way of saying this see http://www.qbyte.org/puzzles/p032s.html
Mitch
Method given here:
http://www.math.com/tables/integrals/more/b%5Ex.htm
Mitch