2^x = x^2 + 7
2^x - x^2 - 7 = 0
Now use the http://www.shodor.org/UNChem/math/Newton/index.html" to find root(s) of f(x) = 0
If the eqn, 2^x=x^2+7, looks simple enough you could try simple substitution.
x=1: 2 = 1 + 7 -- nope, lhs too small
x=2: 4 = 4 + 7 -- nope, lhs too small
x=3...
Start by multiplying both sides by (2-x).
However you must bear in mind that, depending upon the possible values of x, that the term (2-x) could be either positive or negative.
And, when you divide an inequality by a negative number, then you change the direction of the inequality symbol.
Since I can't do this problem, then thanks for saying it seems inpossible as it is.
Factorial function not included, I'm afraid :(, nice one though :)
AFAIK == As Far As I Know.
AFAIAA == As Far As I Am Aware.
I would guess that the order of operands is simply that the use of the operands should follow that order of preference. + - % X ^.
Nice answers. It's just that the integers are only allowed to be used once only - AFAIK!
This is supposed to be a 6th grade problem.
using integers 8 4 2 1 in order, create an equation to equal 18 and another to equal 19. you may use + - % X and exponents (^) in the same order.
Now if the integers and operands (including exponent) are to be in any order then the answers are...
Thanks, I should have mentined that about mu. It is a constant value.
So I was correct after all. :smile:
I guess that's just as well, since I'm already typing up my results based on that conclusion!
Many thanks for the confirmation.
I have a random function f(n) which takes the values +/- 1 with equal probability.
Let the variable X take the sum of the values of f(n) after n steps. Then I can write,
X(T) = \sum_{n=0}^T f(n) where T = 0,1,2,... and X(0) = 0.
And I can write the expectation of X as,
<X> = < \sum_{n=0}^T...
Thanks. Working backwards from that statement makes sense of the previous parts, viz psup(t) = 1/5, 0<=t<=5, zero otherwise.
The mark scheme for this question is,
(a) [5]
(b) [4]
(c) [10]
(d) [6]
For part (c) I'm to use a convolution. I should do,
pret = pser⊕psup
Where I'm...
Is this correct ?
When doing part (b), I got (with some help from elsewhere) that the probability of any given car being suppplied (t<5) is P = t/1000.
I believe that the probability is defined as,
P(t1,t2) = int[t1 to t2] psup(t) dt
where P(t1,t2) is the probability that a car will be...
Thanks. That gives me the right answer.
p(6) works out at 0.11624 which gives 11.18 pro woche.
I've got the answer now, but I still don't understand this stats stuff. I'm pretty good at other maths disciplines, but I just can't seem to follow the "logic" of statistical reasoning.
Anyway...
Here's the full question.
A car rental company signs a contract to take delivery of a new model starting in January 2006, at the rate of n = 1000 cars per year, until the contract ends, after T = 5 years, in January 2011. The cars are removed from service when they are damaged or start to...
Sorry, the pdf I gave is for retiring a car, t years after the start of the program (t=0), where the program is the supply of cars at 1000 cars/year for 5 years. This supply is to be taken as a constant rate.
The pdf that a given car will be retired after t' years of service/use is...
A car rental company receives cars at n = 1000 cars/year for the first 5 years and none thereafter.
The pdf for retiring a car is,
p(t) = (1/5)(1 - exp(-t/3)) , 0<t<5
= (1/5)(exp[-(t-5)/3] - exp(-t/3)) , t > 5
Show that after 6 years cars are being retired from the...