Water's Urgent Math Methods Question:

[Water]

Okay, say for instance it is

square root 2x


Then, would it be a dilation factor of 1/2 from the y axis

or a dilation factor of square root 2 from the x axis.



And, if it is the initial, how so?

[m@tty]

It is either one.

[Water]

I have just seen the light, thankyou  :2funny:

[Water]

New Question

If the point (Loge(3) , (82)(3^1/2)(1/9) )           lies on the curve y = 3e^x/a  + e^ -x/a



The solution said trial and error. Is there a method to find out A though?

[kamil9876]

do u mean ? If so just rewrite it as and you should be able to solve for by plugging in the given point. You end up getting:



then multiply both sides by and you should get a quadratic if you let :



(also don't forget that as well)

[Water]

Of the customers who deal with a car rental company, 45% prefer an automatic car. If there are 35 cars avaliable, what is the probability that the company will not be able to meet the demand for automatics cars in a random group of 80 customers?

[onur369]

your on probability already?

[vea]

Of the customers who deal with a car rental company, 45% prefer an automatic car. If there are 35 cars avaliable, what is the probability that the company will not be able to meet the demand for automatics cars in a random group of 80 customers?

45% of 80 customers=36
So 36 customers want an automatic car.
36-35=1
1 customer is unable to get a car, 1/80 is the probability that the company will not be able to meet the demand for the automatic cars.

I'm not too sure about this, do you have the answer? Or can someone confirm this?

[Water]

your on probability already?

Nah, Onur, I'm working ahead for math methods, trying to finish syllabus.


Ohh, vea the book answer says 0.5432. Using Bionomial probability, since thats the chapter xD

[onur369]

your on probability already?

Nah, Onur, I'm working ahead for math methods, trying to finish syllabus.


Ohh, vea the book answer says 0.5432. Using Bionomial probability, since thats the chapter xD

Good stuff, going to try and do the same on the holidays

[jane1234]

Of the customers who deal with a car rental company, 45% prefer an automatic car. If there are 35 cars avaliable, what is the probability that the company will not be able to meet the demand for automatics cars in a random group of 80 customers?

Wow that is a hard question to wrap your head around. Still a bit sketchy... but this is how I think the book got the answer:

Let X = number of people who want automatic
p = probability that people will want automatic (0.45)
n = 80 customers (this is the number of trials)

Thus your binomial would be written like this X~(80, 0.45)

Now you have 35 cars, but you are trying to find the probability that the company wont meet the demand

So that would mean that the number of people who want autos (X) has to exceed 35 (as you don't have more than 35 cars).

So... type into the calculator binomCdf(80,0.45,36,80)
Otherwise written as Pr(X>35) or Pr (X = > 36) <---- that is meant to be a "greater than or equal to sign"

= 0.5432 !!!

Hope this makes sense.

[Water]

A new maths problem >:


Suppose that the probability that an operating machine breaks down by the end of the day is 0.10, and the probability that a broken machine is repaired by the end of the day is 0.55;
Find:

a) The probability that a machine that breaks down on day 0 will not be working until day 5.


This is what I did 0.10 x (0.45) ^3 x 0.55


But the answer is 0.0266


How did the book do it >;?

[jane1234]

A new maths problem >:


Suppose that the probability that an operating machine breaks down by the end of the day is 0.10, and the probability that a broken machine is repaired by the end of the day is 0.55;
Find:

a) The probability that a machine that breaks down on day 0 will not be working until day 5.


This is what I did 0.10 x (0.45) ^3 x 0.55


But the answer is 0.0266


How did the book do it >;?

Do you have the context of the question? ie what part of the probability chapter it was?

[Water]

Markov Chain Chapter: However, comparing run length for Bernoulli sequences and Markov Chain. Or page 585.

[xZero]

ROFL I remember doing this question last yr

a) the machine is already broken at the start of day1 so we don't count the 0.1, so by the end of day1,2,3 and 4 the machine will not be fixed so P = (0.45)^4. By the end of day5 the machine should be fixed so P = (0.45)^4 x 0.55 and I got 0.0226 (hoping for a typo in your answer :/)