Under my original approach, it is impossible.

I read the question again and noticed that you lose an apple for every mile driven. Although we could get 666.66 apples, we lose out on the final 333.33 apples.

I wouldn't be surprised if we could get 888.88 apples using a different approach. I'll think about it over dinner tonight.

i'm pretty sure the optimal strategy involves the maximum number of trips with full loads. since moving forward inherently reduces the number of full loads you can do, the most you could ever do is 6, 3 from the first point, 2 from the second, 1 from the third. i believe 833 is optimal because it uses 6 full loads without ever having leftover.

I can prove 833 is best possible, or 833 1/3 if fractions are allowed. (Beat that.)

hey i was just wondering if you have just the graph of an exponential equation, if theres an easier way to write an equation for it besides educated trial and error. Thanks.

Is the function of the form y=ab^x?

If so, take two points on the graph, (x_0,y_0), and (x_1,y_1). Then you have that y_1/y_0 = b^{x_1-x_0}, so log(y_1/y_0) = (x_1-x_0)log b, or b = exp(1/(x_1-x_0)log(y_1/y_0)). Then determine a from a = y_0/b^{x_0}.

By choosing x=0 and x=1, this is especially easy. For x=0, f(0)=a. For x=1, f(1)=ab, so f(1)/a = b.

EDIT: Of course, you have to have such specific information, which may not be the case.

One of my friends is doing an experiment related to dance therapy and I need some help choosing what statistics test to choose for analyzing it.

She had a group of 9 students. She made them fill out a questionnaire of 10 questions BEFORE the dance therapy sessions took place. Each question had 5 choices, with points given to each choice from 0-4. 0 was the least stressed out, and 4 was the most stressed out. She then did the sessions, and gave them the same questionnaire again.



That is the actual data. She basically wants to know if the therapy actually made a difference that is statistically significant. Which tests would be the most appropriate to figure this out? I was thinking of a paired t-test, but I studied stats so long ago, that I cannot remember.

Thanks a ton!

Edit: I would need to find out if the group overall experienced a change due to the experiment. Also, I would need to find out if any particular question was experienced a statistically significant change due to the experiment.

So basically you have your pre-therapy results. Lets let this be the true mean.

Now calculate the probality of the second mean given the true mean and the standard deviation of the first one.

I.e. use our favorite phi function and standardize the data.

In a short answer it looks like these results arent very significant. Since the second result is only 1/4 std away from the original mean.

How would I do this? Pretend I am a maths noob

Anyone familiar with MATLAB? I will pay you to help me with my homework. I have done most of it (I think correctly as well), but I am having problem just making sure my code is correct. It seems that my .dat files are not saving correctly or something. I am not exactly sure since I am completely new to programming language. If someone would like to look over my code I'll pay you $10 to make sure everything I've done is correct.

(And the homework is extremely simple!)

Here is my code:

Problem #1:
I can't copy and paste the PDF files but its essentially

1000 - the sum from i = 1 to 10000 multiplied by .1 (and the other problems just changed the numbers)

Write a MATLAB algorithm to compute each of the above repeated subtractions and compare the computed answer with the exact answer (IE: exact answer - computed answer) where the exact answer = 0.

sum=0
for i=1:10000
sum=sum+.1
x1=1000-sum
error=abs(x-0)
end

save A1.dat x1 -ascii

sum=0
for i=1:100000
sum=sum+.1
x2=10000-sum
error=abs(x2-0)
end

save A2.dat x2 -ascii

sum=0
for i=1:1000000
sum=sum+.1
x3=100000-sum
error=abs(x3-0)

end
save A3.dat x3 -ascii

Compute:

(a) A + B (b) 2x + y (c) Ax (d) B(x + 2y) (e) Dy (f) AB (g) BC (h) CD

A=[3 1;3 2]
B=[2 4;5 1]

a2=[A+B]
save A4.dat a2 -ascii

x=[1;0]
y=[0;1]

b2=2*x+y
save A5.dat b2 -ascii

A=[3 1;3 2]
x=[1;0]

c2=A*x
save A6.dat c2 -ascii

B=[2 4;5 1]
x=[1;0]
y=[0;1]

d2=B*(x+2*y)
save A7.dat d2 -ascii

D=[8 0;0 3;2 7]
y=[0;1]

e2=D*y
save A8.dat e2 -ascii

A=[3 1;3 2]
B=[2 4;5 1]

f2=A*B
save A9.dat f2 -ascii

B=[2 4;5 1]
C=[2 1 2;3 3 1]

g2=B*C
save A10.dat g2 -ascii

C=[2 1 2;3 3 1]
D=[8 0;0 3;2 7]

h2=C*D
save A11.dat h2 -ascii

I can send you my .m file if you would like as well.

^^ Solved the problem...seems I needed to add semicolons after each variable that I set.

Thanks@!

Hai guys, my question is Arithmetic series, we have been given first term a=5 and sum of 21 terms is 360.

I need to find common difference (d) and nth(21st) term. I have only worked with this before when we are given d so unsure how to go about this now. Probably really simple but I just can't seem to get it.

This is Q from mock test, so I dont just need to know the answer I need to know the method.

Thanks

Arithmetic series: a, a + d, a + 2d, ...

Sum of first n terms: na + d(1 + 2 + .. + (n - 1)) = na + d(n - 1)(n)/2

So 360 = 21(5) + d(20)(21)/2

actually the sum of first 21 is 630

Also we don't know d so I must be missing something but I cant see how to use this formula

630 = 21(5) + d(20)(21)/2 = 21(5 + 10d)
30 = 5 + 10d
d = 2.5

I see now thx alot man

for trigonometric identities: circular functions of allied angles

i've learned sin (or whatever function)
sin(180-A)= whatever
sin(180+A) = whatever
sin (360-A) = whatever
sin (90+A) = whatever
but i haven't leanred
sin or whatever function
sin(90-A)

so
cot(180-A) = -cotA
cot(180+A) = cotA
cot(-A) = -cotA
cot(90+A) = -tanA

What is cot(270-A) ??

also what is sin(270-A) and cot(270+A) and tan(270+A)

Hi math enthusiasts!

If anyone has tough quiz questions or trivia can you post them here or message me them as I love solving the answer. Not sure if you people can price securities but this question was given to me by a friend. It is easy to work out but I am not sure what level people are on here.

So if you are bored or enjoy these sorts of things, kudos to first correct answer:

Suppose the spot price of jet fuel is £80 per barrel currently. The annual risk free rate is 2.5%, the annual storage cost of jet fuel is s% of its current spot price, and the convenience yield for jet fuel futures is zero. We know that the spot price of jet fuel will be £100 (state 1) with probability .95 and £75 (state 2) with probability .05 next year. The futures price of one-year jet fuel futures is £84.

Now answer the following questions.

(a) What should be s, i.e. the annual storage cost as a percentage of the current spot price?

(b) What should be the risk neutral probability of state 1?

Let me know if you want it tougher!

I posted this in the probability forum before realizing that they get about 3 views per day, so I figured I'd post it here too.

Question 1:
How many ways are there to arrange n white and 5 black marbles so that every black marble is adjacent to at least one other black marble? (Hint: when n=2, the answer should be 9)

Question 2:
If 4 white, 6 red, 9 blue, and 6 black marbles are arranged at random, what is the probability that every black marble is adjacent to at least one other black marble?

for question 1 I assumed that the proper was to do this was to put the black balls into 1 group of 2 and 1 group of three. Now we are arranging n white balls and 2 groups of black balls, so it would follow that the number of ways to arrange this is (n+2)!/n!1!1!. That however is wrong, and I have no idea why.

for question 2, since every blue marble is indistinguishable from every blue marble, etc... the number of ways to arrange that many marbles is (4+6+9+6)!/4!6!9!6! = 3435630198000 = a. If every black marble has to be adjacent to at least one other black marble, we can group two black marbles together and count them as one, so now we have 3 groups of black marbles. The total number of ways to arrange this is (4+6+9+3)!/4!6!9!3! = 29875045200 = b. The answer to this question is now b/a = 0.00869565217. Is this right?

TYIA for any help

When adding components of vectors, does it matter if the vectors' angles being added are measured from different axis? I have three vectors, one is measured from the +x axis going counter-clockwise, one is measured from the -y axis going counter-clockwise, and the last is measured from the -x axis going clock-wise. Thanks for the help.

you can figure it out from what you've listed above by writing 270-A as 90+(180-A) or 90+(180+A).

Your idea is pretty much right. Let's group the black into groups of 2 and 3, and then treat the system in 2 cases:
Case 1: The 2 and 3 black groups are seperated by a white ball.
You have 2 distinguishable black groups and n-1 indisdiguishable white balls (n-1 since one MUST be in between the two black groups). Then you have C(n+1;1,1,n-1) or (n+1)!/(n-1)! or (n+1)n. For n=2, this 3!/1!1!1!=6.
Case 2: All black balls are together. Then you now only have 1 black group, and n indistinguishable white balls. Then you have C(n+1;1,n), or (n+1)!/n! or just n+1. For n=2, then you have 3!/2!=3.

All together, n(n+1)+(n+1)=(n+1)^2

Your solution for "a" is totally correct. And while b would seem the right way to work it at first, it would NOT take into account black balls arranged in 2 groups of 3, so something is missing. It does, however, take into account balls in one group of 6, the 4-2 groups, and the 2-2-2 groups. So we just need to add on on 3-3 groups. (Note: The error in your equation for problem 1 was that you assumed that the groups of 2 and 3 were INdistinguishable, which they are not since you can distinguish a group of 2 objects from 3. In this case, we don't have to worry about that since we have an even number of objects and we can group them nicely).
You might immediately say, "Let's just use (4+6+9+2)!/4!6!9!2!" However, we do NOT want to overcount the one group of 6 since we already accounted for that. So there are two solutions from here -- either subtract the overcounted set (4+6+9+1)!/4!6!9!1! or set up the problem like we did in problem 1, choosing a red ball OR blue OR white to be the middle piece. Clearly the subtraction is easier.
In summary, a is correct, but b= (4+6+9+3)!/4!6!9!3!+(4+6+9+2)!/4!6!9!2!-(4+6+9+1)!/4!6!9!1!

I just noticed you are from the Univ. of Waterloo. I applied there for graduate study. Hope I get to see you there!

I have the following ODE:

a(du/dx) + (1+ax)(d^2u/dx^2) = 0

a is a known positive constant, and 0<x<L, the length of a rod (this is one variation of the steady state heat equation). I don't remember much from my ODE class, but is there a way to solve this by hand?

I tried substituting something like y = u', but I only get the trivial solution.