Can anyone help me out with this math? I getting confused on this, and it seems simple as hell. What would be the general calculation?



Would it simply be 40/0.1 = 400/0.1 = 4000/0.1 = 40,000 colonies/mL? Or would I need to do it once more to get 400,000?

In each step you go down an order of magnitude. Since you add 1 to 9 to form 10 the concentration in each step is 1/10 the one before. So if at the 3rd (D) it was 40 per 0.1 ml it means 400/ml, the second is 4000/ml, the first is 40000/ml and the original must be 400000/ml.

But you plated 0.1mL, so it's already in colonies/mL no? 400,000 I think would be the 4th. Given that it only goes back 3x (so by a factor of 1000).

The problem tells you the one in D has 40 colonies not 40/ml but 40 total. Its 40 in 0.1 hence 400/ml concentration at D.

seems that I can't set this matrix up right.

3 stocks priced $16, $56, $80 with a dividend yield of 7% 2% 2% respectively. You bought 200 shares totaling $8400 worth of the 3 and expect $248 in dividends back.

I set up my matrix as

16 56 80 8400
.07 .02 .02 248
1 1 1 200

where did i go wrong

Your second equation are percentages that sum up to a total amount of money. Switch to money units.

damn, thank you. knew it was something simple.

Ah okay, makes sense.

I was wondering if you could check my answer to this:

A pure culture in exponential growth phase has a bacterial concentration of 6.4 x 10^8 cells/ml. If the bacterium has a generation time of 1 hour, how many hours ago was the cell concentration 8.0 x 10^7 cells/ml?

I got 3hrs ago.

Yes its 8x more now so its 3 doublings or 3 hours if they double each hour.

An absent minded bank clerk mixed up the pounds and pennies when he gave Mr Brawson his money. Eg £7.14 was £14.07

Mr Brawson got pounds instead of pennies and pennies instead of pounds.

Then Mr Brawson bought a 5p Haribo. He discovered that the money he had left was DOUBLE the amount originally requested.

What was the amount he originally requested?

Halp please

He was supposed to get X pence and Y pounds, or 100*Y + X pence.
He actually got 100*X + Y pence.

When you take away 5p, he has 2x what he was supposed to, so

100*X + Y - 5 = 2 * (100*Y + X)

solve for Y and X

And notice also this is a Diophantine equation (ie x,y integers) and in particular also 0<=Y,X<100.

Ok so on this one I take the inverse of my first matrix and * by B.

So i have 3 4
24 6

take the inverse then multiply by
A
B
?

Isn't this just Cramer's Rule?

Didn't learn about that yet. But he gave us something semi similar but didnt explain farther than A inverse * B but when I did that the answer was wrong.

hi, i'm supposed to do this in python

a).Write a function that takes input a,b,c and finds the solution to a^x=b(mod c)
b). use the function u wrote in part a to solve 34091202317940^x = 46461034929471 (mod 61704897745301)

i tried just doing it iteratively by simply doing accumulator*a n times until accumulator = b(mod c) then returning n, however this function overflows for part b, since the numbers are just too big, so i was wondering is there a smarter way to do this?(perhaps some theorem that i need to use?)

a * b (mod c) = [ a (mod c) * b (mod c) ] (mod c)

So at each step you should take your accumulator and mod out by c:

---

Simple code (used dots to get spacing right):

a = 34091202317940
c = 61704897745301
b = 46461034929471

x = 1

current = a

def increment(curr, a, c, x):
....curr *= a
....curr %= c
....x += 1
....return curr, x

while current != b:
....current, x = increment(current, a, c, x)

print x

hi!, thx for the reply, this solved the overflow problem, but unfortunately my function is really really slow on large inputs, i was wondering if there is a way where i can do this more efficiently? (like skip over some cases that i don't need to test?)

large inputs like what?

I ran my code above on a fairly barebones desktop machine running windows, and it finishes super fast.

I could certainly get rid of the function call...

It's not supposed to be easy.

Just need a work check.



Here's how I went about it, I may be wrong.

The simple way is to look at the information and see at what minute has 90% of the original been killed. On the information given it's at 20 minutes (100,000 bacteria).

You can also use the equation:

Decimal Reduction Time = Time Heated/(Log10 of Original - Log10 of Sample After Heating).

DRT = 20/(Log10 of 1M - Log10 of 100k))

DRT = 20/6-5
DRT = 20

Is this correct?

Hi, I have this problem that I'm not sure about

Anybody seen something similar? Any ideas on how to solve it?







Find the number a, such that the function P is a probability function on the set U
--------------------------

I'm not really sure how to tackle this, as I can barely understand what is being asked.

I've tried this:

P(i) is a probability function on the set U, when

When a is the above, the probability of picking a number i from the set U is correct(?)

Where do you get a/7^i = i/(i*N) ? And isn't that just a = 7^i / N ? In any case it looks like you're getting an a that depends on i. But a should be constant.

What you want is for the function values to sum to 1.

i.e. You want Sum(i=1...N) [a/7^i] = 1

Solve for a.

PairTheBoard

Yes, you're right. I accidently solved a/7^i = i/(i*N) <=> a = i/i*N*7^i.

It should be just be (7^i)/N like you wrote.

My idea was that i/i*N, which could be written just 1/N, would generate the respective probabilities of the elements of the set. Such that Sum(i=1,2,...,N) 1/N = 1. Therefore a/7^1 had to equal 1/N (previously i/i*N)

This gives, solving for a: a/7^i = 1/N <=> a = 7^i/N and then the sum(i=1,2,...N) (7^i/N) / 7^i = 1/N = 1
--------------
I'm confused here, what you asked me to do, solving [a/7^i] = 1 <=> a = 7^i. Then wouldn't a be dependant upon i?
This solution, where a = 7^i, would generate the final equition of 7^i/7^i which is just 1/1 = 1. That doesn't seem right because it doesn't represent the probability of choosing a particular element from the set?

I also think that a must definitely depend upon N (like in my example), because it matters how many elements are in the set.

Your intuition is correct. PTB didn't say [a/7^i] = 1. He said:

You need to sum over all i, after which your sum will depend on N.

Sum a/7^i is geometric, so you can sum it easily.