Your Questions About Odds Calculator

Mark asks…

How would I solve this?

Find the difference between the sum of the first 600 multiples of 3 and the sum of the first 600 odd numbers. I cannot just input the numbers into a calculator. I have to show all of my work somehow and solve this. How do i even set this up? to find an answer?

admin answers:

Sum of first 600 multiples of 3 are

3(1 + 2 + 3 + 4 ……………600 )

recall that sum of first n natural numbers = n(n+1)/2

3 (1 + 2 + 3 …….600 ) = 3 (600)(601)/2 = ( 900)(601) ———(1)

sum of first 600 odd numbers = from n = 1 to 600 [ 2n – 1 ]

[ 1 + 3 + 5 + 7 …………….1199 ]

This is arithmetic progression with first term, a = 1 and common difference, d = 2

Sum = Sn = n/2 [ a + an ]

= 600/2 [ 1 + 1199 ] = (300)(1200) ———–(2)

subtract (2) from (1)

(900)(601) – (300)(1200)

= 300 (3*601 – 1200 )

= 300 (1803 – 1200)

= 300*603

= 180900

George asks…

Odds of getting pregnant?

Well, I know this question is over asked but my hubby is coming home for days 15-23 of my crazy 43 day cycle. I haven’t been paying uber attention to when I ovulate (I think my luteal phase is closer to 3 weeks than 2, I may be way off here…) but I know sperm can live two or three days which would cover about a quarter of my cycle. The fertility calculators I’m using are showing me the 6th of 7th of december if they even go past a 36 day cycle, which some don’t but these calculate at a standard 14 day luteal cycle.
So… if I ovulate while he’s home or even shortly after he leaves what are my chances of actually achieving a sustainable pregnancy? We’re both young and pretty healthy with no known issues except my long cycles.
What I’m trying to ask is if I do end up ovulating at the right time and we drown that time frame in bd’ing how likely am I to get pregnant and stay pregnant?

admin answers:

I think you have a good chance

good luck and baby dust to you

prehaps on ovulation test kit might be helpful

Sandra asks…

Save program as app on ti 84 plus?

How can I save a program without archiving? It will get deleted on the 30 sum odd calculators If it is simply archived because it is for a classroom and people will forget to searching them, does anyone have any ideas?
ok so i typed this on my ipod which is why it makes no sense. so to rephrase… we entered in the slope Field program. i then discovered how to archive this program. this is ok but not functional when there are 30 calculators that are used by 150 students. if we simply archive this program it will be deleted due to forgetting to re-archive the program. that is why i am looking to change it to an app so when we clear the ram it will not disappear even if it is not archived.

admin answers:

I can’t fully understand your question. The grammar is not very comprehensible.

Anyway, archived programs will not be deleted after a RAM clear. Archive clears, on the other hand, are different. Even if you could convert the program as an app, the Archive clear would erase all apps.

So, in other words, there is a way to „save“ a program as an application, but you need to do it on the computer with BasicBuilder (Google it). Archiving programs is much easier and convenient.

David asks…

Question for electrical engineer! What is 10 kilowatts multiplied by 250million?

This is a very odd question but it is extraordinarily important for a science project I am doing. I cant find the correct calculator online because I cant figure how to convert it… Wikipedia has failed me!

Here’s what I need to figure out – if there are 250 mechanisms that each create 10 kilowatts of power, how much total energy does that equate to???

admin answers:

2500 kilowatts is correct. 2500 gigawatts = 2,500,000 megawatts power is correct for your headline. Yes you have to specify time to get energy. Neil

Susan asks…

I have discovered how to divide any even number by two with a broken calculator that can only add…?

Please tell me if this works in every case.

Consider the problem, 7236510 / 2

Divide each digit by 2 and drop the remainder, if any

… resulting in 3113200

Write down a 1 followed by zeros for each digit that was odd in the original problem… and divide each of these by 2 (which can be easily done in your head)

1000000 —-> 500000
0010000 —-> 5000
0000100 —-> 50
0000010 —-> 5

Add up the following…


3618255 = 7236510 / 2

admin answers:

Yes, that will always work.
Let n be any particular digit.

If n is even, then n/2 is the resulting digit when divided by 2.
If n is odd, then n/2 = (n-1)/2 + 1/2, so you divide it by 2, dropping the remainder. Then the 1/2 = 0.5, so you add 5 to the answer in the following digit.

You will probably enjoy this alternative to multiplying:
To multiply 37 * 51 (or any other two numbers), divide the smaller number by 2, dropping any remainder. Double the bigger number. Repeat until the smaller number is 1:

37 * 51
18 * 102
9 * 204
4* 408
2 * 816
1 * 1632

Then cross out any line where the smaller number is even. I get:
37 * 51
9 * 204
1 * 1632

Finally, add the larger numbers to get the product:
1887 = 37*51


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