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RackNerd's OFFICIAL LET BLACK FRIDAY! HUNDREDS OF GIVEAWAYS + CRAZY DEALS, MASSIVE (come see)

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Comments

  • chitreechitree Member

    @dustinc said:

    @chitree said:

    BRING BACK @FRANKZ

    @FrankZ -- what's it going to take to have you back!

    COME HOME @FRANKZ

  • dev_vpsdev_vps Member

    @Kevinf100 said:

    @dev_vps said:

    @Kevinf100 said:

    @dev_vps said:

    @Kevinf100 said:

    @dev_vps said:
    @dustinc
    I teach for AP Calculus and JEE Calculus.

    Here is a problem I ask my students
    — Feb 28 is picked for a random year. What is the probability that it falls on a Sunday?
    — Feb 29 is picked for a random year. Of course it has to be a leap year. What is the probability that it falls on a Sunday?

    The answer for the first question is 1/7.
    What is the answer for the second question?

    Adding more days to the year doesn't change the calendar days. So the answer is still 1/7. The same reason why when you flip a coin, every flip is a 1/2 you will get heads. The amount doesn't effect probably. If heads = h and tails = t, landing hththt is the same probably as hhhhhh, the more you flip tho, the more likely it will be 1/2. So even if I pick 7 random years and it turned out it it was sunday 6 out of 7 times, doesn't mean the odds was 6/7.

    Can I skip class for the rest of the day now?

    Your logic is correct. The answer is close to 1/7, but it is not same as 1/7

    Hint — leap year does NOT occur after every four years

    Is it > 1/7 but < 1/8 due to the year could also not be a leap year, thus Feb 29 never can happen?

    Leap year occurs 97 times in every 400 years.

    (97 / 400) * (1 / 7) = 0.034 or 3.4%

    Answer would have been 1/7 if leap years were equally distributed, but they are 97 (in place of 100)

    The answer is (97/100) x (1/7)

    As I said before answer is close to 1/7 and in the original post you did put correct logical reasoning.

    ** chatGPT may give wrong answer for this

  • codelockcodelock Member

    @chitree said:

    @dustinc said:

    @chitree said:

    BRING BACK @FRANKZ

    @FrankZ -- what's it going to take to have you back!

    COME HOME @FRANKZ

    Frankz has been missing for so long

  • chitreechitree Member

    OMG!!! RACKNERD'S BLACK FRIDAY IS LIVE, GO HERE: https://www.racknerd.com/BlackFriday/ 💯

  • chitreechitree Member

    I-LOVE-RACKNERD-GIVEAWAYS !!

  • noob404noob404 Member

    @CTRL said:
    Oh boy good thing I checked early this year. Missed out on all the giveaways last black friday. Would be nice to snag a VPS. B)

    Stay and you shall be rewarded.

  • noob404noob404 Member

    @chitree said:

    OMG!!! RACKNERD'S BLACK FRIDAY IS LIVE, GO HERE: https://www.racknerd.com/BlackFriday/ 💯

    Welcome back, dude. Busy this year?

  • CTRLCTRL Member

    @noob404 said:

    @CTRL said:
    Oh boy good thing I checked early this year. Missed out on all the giveaways last black friday. Would be nice to snag a VPS. B)

    Stay and you shall be rewarded.

    That's what I'm hoping for lol

  • Kevinf100Kevinf100 Member
    edited November 2024

    @dev_vps said:

    @Kevinf100 said:

    @dev_vps said:

    @Kevinf100 said:

    @dev_vps said:

    @Kevinf100 said:

    @dev_vps said:
    @dustinc
    I teach for AP Calculus and JEE Calculus.

    Here is a problem I ask my students
    — Feb 28 is picked for a random year. What is the probability that it falls on a Sunday?
    — Feb 29 is picked for a random year. Of course it has to be a leap year. What is the probability that it falls on a Sunday?

    The answer for the first question is 1/7.
    What is the answer for the second question?

    Adding more days to the year doesn't change the calendar days. So the answer is still 1/7. The same reason why when you flip a coin, every flip is a 1/2 you will get heads. The amount doesn't effect probably. If heads = h and tails = t, landing hththt is the same probably as hhhhhh, the more you flip tho, the more likely it will be 1/2. So even if I pick 7 random years and it turned out it it was sunday 6 out of 7 times, doesn't mean the odds was 6/7.

    Can I skip class for the rest of the day now?

    Your logic is correct. The answer is close to 1/7, but it is not same as 1/7

    Hint — leap year does NOT occur after every four years

    Is it > 1/7 but < 1/8 due to the year could also not be a leap year, thus Feb 29 never can happen?

    Leap year occurs 97 times in every 400 years.

    (97 / 400) * (1 / 7) = 0.034 or 3.4%

    Answer would have been 1/7 if leap years were equally distributed, but they are 97 (in place of 100)

    The answer is (97/100) x (1/7)

    As I said before answer is close to 1/7 and in the original post you did put correct logical reasoning.

    ** chatGPT may give wrong answer for this

    ** Didn't have to use chatGPT, I took a stats class!
    Wait where did you get 97/100?

  • codelockcodelock Member

    @Kevinf100 said:

    @dev_vps said:

    @Kevinf100 said:

    @dev_vps said:

    @Kevinf100 said:

    @dev_vps said:

    @Kevinf100 said:

    @dev_vps said:
    @dustinc
    I teach for AP Calculus and JEE Calculus.

    Here is a problem I ask my students
    — Feb 28 is picked for a random year. What is the probability that it falls on a Sunday?
    — Feb 29 is picked for a random year. Of course it has to be a leap year. What is the probability that it falls on a Sunday?

    The answer for the first question is 1/7.
    What is the answer for the second question?

    Adding more days to the year doesn't change the calendar days. So the answer is still 1/7. The same reason why when you flip a coin, every flip is a 1/2 you will get heads. The amount doesn't effect probably. If heads = h and tails = t, landing hththt is the same probably as hhhhhh, the more you flip tho, the more likely it will be 1/2. So even if I pick 7 random years and it turned out it it was sunday 6 out of 7 times, doesn't mean the odds was 6/7.

    Can I skip class for the rest of the day now?

    Your logic is correct. The answer is close to 1/7, but it is not same as 1/7

    Hint — leap year does NOT occur after every four years

    Is it > 1/7 but < 1/8 due to the year could also not be a leap year, thus Feb 29 never can happen?

    Leap year occurs 97 times in every 400 years.

    (97 / 400) * (1 / 7) = 0.034 or 3.4%

    Answer would have been 1/7 if leap years were equally distributed, but they are 97 (in place of 100)

    The answer is (97/100) x (1/7)

    As I said before answer is close to 1/7 and in the original post you did put correct logical reasoning.

    ** chatGPT may give wrong answer for this

    ** Didn't have to use chatGPT, I took a stats class!
    Wait where did you get 97/100?

    But you can use chatgpt now :lol:

  • ccap_ggoccap_ggo Member

    RACKNERD-HAS-THE-BEST-BLACK-FRIDAY-GIVEAWAYS!

  • mbjonesmbjones Member

    Almost to 100

  • noob404noob404 Member

    I am gonna take some time off.

  • ccap_ggoccap_ggo Member

    RACKNERD-HAS-THE-BEST-BLACK-FRIDAY-GIVEAWAYS!FOUR

  • ccap_ggoccap_ggo Member

    RACKNERD-SERVICE-EXTENSION-GIVEAWAY

  • ccap_ggoccap_ggo Member

    RackNerdGang

  • codelockcodelock Member

    3 pages to go go go go

  • noob404noob404 Member

    GO go go! So close to 100!

  • codelockcodelock Member

    @noob404 said:
    GO go go! So close to 100!

    Indeed very close to page 100

  • xpress7xpress7 Member

    GOOOOOOO

  • noob404noob404 Member

    Not many people online around this time of the day.

  • noob404noob404 Member

    But, the party has to go on!

    RACKNERD PARTY LIT!

  • BasToTheMaxBasToTheMax Member

    Hi everyone

  • nyamenknyamenk Member

    Chef where are you?

  • xpress7xpress7 Member

    @nyamenk said:
    Chef where are you?

    Sleeping, probably.

  • CTRLCTRL Member

    @codelock said:

    @noob404 said:
    GO go go! So close to 100!

    Indeed very close to page 100

    I'm helping. ;)

  • noob404noob404 Member

    @nyamenk said:
    Chef where are you?

    Prepping for the big reveal in a few days I believe.

  • OKHaoOKHao Member

    Hi. I would like to double the bandwidth.
    Order Number: 4919730850
    Thanks!

  • MrEdMrEd Member

    Good morning! ;) I managed to snag the 1 month extension on my phone! :D Anyone willing to trade the RAM UPGRADE to it? :D

    Anyways, I am at work, so if Chef has something prepared, it would be great time for that :D

  • DeadlyChemistDeadlyChemist Member

    20 pages later i woke up
    And probably missed some goodies, but more to come

This discussion has been closed.