what's a good pair of noise canceling headphones that won't give me a headache?

what's a good pair of noise canceling headphones that won't give me a headache?

POSIWID: The Purpose Of A System Is What It Does Shirt $21.68

UFOs Are A Psyop Shirt $21.68

POSIWID: The Purpose Of A System Is What It Does Shirt $21.68

  1. 2 weeks ago
    Anonymous

    50% duh

  2. 2 weeks ago
    Anonymous

    sum 0 to inf 0.3*0.2^n = 0.375

    • 2 weeks ago
      Anonymous

      (-0.6 + sqrt(0.6)) / 0.4 = 43.6%

      this would only be true if it's a single chest that can appear within the 20% prob, since it's two chests that overlap, you can't use infinite summation, so your options are quadratic formula or monte carlo simulation

      • 2 weeks ago
        Anonymous

        I misread it you're right

  3. 2 weeks ago
    Anonymous

    >what's a good pair of noise canceling headphones that won't give me a headache?
    go to raves and get tinnitus for free

  4. 2 weeks ago
    Anonymous

    >noise canceling headphones that won't give me a headache
    reconsider whether you actually need active noise canceling

  5. 2 weeks ago
    Anonymous

    wtf is this moronic and cringe gatekeeper problem?
    How do you even write a program for this?
    This is an easy-tier high school math problem.
    You should have posted it on IQfy, it's more relevant there.

    • 2 weeks ago
      Anonymous

      >wtf is this moronic and cringe gatekeeper problem?
      >How do you even write a program for this?
      You're wrong and it says you can write in any language you choose.

      • 2 weeks ago
        Anonymous

        Yeah I though there's only 1 chest

    • 2 weeks ago
      Anonymous

      probabilistic answer (in py3 because you seem moronic)

      ```
      import random

      data = []

      def chest(data):
      possible = [1,1,1,1,1,2,2,2,3,3]
      c = random.choice(possible)
      if c == 1:
      data.append(0)
      if c == 2:
      data.append(1)
      if c == 3:
      chest(data)

      for i in range(10000000):
      chest(data)

      result = data.count(1)/ (data.count(1) + data.count(0))
      print(result)
      ```

      • 2 weeks ago
        Anonymous

        >How do you even write a program for this?
        monte carlo simulation
        essentially, write a function that acts like the chest:
        generate a random number X between 0 and 1
        if X is between 0 and 0.5 return false
        if X is between 0.5 and 0.8 return true
        if X is between 0.8 and 1 return the function itself two times (only one of the chests needs to return true, but it can continue recursively to find at least one returning true if more chests are found)
        then, call the function 100,000 times and find the proportion of true that was returned

        Simulating it is cringe
        Isn't there some dp solution for this?

        • 2 weeks ago
          Anonymous

          >Simulating it is cringe
          why?

          [...]
          [...]
          Ok I figured it out
          How about this?
          #include <bits/stdc++.h>
          #define MAXN 2000LL
          using namespace std;
          typedef long long ll;
          double dp[2][MAXN];
          int main() {
          ios::sync_with_stdio(false); cin.tie(NULL);
          dp[0][1] = 0.8;
          double ans = 0;
          for(ll i = 0; i < 2000; i++) {
          for(ll j = 2; j < MAXN - 1; j++)
          dp[1-(i&1)][j] = dp[i&1][j - 1] * 0.2 + dp[i&1][j + 1] * 0.8;
          dp[1-(i&1)][1] = dp[i&1][2] * 0.8;
          dp[1-(i&1)][MAXN - 1] = dp[i&1][MAXN - 2] * 0.2;
          double expected = 0;
          for(ll i = 1; i < MAXN; i++)
          expected += dp[1-(i&1)][i] * i;
          double bad = pow(0.5, expected);
          ans += 1 - bad;
          }
          ans = 2 - ans;
          cout << ans << 'n';
          return 0;
          }

          This gets me 0.438475
          Is that corect?

          you're close, but the exact solution is 0.4364917, my simulation is giving me probability: 0.4364493 with one million iterations

          no idea but you got the same answer as

          [...]
          >(-0.6 + sqrt(0.6)) / 0.4 = 43.6%

          would probably be closer if you ran more iterations. would be curious to know how they got that math

          quadratic formula, P = 0.3 + 0.2(2P - P^2), set to 0 and find the positive root

          • 2 weeks ago
            Anonymous

            to clarify for other anons, this formula comes from modeling it as a branching process and finding the fixed point. If P is chance of obtaining at least one gold bar, by criteria in the problem we have p = 0.3 + 0.2*(1 - (1-p)(1-p)).

          • 2 weeks ago
            Anonymous

            yeah that works, I was thinking of the union of two events when I solved it P(AuB) = P(A) + P(B) - P(AnB), since A and B are the same event, it can be shortened to 2P - P^2

      • 2 weeks ago
        Anonymous

        >How do you even write a program for this?
        monte carlo simulation
        essentially, write a function that acts like the chest:
        generate a random number X between 0 and 1
        if X is between 0 and 0.5 return false
        if X is between 0.5 and 0.8 return true
        if X is between 0.8 and 1 return the function itself two times (only one of the chests needs to return true, but it can continue recursively to find at least one returning true if more chests are found)
        then, call the function 100,000 times and find the proportion of true that was returned

        [...]
        Simulating it is cringe
        Isn't there some dp solution for this?

        Ok I figured it out
        How about this?
        #include <bits/stdc++.h>
        #define MAXN 2000LL
        using namespace std;
        typedef long long ll;
        double dp[2][MAXN];
        int main() {
        ios::sync_with_stdio(false); cin.tie(NULL);
        dp[0][1] = 0.8;
        double ans = 0;
        for(ll i = 0; i < 2000; i++) {
        for(ll j = 2; j < MAXN - 1; j++)
        dp[1-(i&1)][j] = dp[i&1][j - 1] * 0.2 + dp[i&1][j + 1] * 0.8;
        dp[1-(i&1)][1] = dp[i&1][2] * 0.8;
        dp[1-(i&1)][MAXN - 1] = dp[i&1][MAXN - 2] * 0.2;
        double expected = 0;
        for(ll i = 1; i < MAXN; i++)
        expected += dp[1-(i&1)][i] * i;
        double bad = pow(0.5, expected);
        ans += 1 - bad;
        }
        ans = 2 - ans;
        cout << ans << 'n';
        return 0;
        }

        This gets me 0.438475
        Is that corect?

        • 2 weeks ago
          Anonymous

          no idea but you got the same answer as

          (-0.6 + sqrt(0.6)) / 0.4 = 43.6%

          this would only be true if it's a single chest that can appear within the 20% prob, since it's two chests that overlap, you can't use infinite summation, so your options are quadratic formula or monte carlo simulation

          >(-0.6 + sqrt(0.6)) / 0.4 = 43.6%

          would probably be closer if you ran more iterations. would be curious to know how they got that math

        • 2 weeks ago
          Anonymous

          In J this is just
          chest =. 0:`1:`([: ($:@? +. $:@?) 100"_) @. (49 79&I.)
          (+/ % #) chest"0 ? 10000000 # 100
          0.436472

          • 2 weeks ago
            Anonymous

            chest =. 0:`1:`([: +.&($:@?)~ 100"_) @. (49 79&I.)

            improoved

          • 2 weeks ago
            Anonymous

            I wasn't simulating it.

    • 2 weeks ago
      Anonymous

      >How do you even write a program for this?
      monte carlo simulation
      essentially, write a function that acts like the chest:
      generate a random number X between 0 and 1
      if X is between 0 and 0.5 return false
      if X is between 0.5 and 0.8 return true
      if X is between 0.8 and 1 return the function itself two times (only one of the chests needs to return true, but it can continue recursively to find at least one returning true if more chests are found)
      then, call the function 100,000 times and find the proportion of true that was returned

    • 2 weeks ago
      Anonymous

      Because the modern IQfy lurker does not belong on IQfy. Because they can't do basic math, and are thus reduced to coping with ineffective subpar method. Because they are dumb. Because they are dysgenic.

  6. 2 weeks ago
    Anonymous

    uooooh chest erotic!

  7. 2 weeks ago
    Anonymous

    30%

    • 2 weeks ago
      Lucretia simp

      Imoutos are so smart

  8. 2 weeks ago
    Anonymous

    Why would I need a program?
    x=0.3+0.2*x
    x=0.375

    • 2 weeks ago
      Anonymous

      see

      (-0.6 + sqrt(0.6)) / 0.4 = 43.6%

      this would only be true if it's a single chest that can appear within the 20% prob, since it's two chests that overlap, you can't use infinite summation, so your options are quadratic formula or monte carlo simulation

      • 2 weeks ago
        Anonymous

        Sorry, didn't read.
        Ok then it would be
        x=0.3+0.2*(1 - (1-x)^2)
        x≈0.44

  9. 2 weeks ago
    Anonymous

    cleary around 36%

  10. 2 weeks ago
    Anonymous

    const NUMBER_OF_SIMULATIONS = 100000;

    function openChest() {
    const random = Math.random();

    if (random < 0.3) return true;
    if (random < 0.5) return openChest() || openChest();
    return false;
    }

    let gold = 0;

    for (let i = 0; i < NUMBER_OF_SIMULATIONS; i++) {
    if (openChest()) gold++;
    }

    console.log(gold / NUMBER_OF_SIMULATIONS);

    • 2 weeks ago
      Anonymous

      >noooooooooooooo you can't apply the actual notion of probability

  11. 2 weeks ago
    Anonymous

    Holy shit this broke ChatGPT
    It turned red after recursively giving me the wrong answer
    It said the answer was 0.6 something and I told it it has to be lower than 50% and it broke
    It finally gave the right answer without explaining how and its talking gibberish when I asked it to explain the mistake

    • 2 weeks ago
      Anonymous
  12. 2 weeks ago
    Anonymous

    you need to wear them for a couple of days until you get used to
    bose and sennheisers have larger earcups

  13. 2 weeks ago
    Anonymous

    p=0.3 + 0.2*(p + p)
    p=0.5
    I know this is wrong, but why?

    • 2 weeks ago
      Anonymous

      0.5 is just the one chest that contains nothing
      so it definitely has to be smaller than 0.5 and then some

    • 2 weeks ago
      Anonymous

      >at least one gold bar
      p + p is the expected value of the number of gold bars you get after opening two chests.

    • 2 weeks ago
      Anonymous

      you can only add the two probabilities like that if they are mutually exclusive, but in this case, the two chests overlap, therefore, you have to subtract the intersection (P^2)

    • 2 weeks ago
      Anonymous

      you can only add the two probabilities like that if they are mutually exclusive, but in this case, the two chests overlap, therefore, you have to subtract the intersection (P^2)

      too add on to this, I'll say that if you take your example and add a third p, 0.3 + 0.2*(p + p + p), you get p = 0.75, which obviously can't be right, because you have a 50% chance of nothing and 75% chance of gold bar, the probabilities have to equal one, so that is a clue that the probabilities of the chests overlap and something needs to be subtracted, which in this case is their intersection

  14. 2 weeks ago
    Anonymous

    OP's pic is too easy, here's a more fun one.

    • 2 weeks ago
      Anonymous

      open_chest <- function() {
      outcome <- sample(c("empty", "gold", "double"), 1, prob = c(0.5, 0.3, 0.2))
      if (outcome == "empty") {
      return(0)
      } else if (outcome == "gold") {
      return(1)
      } else {
      return(open_chest() + open_chest())
      }
      }

      calculate_probability <- function(num_simulations, num_bars) {
      count <- 0
      for (i in 1:num_simulations) {
      if (open_chest() == num_bars) {
      count <- count + 1
      }
      }
      probability <- count / num_simulations
      return(probability)
      }

      num_simulations <- 100000
      num_bars <- 1
      probability <- calculate_probability(num_simulations, num_bars)
      cat("Probability of obtaining", num_bars, "gold bar(s):", probability, "n")

      • 2 weeks ago
        Anonymous

        What is the scuffed language? At first I thought javascript but I don't think javascript uses <- for assignment.

        • 2 weeks ago
          Anonymous

          R

    • 2 weeks ago
      Anonymous

      For n = 1
      x = 0.3 + 0.2 *(1 - (1-x)(1-x) - x*x)
      x ≈ 0.395644

      No idea how to do it for bigger ns and my simulation gives me 0.387 instead.

  15. 2 weeks ago
    Anonymous

    First, let's calculate the probability of NOT getting a gold bar. There are two ways this can happen:

    1. The chest contains nothing (50% chance).
    2. The chest contains two smaller chests, and neither of them contains a gold bar (0.2 * 0.7^2 = 0.098 chance - we multiply by 0.7 twice because each smaller chest has a 70% chance (100% - 30%) of NOT containing a gold bar).

    So, the total probability of NOT getting a gold bar is 0.5 + 0.098 = 0.598.

    Now, to find the probability of getting AT LEAST one gold bar, we simply subtract the probability of NOT getting one from 1 (certainty):

    Probability (at least one gold bar) = 1 - Probability (no gold bar)
    = 1 - 0.598
    = 0.402

    Therefore, there is a 40.2% chance of obtaining at least one gold bar from this treasure chest.

  16. 2 weeks ago
    Anonymous

    I got 3/8. Someone wanna check my math?

    • 2 weeks ago
      Anonymous

      never mind, i am a moron and can't even read the question

  17. 2 weeks ago
    Anonymous

    Been using pic related for almost a year. Very happy with them.
    My hearing is pretty sensitive, so strong active noise canceling was a priority.

  18. 2 weeks ago
    Anonymous

    Who needs to write a program for a simple geometric summation problem?
    n=1 gives .3
    n=2...,K gives you chance of not obtaining a gold bar (.70)^2 or 49
    0.2 * sum n->K (49/100) * (2/10)^n = (49/100) / (80/100) = 0.2 * .6125 = .1225
    so in total .4225 or 42.25%

    • 2 weeks ago
      Anonymous

      Wrong it's ~0.43649167
      The closed form is
      x=0.3+0.2*1-(1-x)^2 (right associative)

    • 2 weeks ago
      Anonymous

      For the same reason you can't write 1..k you can't write 2..k, so yours is just a second order approximation of the answer

  19. 2 weeks ago
    Anonymous

    I hate math and exams
    I'm also too dumb to solve this

  20. 2 weeks ago
    Anonymous

    50%
    You either get a gold bar or you don't.

  21. 2 weeks ago
    Anonymous

    100% chance of having sexo with Kurisu

  22. 2 weeks ago
    Anonymous

    I ran 10^9 times and got 0.4364961

  23. 2 weeks ago
    Anonymous

    About tree fiddy

Your email address will not be published. Required fields are marked *