#selinaconcisemathematicsclass8probability Suchergebnisse
1% = 0.77 10% = 7.7 3% = 3 x 0.77 = 2.31 13% = 7.7 + 2.31 = 10.01 39% = 3 x 10.01 = 30.03
The balance of probabilities is 51% probable.
Per 8-oz glass: ≈ 1 in 8–16 billion So Neil... you're say'n there's a chance. "Might" Dr. Stone.
(2^8/2^4)/(2^8/2) =2^(8-4)/2^(8-1) =2^4/2^7 =2^(4-7) =2^-3=1/8 Answer: 1/8
If it really is 8 3/4 — which I peg at approximately 1.4% odds — then to me, yes, that would be too big. I bet it’s more like 6 1/4.
Result: n=8 => (n*n)/(n+n) = 4 => (n*n)/2n = 16/4 => n*n = 64 n+n = 16 => (8*8)/(8+8) = 4 🙂
Python def calculate_probability(): import math # Constants MIN_VAL = 300 MAX_VAL = 2100 # The 6 target signature values S_VALUES = {392, 914, 602, 2028, 1522, 1587} TARGET_SUM = 7045 NUM_DRAWS = 6 MIN_HITS = 4 # Total possible…
7/9=0.78 6/7=0.86 22/25=0.88 11/13=0.85 22/25 > 6/7 > 11/13 > 7/9 (b)
Ce qui nous fait donc une probabilité de 0.04×0.5×0.05=0.001 que cet événement se produise ? Soit 0.1%. Y a-t-il un prof de maths dans la salle ?
If someone ever tells you there is a non-zero chance of something, that means there's an 80% chance it will happen.
Using Binomial expansion (√2-1)⁸=[(√2-1)⁴]²=(17-12√2)²=577-408√2.
(√2-1)⁸ = (((√2-1)²)²)² = ((2+1 - 2√2)²)²= ((3 - 2√2)²)²= (3²+2²•2 - 2•3•2√2)² = (17 - 12√2)² = 17²+12²•2 - 2•17•12•√2 = 577 - 408√2 = ~0.00086655
~ (4/10)^8 = 4^8 * 10^(-8) = (2^16) * 10^(-8) =~ 64k/ (10^8) { =65536*(10^-8) } =~ 0.6/1000
Em termos probabilísticos, isso é praticamente impossível. A probabilidade de um aluno errar uma questão no Enem é 4/5= 0,8. Logo, esse valor deveria ser multiplicado por ele mesmo por 90 vezes. O que daria um valor muito baixo: 0, 000000002. Ou seja, é quase impossível.
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