Hi!
I can help you with joy!
Evaluate:
when 
First, add x's:
Now, plug in the value of x (3)
Multiply:`
(Answer)
Hope it helps!
Ask me if you have any doubts.
<em>Answered by</em>
<em>~</em>Silent~
Answer:
i have no clue how to do thats,
Answer:
Our answer is 0.8172
Step-by-step explanation:
P(doubles on a single roll of pair of dice) =(6/36) =1/6
therefore P(in 3 rolls of pair of dice at least one doubles)=1-P(none of roll shows a double)
=1-(1-1/6)3 =91/216
for 12 players this follows binomial distribution with parameter n=12 and p=91/216
probability that at least 4 of the players will get “doubles” at least once =P(X>=4)
=1-(P(X<=3)
=1-((₁₂ C0)×(91/216)⁰(125/216)¹²+(₁₂ C1)×(91/216)¹(125/216)¹¹+(₁₂ C2)×(91/216)²(125/216)¹⁰+(₁₂ C3)×(91/216)³(125/216)⁹)
=1-0.1828
=0.8172
Answer:7
Step-by-step explanation:
This question is on the ukmt junior maths challenge which I'm doing now, out of the options given I got 7, because 2+4+6+5+(5×20)+(7×7)= 166, (then dividing that by the number of numbers to get the mean) so 166÷31= 5.354.....
Also the 3 has a recurring symbol above it
The 7 is times seven so uh yeah<3
