Answer:
0.081 m high
Step-by-step explanation:
324 m/4000 = 0.081 m
Answer:
it's most likely D because 80 plus 80 equals 160 plus 20 is 180
Answer:
The probability that all four aces will be received by the same player is approximately 0.01.
Step-by-step explanation:
Of the 52 cards the four aces can be selected in ways.
If any one of the players receives all the four aces then that player can be selected in ways.
Now for the selected player to receive all the four aces, the four ace cards must be placed among the 13 cards the player receives. This can be done in ways.
Then the total number of ways such that all the four aces is received by one player is .
Then the probability that all four aces will be received by the same player is:
Thus, the probability that all four aces will be received by the same player is approximately 0.01.
The probability that all four aces will be received by the same player in a game of 52 card deck distributed among four players is given by the formula P(E) = 4 * C(48, 9) / C(52, 13), derived through principles of combinatorics.
This is a probability problem that can be solved using combination and permutation principles in mathematics. Specifically, the subject involves combinatorics.
First, we note that each of the four players is dealt 13 cards from a 52 card deck. Hence, for a single player, the total number of ways 13 cards can be chosen from 52 is given by the combination formula C(n, r) = n! / [(n-r)!r!], where ! denotes factorial. In our case, n=52, and r=13.
The total number of ways to deal these 13 cards is thus C(52, 13).
Next, we need to consider the specific case where all four aces end up with one player. This means this player has 9 other cards that are not aces, from the remaining 48 cards (52 total cards - 4 aces). This can happen in C(48, 9) ways.
Therefore, the probability that a specific player gets all four aces is P(E) = C(48, 9) / C(52, 13).
To find the probability that any of the four players gets all four aces, we multiply this result by 4, because there are four players.
So, the final probability is P(E) = 4 * C(48, 9) / C(52, 13).
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