Since there is no repetition allowed, there are 10 possibilities for the 1st digit, 9 for the 2nd, 8 for the 3rd, and 7 for the 4th. This gives a total of (10)(9)(8)(7) = 5040 four-digit codes.
For all odd digits to be used, there are 5 possibilities for the 1st digit (1,3,5,7,9), 4 for the 2nd, 3 for the 3rd, 2 for the 4th. This gives a total of (5)(4)(3)(2) = 120 codes that only use odd digits.
Therefore there are 5040 - 120 = 4920 codes that do not consist of all odd digits. The probability is 4920/5040 = 41/42.
Answer: 0.003757(approx).
Step-by-step explanation:
Total number of combinations of selecting r things out of n things is given by:-
Total cards in a deck = 52
Total number of ways of choosing 8 cards out of 52 = 
Total number of ways to choose 5 clubs and 3 cards with one of each remaining suit = 
[since 1 suit has 13 cards]
The required probability = 
Hence, the required probability is 0.003757 (approx).
$76.5
51/6 = 8.5
8.5 x 9 = 76.5
Answer:
parallel
Step-by-step explanation:
all the details are in the attached picture.
Answer:
2kg
Step-by-step explanation:
Let the sum of the other 2 bricks be R. Given that the heaviest brick weighs 2/3 times as much as the other 2 bricks in total, then the weight of the heaviest brick H in terms of the weight of the other 2 will be
H = 2/3 * R
= 2R/3
Given that the 3 bricks weigh 5kg in total then,
2R/3 + R = 5
Multiply through the equation by 3
2R + 3R = 15
5R = 15
Divide both sides by 5
R = 15/5
= 3
The weight of the heaviest brick is 2R/3. Since R = 3, the heaviest block weighs
= 2 * 3/3
= 2kg
