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:
7 is answer
Step-by-step explanation:
Answer:
A. 20000
Step-by-step explanation:
45-36 = 9 so you add 1,000 * 9 to 11,000, which gets you 20,000
Answer:
B
Step-by-step explanation:
Also the answer must be under 9 so that rules out option D
Because We know CE = 2 and AD is visibly larger than CE, 6We can rule out option A
And that brings us down to B and C
CB is equal to 6, And I'm no genius, but I'm pretty sure AD is DEFINATLY not equal to 6.
so the answer is B
Answer: The total number of logs in the pile is 6.
Step-by-step explanation: Given that a stack of logs has 32 logs on the bottom layer. Each subsequent layer has 6 fewer logs than the previous layer and the top layer has two logs.
We are to find the total number of logs in the pile.
Let n represents the total number of logs in the pile.
Since each subsequent layer has 6 fewer logs then the previous layer, so the number of logs in each layer will become an ARITHMETIC sequence with
first term, a = 32 and common difference, d = -6.
We know that
the n-th term of an arithmetic sequence with first term a and common difference d is

Since there are n logs in the pile, so we get

Thus, the total number of logs in the pile is 6.