Answer:
The number is 17
Step-by-step explanation:
Answer:
The probability that a customer pays late each month is P(B) = 0.27
Step-by-step explanation:
Let P( A ) be the probability that the customer pays on time and the value is 0.55
Let P( B ) be the probability that a customer pays late each month
So
The probability that a customer pays late or on-time each month is P(A u B) and the value is 0.82
The probability that a customer pays on-time and late each month is P(A n B) and the value is zero ( 0 ) given that it is impossible
Now The probability that a customer pays late or on-time each month is mathematically represented as
P(A u B) = P(A) + P(B) - P(A n B)
=> 0.82 = 0.55 + P( B ) - 0
=> P(B) = 0.27
You have cards with the letters A, B, C, D, E, F, G, H, I, J, K, L, M, N, O, P. Event U is choosing the cards A, B, C
Thepotemich [5.8K]
Answer:
If you are asking how many events use P: 1 event is using the letter P (event W)
Step-by-step explanation:
Why event U cannot be using P: This event is only using the letters A, B, C, and D.
Why event V cannot be using P: This event is only using vowels (A, E, I, O, U)
Answer:
if repetition is allowed, 
if repetition is not allowed.
Step-by-step explanation:
For the first case, we have a choice of 26 letters <em>each step of the way. </em>For each of the 26 letters we can pick for the first slot, we can pick 26 for the second, and for each of <em>those</em> 26, we can pick between 26 again for our third slot, and well, you get the idea. Each step, we're multiplying the number of possible passwords by 26, so for a four-letter password, that comes out to 26 × 26 × 26 × 26 = possible passwords.
If repetition is <em>not </em>allowed, we're slowly going to deplete our supply of letters. We still get 26 to choose from for the first letter, but once we've picked it, we only have 25 for the second. Once we pick the second, we only have 24 for the third, and so on for the fourth. This gives us instead a pretty generous choice of 26 × 25 × 24 × 23 passwords.
