Using the fundamental counting theorem, we have that:
- 648 different area codes are possible with this rule.
- There are 6,480,000,000 possible 10-digit phone numbers.
- The amount of possible phone numbers is greater than 400,000,000, thus, there are enough possible phone numbers.
The fundamental counting principle states that if there are p ways to do a thing, and q ways to do another thing, and these two things are independent, there are ways to do both things.
For the area code:
- 8 options for the first digit.
- 9 options for the second and third.
Thus:

648 different area codes are possible with this rule.
For the number of 10-digit phone numbers:
- 7 digits, each with 10 options.
- 648 different area codes.
Then

There are 6,480,000,000 possible 10-digit phone numbers.
The amount of possible phone numbers is greater than 400,000,000, thus, there are enough possible phone numbers.
A similar problem is given at brainly.com/question/24067651
Answer:
4
Step-by-step explanation:
36/9 =4
When you see an x put 9 ;)
9514 1404 393
Answer:
B 2/6
Step-by-step explanation:
2 of the 6 possible outcomes are ones that are of interest. A "fair" die means the mutually-exclusive outcomes have equal probability, so ...
P(4 or 5) = P(4) +P(5) = 1/6 + 1/6 = 2/6
First, you can combine all of the terms involving x:
0.3x + 0.03x + 0.003x + 0.0003x + 0.00003x = 3
0.33333x = 3
0.33333x / 0.33333 = 3 / 0.33333
x = 9.0000900009…
x ≈ 9
Hope this helps! :)