contestada

Current rules for telephone area codes allow the use of digits​ 2-9 for the first​ digit, and​ 0-9 for the second and third​ digits, but the last two digits cannot both be 1​ (to avoid confusion with area codes such as​ 911). How many different area codes are possible with these​ rules? That same rule applies to the exchange​ numbers, which are the three digits immediately preceding the last four digits of a phone number. Given both of those​ rules, how many​ 10-digit phone numbers are​ possible? Given that these rules apply to the United States and Canada and a few​ islands, are there enough possible phone​ numbers? (Assume that the combined population is about​ 400,000,000.)

Respuesta :

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:

[tex]8 \times 9 \times 9 = 648[/tex]

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

[tex]648 \times 10^7 = 6,480,000,000 [/tex]

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 https://brainly.com/question/24067651

gotttaluhan

Answer: Number of possible area codes: 792

Amount of​ 10-digit phone numbers: 6,273,640,000

Step-by-step explanation:

Otras preguntas