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Phone numbers consist of a three-digit area code followed by seven digits. if the area code must have a 0 or1 for the second digit, and neither the area code nor the seven-digit number can start with 0 or 1, how many different phone numbers are possible? how did you come up with your answer? a. 8 times 10 times 10 times 8 times 10 times 10 times 10 times 10 times 10 times 10 = 6,400,000,000 b. 8 times 2 times 10 times 8 times 10 times 10 times 10 times 10 times 10 times 10 = 1,280,000,000 c. 8 times 2 times 10 times 8 times 10 times 10 times 10 times 10 times 10 = 128,000,000 d. 8 times 2 times 10 times 8 times 10 times 10 times 10 times 10 times 10 times 10 times 10 = 12,800,000,000

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shubhamchouhanvrVT

The possible number of different phone numbers will be 128000000

What is probabilty?

The extent to which an event is likely to occur, measured by the ratio of the favourable cases to the whole number of cases possible.

In this exercise we have to use the knowledge of probability to find the phone number, so:

1 280 000 000 different ways.

First we have to establish some information about this exercise such as:

  • The first slot cannot start with 0 or 1, which leaves us 2, 3, 4 ..., 9 to fill.
  • This means we have 8 different arrangements for the first slot.
  • The second slot have a 0 or 1 for the second digit, which leaves us with 2 different arrangements.
  • There are no restrictions for the third slot, so we would have 10 different arrangements.

Thus, now calculating the possibilities of occurrence, we have:

[tex]8\times 2\times 10=160[/tex]

[tex]8\times 10^6=8000000[/tex]

[tex]160\times 8000000=1280000000[/tex]

See more about probability follow

brainly.com/question/795909

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