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Anastaziya [24]
3 years ago
5

Suppose you are choosing a 6 digit personal access code. This code is made up of 4 digits chosen from 1 to 9, followed by 2 lett

ers chosen from A to Z. Any of these digits or letters can be repeated. Use the counting principle to find the total number of personal access codes that can be formed
Mathematics
1 answer:
iragen [17]3 years ago
5 0

Answer:

4435236

Step-by-step explanation:

Total digits of code = 6

Code options for first 4 digits = any of 1 - 9 = 9 options

Code option for last 2 digits = A - Z = 26 options

Code number 1 = 9 possible values

Code number 2 = 9 possible values

Code number 3 = 9 possible values

Code number 4 = 9 possible values

Code number 5 = 26 possible values

Code number 6 = 26 possible values

Hence, total number of possible access codes :

9 * 9 * 9 * 9 * 26 * 26

9^4 * 26^2

= 4435236 possible ways

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(3 points) Determine whether each of these functions f : {a, b, c, d} → {a, b, c, d} is one-to-one and whether each of these fun
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f is onto if for every y in the codomain of f, then there is an element x in the domain of f such that f(x) = y. That is, given an element y in the codomain of f, there exists and element in the domain that is linked to y.

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8 0
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Please help me on this geometry i’ll mark u the brainliest
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Answer:

<h2>               XN = 6</h2>

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