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Katarina [22]
3 years ago
11

A message source M of a digital communication system outputs a word of length 8 characters, with the characters drawn from the t

ernary alphabet {0,1,2}, and all such words are equally probable. What is the probability that M produces a word that looks like a byte (i.e., no appearance of ‘2’)?
Mathematics
1 answer:
marysya [2.9K]3 years ago
7 0

Answer:

The probability that M produces a word that looks like a byte is (\frac{2}{3})^8.

Step-by-step explanation:

It is given that a message source M of a digital communication system outputs a word of length 8 characters.

The characters drawn from the ternary alphabet {0,1,2}, and all such words are equally probable.

Total possible outcomes is

3\times 3\times 3\times 3\times 3\times 3\times 3\times 3=3^8

We need to find the probability that M produces a word that looks like a byte (i.e., no appearance of ‘2’). It means only 0 and 1 are included in the word.

Total favorable outcomes is

2\times 2\times 2\times 2\times 2\times 2\times 2\times 2=2^8

The formula for probability is

p=\frac{\text{Favorable outcomes}}{\text{Total outcomes}}

p=\frac{2^8}{3^8}

p=(\frac{2}{3})^8

Therefore the probability that M produces a word that looks like a byte is (\frac{2}{3})^8.

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Step-by-step explanation:

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Step-by-step explanation:

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2 years ago
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Hi.
NeX [460]

Answer:

(a)

1 sig: 0.005

2 sig: 0.0048

3 sig: 0.00482

(b)

1 sig: 50

2 sig: 50.

3 sig: 50.0

(c)

1 sig: 0.0010

2 sig: 0.00098

3 sig: 0.000981

Step-by-step explanation:

Significant Figures Rules:

  1. Any non-zero digit is significant.
  2. Any trailing zeros after the decimal is significant.
  3. Any zeros between 2 significant digits are significant.
  4. Zeroes before significant numbers in the decimal place are NOT significant; they are placeholders.

(a)

0.004816 - the zeros are placeholders, so they do not count as sig figs.

(b)

50.00168 - the zeros are between 2 significant figures, so they do count as sig figs.

(c)

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Let x = amount of money you have in your account.

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The equation is:

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Now we solve the equation by subtracting 125 from both sides.

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