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Vedmedyk [2.9K]
3 years ago
10

Using parity, the probability for detecting an error, given that one has occurred, is:a. about 50% for either even or odd parity

b. about 70% for even parity and 30% for odd parityc. about 30% for even parity and 70% for odd parityd. about 0% for either even or odd paritye about 100% for either even or odd parity
Mathematics
2 answers:
marishachu [46]3 years ago
4 0

Answer:

a. about 50% for either even or odd parity

Step-by-step explanation:

Parity comes in two flavors, even and odd.

Even parity is the exclusive-or of all the data bits while odd parity is its inverse.

The set of even integers is called parity 0.

The set of odd integers is called parity 1.

This implies that the probability for detecting an error is 50% because the even and odd parity both have a chance of 50%.

nevsk [136]3 years ago
3 0

Answer:

a. about 50% for either even or odd parity

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- Monica goes for a 60 minute bike ride each day in the summer. Let a rep
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Answer:

m = 60a

a = Independent Variable

m = Dependent Variable.

For a = 1 day, m = 60 minutes

For a = 2 days, m = 120 minutes

For a = 3 days, m = 180 minutes

For a = 4 days, m = 240 minutes

For a = 5 days, m = 300 minutes

For a = 6 days, m = 360 minutes

For a = 7 days, m = 420 minutes

Step-by-step explanation:

Monica goes for a 60-minute bike ride each day in the summer.

If a represents the number of days Monica rides her bike and m represents the total number of minutes she spends riding her bike, then a represents the independent variable and m represents the dependent variable.

The relation between a and m can be written as an equation, m = 60a.

Now, for a = 1 day, m = 60 minutes

For a = 2 days, m = 60 × 2 = 120 minutes

For a = 3 days, m = 60 × 3 = 180 minutes

For a = 4 days, m = 60 × 4 = 240 minutes

For a = 5 days, m = 60 × 5 = 300 minutes

For a = 6 days, m = 60 × 6 = 360 minutes

For a = 7 days, m = 60 × 7 = 420 minutes

Therefore, this is the table to show how many minutes Monica would ride her bike over the course of 1 week. (Answer)

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