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wel
2 years ago
12

Write 1011.1012 in normalized exponential form.

Mathematics
1 answer:
just olya [345]2 years ago
3 0

1011.101_2 in a normalized exponential form is written as 0.1011101 \times 2 ^4

<u>Given the following data:</u>

  • 1011.101_2

<h3>A normalized exponential form.</h3>

In Computer technology, binary numbers like decimal numbers in mathematics can be written in an exponential form and it involves the use of powers of two (2) rather than powers of ten (10).

Additionally, all non-zero binary numbers have a unique normalized exponential form and their binary point must always appear before the first one (1) bit, which yields a unique integer (exponent) and a unique mantissa respectively.

Thus, the normalized exponential form for the given binary number is written as follows:

1011.101_2 = 0.1011101 \times 2 ^4

<u>Note:</u>

  • Exponent = 4.
  • Mantissa = 0.1011101.

Read more on binary numbers here: brainly.com/question/5381889

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3 years ago
How do I do this because it's really confusing haha
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6 0
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A peculiar die has the following properties: on any roll the probability of rolling either a 2, a 3, or a 5 is 1/2, just as it i
igor_vitrenko [27]

Answer:

A. P( 2∪3∪1 )=1/2

B. P( 2∪3∪4∪5∪6 )= 13/16

Step-by-step explanation:

A. At first we need to get all the information that we can from the question, so we focus on the probabilities given P(2∪3∪5)=1/2 and P(4∪5∪6)=1/2 and we need to use P(5)=3/16.

Notice that when you have a dice, it´s impossible to get two values at the same try. Because of this and the property of probability for events:

P(2∪3∪5)=P(2) + P(3) + P(5) - P(2∩3)(=0) - P(3∩5)(=0) - P(2∩5)(=0) + P(2∩3∩5)(=0)

P(2∪3∪5)=P(2) + P(3) + P(5)

1/2=P(2) + P(3) + 3/16

P(2)+P(3)=5/16

If we do the same with P(4∪5∪6), we get:

P(4)+P(6)=5/16

Now, with this new information we can find P(1):

P(1)= 1 - P(2∪3∪4∪5∪6)

P(1)= 1 - (P(2) + P(3) + P(4) + P(5) + P(6)) (The probability of two of them happening at the same time is zero)

P(1)= 1 - (5/16 + 5/16 + 3/16)

P(1)= 1 - 13/16

P(1)= 3/16

And with it, we can find P(2∪3∪1):

P(2∪3∪1)=P(2)+P(3)+P(1)

P(2∪3∪1)=5/16+3/16

P(2∪3∪1)=1/2

B. Notice if you want to roll anything but a 1, then the probability you are looking for is "P(2∪3∪4∪5∪6)" but we found that in part A

P(2∪3∪4∪5∪6)=13/16

4 0
3 years ago
1 2/3 −3/4= _ −_=<br> Rewrite using twelfths.
Triss [41]
See the explanation I made for you!
Have a nice day!

3 0
3 years ago
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