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Sagot :
Sure, let's convert the binary number [tex]\( 1011_2 \)[/tex] to its decimal equivalent step by step.
A binary number is a number expressed in base-2 numeral system. Each binary digit (bit) is a power of 2. To convert the binary number [tex]\( 1011_2 \)[/tex] to decimal (base 10), you multiply each bit by 2 raised to the power of its position (counting from right and starting at 0) and then sum the results.
Here is the binary number [tex]\( 1011_2 \)[/tex]:
```
1 0 1 1
```
We can write it as:
[tex]\( 1 \times 2^3 + 0 \times 2^2 + 1 \times 2^1 + 1 \times 2^0 \)[/tex]
Let's calculate it step by step:
1. The leftmost bit (1) is in the [tex]\(2^3\)[/tex] position:
[tex]\(1 \times 2^3 = 1 \times 8 = 8\)[/tex]
2. The next bit (0) is in the [tex]\(2^2\)[/tex] position:
[tex]\(0 \times 2^2 = 0 \times 4 = 0\)[/tex]
3. The next bit (1) is in the [tex]\(2^1\)[/tex] position:
[tex]\(1 \times 2^1 = 1 \times 2 = 2\)[/tex]
4. The rightmost bit (1) is in the [tex]\(2^0\)[/tex] position:
[tex]\(1 \times 2^0 = 1 \times 1 = 1\)[/tex]
Now, sum all these values:
[tex]\(8 + 0 + 2 + 1\)[/tex]
Which gives us:
[tex]\(11\)[/tex]
So, [tex]\( 1011_2 \)[/tex] converted to decimal (base 10) is [tex]\( \boxed{11} \)[/tex].
A binary number is a number expressed in base-2 numeral system. Each binary digit (bit) is a power of 2. To convert the binary number [tex]\( 1011_2 \)[/tex] to decimal (base 10), you multiply each bit by 2 raised to the power of its position (counting from right and starting at 0) and then sum the results.
Here is the binary number [tex]\( 1011_2 \)[/tex]:
```
1 0 1 1
```
We can write it as:
[tex]\( 1 \times 2^3 + 0 \times 2^2 + 1 \times 2^1 + 1 \times 2^0 \)[/tex]
Let's calculate it step by step:
1. The leftmost bit (1) is in the [tex]\(2^3\)[/tex] position:
[tex]\(1 \times 2^3 = 1 \times 8 = 8\)[/tex]
2. The next bit (0) is in the [tex]\(2^2\)[/tex] position:
[tex]\(0 \times 2^2 = 0 \times 4 = 0\)[/tex]
3. The next bit (1) is in the [tex]\(2^1\)[/tex] position:
[tex]\(1 \times 2^1 = 1 \times 2 = 2\)[/tex]
4. The rightmost bit (1) is in the [tex]\(2^0\)[/tex] position:
[tex]\(1 \times 2^0 = 1 \times 1 = 1\)[/tex]
Now, sum all these values:
[tex]\(8 + 0 + 2 + 1\)[/tex]
Which gives us:
[tex]\(11\)[/tex]
So, [tex]\( 1011_2 \)[/tex] converted to decimal (base 10) is [tex]\( \boxed{11} \)[/tex].
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