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1 year ago

What is the difference between decimal addition and binary addition?

1 answer:

7 0

The only difference between binary addition and regular addition (decimal addition) is that binary addition uses a value of 2 rather than a value of 10.

<h3>What is binary addition?</h3>

The binary addition uses a value of 2.

If you add 8 plus 2 in decimal addition, you obtain 10, which you write as 10. In the sum, this results in a digit of 0 and a carry of 1.
Similar things happen in binary addition whenever you add 1 and 1. The outcome is always 2 but because 2 is expressed as 10 in binary, we obtain a digit 0 as well as a carry of 1 after adding 1 + 1 in binary.

Thus, in binary:

0 + 0 = 0

0 + 1 = 1

1 + 0 = 1

1 + 1 = 10 (which is 0 carry 1)

<h3>What is decimal addition?</h3>

Decimal numbers with digits 0 through 9 are also referred to as denary.

In every aspect of our daily lives, we employ decimal numerals.
Computers can't operate with decimal numbers like humans do.
Computers can only operate with binary numbers, as was previously stated.
To save decimal numbers on computers, binary numbers must be transformed.
They are changed from binary into decimal numerals whenever they are shown on the screen.

To know more about binary addition, here

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4 years ago

A high school principal wishes to estimate how well his students are doing in math. Using 40 randomly chosen tests, he finds tha

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99% confidence interval for the population proportion of passing test scores is [0.5986 , 0.9414].

Step-by-step explanation:

We are given that a high school principal wishes to estimate how well his students are doing in math.

Using 40 randomly chosen tests, he finds that 77% of them received a passing grade.

Firstly, the pivotal quantity for 99% confidence interval for the population proportion is given by;

P.Q. = $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ~ N(0,1)

where, $\hat p$ = sample proportion of students received a passing grade = 77%

n = sample of tests = 40

p = population proportion

<em>Here for constructing 99% confidence interval we have used One-sample z proportion test statistics.</em>

So, 99% confidence interval for the population proportion, p is ;

P(-2.5758 < N(0,1) < 2.5758) = 0.99 {As the critical value of z at 0.5%

level of significance are -2.5758 & 2.5758}

P(-2.5758 < $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < 2.5758) = 0.99

P( $-2.5758 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < ${\hat p-p}$ < $2.5758 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ) = 0.99

P( $\hat p-2.5758 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < p < $\hat p+2.5758 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ) = 0.99

<u>99% confidence interval for p</u> = [ $\hat p-2.5758 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ , $\hat p+2.5758 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ]

= [ $0.77-2.5758 \times {\sqrt{\frac{0.77(1-0.77)}{40} } }$ , $0.77+2.5758 \times {\sqrt{\frac{0.77(1-0.77)}{40} } }$ ]