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Q: What is the least significant digit of 9200?

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9.2x10^3 has 2 significant figures, or 9200 (also 2 sig figs)

Yes A whole number is even if its least significant digit is even. The least significant digit of a whole number is the digit in the "ones place". In other words, if the digit at the far right is 0, 2, 4, 6 or 8, the number is even.

When the last digit of a significant digit is 5, you always round up.

40 percent of 9200 = 368040% of 9200= 40% * 9200= 40%/100% * 9200= 40 * 92= 3680

Add the least significant digits (carry any thing greater than the base)Add the carry and the next least most significant digits, (carry anything greater than the base)Add the third most significant digits, (carry anything greater than the base)Add the carry to the most significant digit.Done

unsigned lowest_digit (unsigned num) { unsigned lowest = 9; // lowest digit until proven otherwise... while (num) { unsigned digit = num%10; // extract least-significant digit if (digit

All digits are significant in this case. Any digit after a decimal point is a significant digit.

When you divide 1387 by 2 you get a remainder of 1, so it is odd. A number ending (least significant digit) in a odd number digit like 1, 3, 5, 7, and 9 is odd. A number ending (least significant digit) in a even number digit like 0, 2, 4, 6, and 8 is even.* * * * *Nearly correct!An integer ending in 1, 3, 5, 7 and 9 is odd and an integer ending in 0, 2, 4, 6, 8 is even. The concept of odd and even does not apply to decimal fractions.When you divide 1387 by 2 you get a remainder of 1, so it is odd. A number ending (least significant digit) in a odd number digit like 1, 3, 5, 7, and 9 is odd. A number ending (least significant digit) in a even number digit like 0, 2, 4, 6, and 8 is even.

The first significant digit is the 5, which is in the ten-thousandth's position.

But according to the rules of significant figures, the least number of significant figures in any number of the problem determines the number of significant figures in the answer which, in this case, would be 11.

2500 is the least whole number that rounds to 3000 because the last significant digit is 5.

Imagine a decimal number, for example, 123 (hundred and twenty-three). Each digit has a corresponding place-value; the right-most digit has the place-value 1, the next digit (counting from the right) has the place-value 10, the next digit hast eh place-value 100. The right-most position (where the digit "3" is in this example) is in the position of least value - the least significant position. When several bits represent an integer, the situation is the same, except that the numbers are in base-2 instead of base-10 (each position is worth twice as much as the position to the right). But you still have the concept of place-value, and the digit that represents the 1's position is the "least significant bit".

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