9
having 90 or99
Any number that has 4 or 5 digits. A digit in a number is one number place. 1 has 1 digit, 11 has 2 digits, 111 has 3 digits, and so on.
If the digits may not be repeated, there are 6x5x4x3x2x1 = 720 numbers, If digits can be repeated, the answer is 6x6x6x6x6x6 = 46,656 numbers If zero is ong the available digits but is excluded from the first place. 5x6x6x6x6x6=38,880 numbers
They are written as numbers usually are. The place value of the digit immediately to the left of the decimal point is ones and the place value of all other digits is ten times the value of the digit to their right.
There are a total of 900 3-digt numbers (100 thru 999); Of these there are 9x9x8 = 648 numbers that do not contain repeating digits (first position: 9 possibilities, second place 9 possibilities; the 10 digits minus the one used in the first place. and third place 8 possibilities; 10 digits minus the two used in the first and second places) So, there should be 900-648=252 3-digit numbers with at least one repeating digit.
The positional place values of digits in negative numbers are in ascending order from least to greatest as for example in the number -987 the least value digit is 9 and the greatest value digit is 7 because -900 < -80 < -7 The positional place values of digits in positive numbers are in descending order from greatest to least as for example in the number 987 the least value digit is 7 and the greatest value digit is 9 because 900 > 80 > 7
No. A number with multiple digits does not have a place value. A single digit in a multi-digit number has a place value.
This is only possible if one of the digits is equal to zero. There are 90 3-digit numbers with a zero in the 10's place, and 90 3-digit numbers with a zero in the 1's place - and 9 numbers that have both a zero in the 10's place and a zero in the 1's place; these would be counted double if you just add the first two. So, you get: 90 + 90 - 9 such numbers.
Any number that has 4 or 5 digits. A digit in a number is one number place. 1 has 1 digit, 11 has 2 digits, 111 has 3 digits, and so on.
200
If the digits may not be repeated, there are 6x5x4x3x2x1 = 720 numbers, If digits can be repeated, the answer is 6x6x6x6x6x6 = 46,656 numbers If zero is ong the available digits but is excluded from the first place. 5x6x6x6x6x6=38,880 numbers
They are written as numbers usually are. The place value of the digit immediately to the left of the decimal point is ones and the place value of all other digits is ten times the value of the digit to their right.
There are a total of 900 3-digt numbers (100 thru 999); Of these there are 9x9x8 = 648 numbers that do not contain repeating digits (first position: 9 possibilities, second place 9 possibilities; the 10 digits minus the one used in the first place. and third place 8 possibilities; 10 digits minus the two used in the first and second places) So, there should be 900-648=252 3-digit numbers with at least one repeating digit.
3,612,345,678
The positional place values of digits in negative numbers are in ascending order from least to greatest as for example in the number -987 the least value digit is 9 and the greatest value digit is 7 because -900 < -80 < -7 The positional place values of digits in positive numbers are in descending order from greatest to least as for example in the number 987 the least value digit is 7 and the greatest value digit is 9 because 900 > 80 > 7
Answer: 2058 Explanation: Consider the thousandths place. You can place digits 1,2,3,4,5,6 (not 0 as the number would then become a 3-digit). Now consider hundredths place. You can place all the allowed digits ( ie, 0,1,2,3,4,5,6). In the same way you can place all allowed digits in tens and ones place. No. of digits which can be placed in thousandths place: 6 No. of digits which can be placed in hundredths place: 7 No. of digits which can be placed in tens place: 7 No. of digits which can be placed in ones place: 7 Total possible numbers which can be formed: 6 X 7 X 7 X 7 = 2058
the 3 digits is the ones place the 5 digits is the tenths place the 9 digits is the hundredths place
To round any number, look at the digit to the right of the place you are rounding to. If the digit is 5 or more, change the digit in the place you are rounding to to the next higher digit. If the digit to the right of the place you are rounding to is less than 5, leave the digit in the place you are rounding to as it is. Change all digits to the right of the place you rounded to to zeros.