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What is 392 rounded to the nearest ten?
390
What is 392.049 rounded to 1 decimal place?
392.0 or commonly known as 392
How many two digits numbers are possible and the first digit can not be zero?
There are 90 two-digit numbers possible. The first digit can range from 1 to 9 (9 options), while the second digit can range from 0 to 9 (10 options). Therefore, the total number of two-digit numbers is 9 (first digit) multiplied by 10 (second digit), which equals 90.
How many three digits are there in which all the digit are distinct?
There are the digits 1 through 9 for the first digit. Then, we have 0 through 9 for the second digit - excluding the first digit. For the third digit, we have 0 through 9 excluding the two previous digits
How many 3 digits no are there such that each digit is distinct to each other?
For the first digit, you have 9 options (1-9; the 3-digit number would not start with 0). For the second digit, you also have nine options (any digit from 0-9, except the digit used for the first digit). For this third digit, you have only eight options, since two digits are "forbidden", regardless of what digits you used first. That gives a total of 9 x 9 x 8 combinations.
How many two digit number are there in which the tens digit is greater than the ones digit?
If the first digit is 9, you have 9 options (0-8) for the second digit. If the first digit is 8, you have 8 options (0-7) for the second digit. Etc. This leaves you with the arithmetic series: 0 + 1 + 2 + 3 + ... + 9.
How many three digit numbers can be formed that do not contain the number zero and no number can repeat?
Let's look at this digit-by-digit: The first digit can be any number 1-9: 9 choices The second digit can be any number 1-9 except the one that the first digit is: 8 choices The third digit can be any number 1-9 except the ones chosen by the first and second digits: 7 choices 9*8*7 = 504 total numbers
How many two-digits numbers are there in which the tens digit is greater than the ones digit?
If the first digit is 9, you have 9 options (0-8) for the second digit. If the first digit is 8, you have 8 options (0-7) for the second digit. Etc. This leaves you with the arithmetic series: 0 + 1 + 2 + 3 + ... + 9.
How many two digit numbers can be formed with the digits 0123456789 if repetition of digits are not allowed?
To form a two-digit number using the digits 0-9 without repetition, the first digit (the tens place) can be any digit from 1 to 9 (9 options), since a two-digit number cannot start with 0. The second digit (the units place) can then be any of the remaining 9 digits (including 0 but excluding the first digit). Therefore, the total number of two-digit numbers that can be formed is 9 (choices for the first digit) multiplied by 9 (choices for the second digit), resulting in 81 possible two-digit numbers.
How many natural numbers between 100 and 1000 have 3 different digits?
The first digit can be any one of 9. (any digit except zero)For each of these . . .The second digit can be any one of 9. (any digit except whatever the first one is)For each of these . . .The third digit can be any one of 8. (any digit except whatever the first 2 are)Total possibilities = 9 x 9 x 8 = 648
How many 4 digit numbers have 6 as their second digit?
There are 9 digits that can be the first digit (1-9); for each of these there is 1 digit that can be the second digit (6); for each of these there are 10 digits that can be the third digit (0-9); for each of these there are 10 digits that can be the fourth digit (0-9). → number of numbers is 9 × 1 × 10 × 10 = 900 such numbers.
How many 9 digit numbers are palindromes?
A 9-digit palindrome has the structure where the first five digits determine the last four digits in reverse order. The first digit must be from 1 to 9 (to ensure it's a 9-digit number), giving us 9 options. The next four digits (the second to fifth digits) can each be any digit from 0 to 9, providing 10 options each. Therefore, the total number of 9-digit palindromes is (9 \times 10^4 = 90,000).
