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Out of the following choices where is the insignificant digit in 0.09040?
Wiki User
∙ 7y ago
The zero at the start: before the decimal point.
The final zero (after the 4) IS significant if it is an indication that the number is accurate to 5 decimal places that is, it lies in the interval [0.090395 and 0.090405] rather than [0.09035 to 0.09045].
Wiki User
∙ 7y ago
Wiki User
∙ 7y agoLeading zeros are not significant.
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How many 5 digit numbers are there with distinct digits?
There are 30,240 different 5-digit numbers. Math: 10*9*8*7*6 1st digit has 10 possible choices (0-9) 2nd digit has 9 possible choices (one of the digits was used in the 1st digit) 3rd digit has 8 possible choices 4th digit has 7 possible choices 5th digit has 6 possible choices
If automobile license plates consist of two letters followed by four digits how many different possible license plates are possible if letters and numbers can be repeated?
If the identifying information on each license plate consists of (letter-1)(letter-2)(digit-1)(digit-2)(digit-3)(digit-4), and repetition is allowed: letter-1 has 26 choices, and for each one ... letter-2 has 26 choices, and for each one ... digit-1 has 10 choices, and for each one ... digit-2 has 10 choices, and for each one ... digit-3 has 10 choices, and for each one ... digit-4 has 10 choices. Total number of choices = 262 x 104 = ( 26 x 26 x 10 x 10 x 10 x 10 ) = 676 x 10,000 = 6,760,000
How many 5 digit numbers can be formed using 0 1 2 3 4 without repetition?
Assuming a 5 digit number cannot start with 0, ie it must be at least 10,000, then there are 96 possible numbers: there are 4 choices for the first digit, leaving a choice of 4 for the second digit for each of the choices for the first digit, leaving a choice of 3 for the third digit for each of the choices for the first two digits, leaving a choice of 2 for the fourth digit for each of the choices for the first three digits, leaving a choice of 1 for the fifth (and final) digit for each of the choices for the first four digits, Making a total number of choices 4 x 4 x 3 x 2 x 1 = 96
How many 3-digit numbers can be made using the digits 1 5 7 4 6 if you can only use each digit once in ever number?
You have 5 choices for the 1st digit of the number. 4 choices for the 2nd digit and 3 choices for the 3rd and final digit. This means that there are 5 x 4 x 3 = 60 3-digit numbers that can be made from 5 different digits subject to each digit only being used once.
How many three digit numbers can you write using the digits 1 2 3 4 5 without repeating?
The first digit can be any one of 5 choices. For each of those, the second digit can be any one of 4 choices. For each of those, the third digit can be any one of 3 choices. So the total number of permutations without repetition is 5 x 4 x 3 = 60.
How many 5 digit numbers are there with distinct digits?
There are 30,240 different 5-digit numbers. Math: 10*9*8*7*6 1st digit has 10 possible choices (0-9) 2nd digit has 9 possible choices (one of the digits was used in the 1st digit) 3rd digit has 8 possible choices 4th digit has 7 possible choices 5th digit has 6 possible choices
When recording measurements obtained by the use of scientific equipment the last digit is?
The last digit is insignificant figure or uncertain .
How many 4 digit whole numbers can you make with only the digits 2 and3?
16. You have two choices for the first digit (2 or 3). With each of them you have 2 choices for the second digit. For each 2-digit number so far you have two choices for the third digit and, finally, for each of these there are 2 choices for the 4th digit. All in all, 2*2*2*2 = 16 choices.
How many 3 digit numbers are possible from the digits 0123456789?
Let's solve it a step at a time. For the first digit, how many choices do you have? 9 You can choose 1..9 but not 0, so that's nine choices for the most significant digit. For the second digit, how many choices do you have? 10 It can be 0..9. For the third digit, you also have 10 choices. Choosing one digit doesn't limit your choices for other digits and mirrored numbers (e.g. 123 and 321) are different, so all choices make a unique number. So the total is the product of our three choices: 9x10x10
How many three digit positive integers are there?
There are 9 choices for the first digit (1,2,3,4,5,6,7,8,9) No zero because numbers don't start with zeroes. 10 choices for the second digit (0,1,2,3,4,5,6,7,8,9). 10 choices for the third digit (0,1,2,3,4,5,6,7,8,9) Okay now 9x10x10=900. Answer:900
How many four-digit numbers are there?
9000. The first digit can be any one of 1-9 (9 choices). The second digit can be any one of 0-9 (10 choices). The third digit can be any one of 0-9 (10 choices). The fourth digit can be any one of 0-9 (10 choices). So, in all, 9*10*10*10 = 9000
Find the digit represented by each of the letters a b c Each letter must represent a unique digit and a cannot equal 0?
These numbers are selections from the numbers from 100 to 999. That's 9 choices for the first digit. Each time, the second digit has 9 choices (0 to 9 excluding the hundreds digit), and then the last digit has 8 choices. Total is then 9x9x8 = 648
How many 3-digit numbers can be made using the digits 4 6 7 9?
If the digits may be repeated, there are four choices for the first digit, four choices for the second, and four choices for the third. Thus there are 4x4x4 = 64 possible three-digit numbers with those digits. If the digits may not be repeated, there are still four choices for the first digit, but then only three remaining choices for the second, and then just two choices for the third. In this case there are only 4x3x2 = 24 possibilities.
If automobile license plates consist of two letters followed by four digits how many different possible license plates are possible if letters and numbers can be repeated?
If the identifying information on each license plate consists of (letter-1)(letter-2)(digit-1)(digit-2)(digit-3)(digit-4), and repetition is allowed: letter-1 has 26 choices, and for each one ... letter-2 has 26 choices, and for each one ... digit-1 has 10 choices, and for each one ... digit-2 has 10 choices, and for each one ... digit-3 has 10 choices, and for each one ... digit-4 has 10 choices. Total number of choices = 262 x 104 = ( 26 x 26 x 10 x 10 x 10 x 10 ) = 676 x 10,000 = 6,760,000
There are 60 possibilities that can be made for a three digit number using the numbers 2 4 6 8 and 9 how many of these three digit numbers are smaller than 500?
If this is a homework assignment, please consider trying it yourself first, otherwise the value of the reinforcement to the lesson offered by the homework will be lost on you.There are 60 possibilities that can be made for a three digit number using the numbers 2 4 6 8 and 9. (5 choices for the first digit, 4 choices for the second digit, and 3 choices for the third digit - multiply 5, 4, and 3, and you get 60.)To determine how many of these three digit numbers are smaller than 500, consider that, given 2 4 6 8 and 9, the first digit would have to be 2 or 4. That gives 2 choices for the first digit, 4 choices for the second digit, and 3 choices for the third digit. Multiply 2, 4, and 3, and you get 24.
How many 5 digit numbers can be formed using 0 1 2 3 4 without repetition?
Assuming a 5 digit number cannot start with 0, ie it must be at least 10,000, then there are 96 possible numbers: there are 4 choices for the first digit, leaving a choice of 4 for the second digit for each of the choices for the first digit, leaving a choice of 3 for the third digit for each of the choices for the first two digits, leaving a choice of 2 for the fourth digit for each of the choices for the first three digits, leaving a choice of 1 for the fifth (and final) digit for each of the choices for the first four digits, Making a total number of choices 4 x 4 x 3 x 2 x 1 = 96
What are all the ways to combine 1 2 3 4 5 6 7 8 9 0 in 4 digits?
Digit 1 = 10 choices (0,1,2,3,4,5,6,7,8,9) Digit 2 = 10 choices (0,1,2,3,4,5,6,7,8,9) Digit 3 = 10 choices (0,1,2,3,4,5,6,7,8,9) Digit 4 = 10 choices (0,1,2,3,4,5,6,7,8,9) Number of combinations = 10 x 10 x 10 x 10 = 104 =10000
