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If 21,63..... Is rounded to 21,630, what are the possible digits in the ones place

This is a 4th grade place value question

Research, Knowledge and Information :


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If 2,163 is rounded to 21 ,630, what are the possible digits in the ones place? amswer i6 Tm ( roof ! ) Problem Which number has more factors, 20 or 24?
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CHAPTER 1 Place Value - MHSchool - Macmillan/McGraw-Hill


0021_0023_C01L01_103031.indd 21 3/9/10 10:04 PM. ... Line up the ones place. Compare the digits in the ... REASONING List all of the possible digits that can ...
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Rounding Off Numbers - Basic Math Explained


Learn about rounding off numbers easily by using a number ... (Both possible rounded off numbers plus the number exactly in the ... Rounding Off; Numbers, Digits, Place;
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Rounding Numbers - Math is Fun - Maths Resources


Rounding Numbers What is "Rounding" ? ... 73 rounded to the nearest ten is 70, because 73 is closer to 70 than to 80. ... 0.0165 rounded to 2 significant digits is 0.017.
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How to Round a Number - WebMath


How to Round a Number. ... A number can be rounded to any place value you want. If you type in a number you wish to round below, ...
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Chapter 1 Whole Numbers - The First Step to Success


Chapter 1 Whole Numbers 1.1 Place Value and Rounding ... we replace the tens and ones digits with 0. The rounded number ... 21. 1,824,560 22. 6,340,625 23 ...
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Rounding and Significant Digits | Purplemath


Another consideration in rounding is when you are required to round to "an appropriate number of significant digits". ... rounded something off to ... the ones place ...
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Rounding to the Nearest Hundred - Free lesson for 2nd or 3rd ...


A complete lesson on rounding to the nearest hundred, ... on the number line are rounded to 800. ... the tens and ones digits change to zeros.
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Place value. Positional numeration -- A complete course in ...


Place value Positional ... For, in each class of three digits, there are Ones, ... The Tens place is the middle one (Ones, Tens, Hundreds). There are 8 Ten millions.
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Suggested Questions And Answer :


what is 0.3 repeating rounded to the nearest thousand

0.3333333333 (3's forever) rounded to the nearest thousand: 0.3333333333 = 0000.33333333333 The left-most 0 is in the thousands place.  The next digit over is the 0 in the hundreds place.  Because that digit (0) is less than 5, the 0 in the thousands place does not round up to 1.  We have a 1 in the thousands place and everything to the right of that just becomes 0s. Result:  0000.333333333 becomes 0000.00000000 or just 0. Answer:  0 . If you meant "round to the nearest thousandth," then: 0.33333333 (3's forever) Counting from the left, the third 3 is in the thousandths place.  One digit to the right, in the ten-thousandths place, is another 3.  Because that digit (3) is less than 5, the 3 in the thousandths place does not round up to 4. Answer:  0.333
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Eight horses are entered in a race...(Have more)

1(a) How many different orders are possible for completing the race? 8*7*6*5*4*3*2*1=8!=40,320 (b) In how many different ways can first, second, and third places be decided? (Assume there is no tie.) (8*7*6)=P(8,3)=336 or 6*C(8,3) where 6=3*2*1 the number of ways of arranging three items 2.)Telephone numbers consist of seven digits; the first digit cannot be 0 or 1. How many telephone numbers are possible? 8*10^6=8,000,000  3.)In how many ways can five people be seated in a row of five seats? 5*4*3*2*1=5!=120 4.)In how many ways can five different mathematics books be placed next to each other on a shelf? 5!=120 5.)In a family of four children, how many different boy-girl birth-order combinations are possible? (The birth orders BBBG and BBGB are different.) 16=2*2*2*2 from BBBB to GGGG 6.)Two cards are chosen in order from a deck. In how many ways can this be done if (a) the first card must be a spade and the second must be a heart?  13*13=169 (b) both cards must be spades? 13*12=156 7.)A company’s employee ID number system consists of one letter followed by three digits. How many different ID numbers are possible with this system? 26*10*10*10=26,000 Continued in comment...
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using digits 0-9 only once, what combination of 3 digit numbers add to 1000, one digit will be spare

You have to have 3 sets of 3 digits each.  The first set totals 10 (for the ones place).  The second and third sets total 9 (for the tens and hundreds place).  The 1 carries over from the total of 10 in the ones place, moving the 9 up to 10 in the tens place and the 9 up to 10 in the hundreds place. Ways to make 10: 019, 028, 037, 046, 127, 136, 145, 235 Ways to make 9:  018, 027, 036, 045, 126, 135, 234 If 10 is 019, 9 leaves 234, not enough ways If 10 is 028, 9 leaves 135, not enough ways If 10 is 037, 9 leaves 126, not enough ways If 10 is 046, 9 leaves 135, not enough ways If 10 is 127, 9 leaves 036, 045, enough ways but 036 and 045 share 0 If 10 is 136, 9 leaves 027, 045, enough ways but 027 and 045 share 0 If 10 is 145, 9 leaves 027, 036, enough ways but 027 and 036 share 0 If 10 is 235, 9 leaves 018, not enough ways. We checked all of the possible combinations of 3 digits totaling 10 and it looks like none of them leave any combinations to make two sets of 3 digits totaling 9 each. Answer:  No solution.
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10 vedic maths rules for class IX

2 instead of 5: 34/5 can be calculated by multiplying 34 by 2 instead of dividing by 5. 34*2=68. We just move the decimal point one place to the left: 34/5=6.8. 124/5=24.8 because 124*2=248. Move the decimal point: 248 becomes 24.8. 34*5 is the same as 34/2=17 but we add a zero to make 17 into 170. 73*5 is the same as 73/2=36.5 then move the decimal point one place to the right (or add zero): 36.5 becomes 365=73*5. So we only need to know how to multiply and divide by 2 to divide and multiply by 5. We just move the decimal point. Divisibility by 9 or remainder after dividing by 9. All multiples of 9 contain digits which added together give 9. As we add the digits together, each time the result goes over 9 we add the digits of the result together and use that result and continue in this way up to the last digit. Is 12345 divisible by 9? Add the digits together 1+2+3=6. When we add 4 we get 10, so we add 1 and zero=1 then we add 5 to get 6. The number is not exactly divisible by 9, but the remainder is 6. We can also ignore any 9's in the number. Now try 67959. We can ignore the two 9's. 6+7=13, and 1+3=4; 4+5=9, so 67959 is divisible by 9. Multiplying by 11. Example: 132435*11. We write down the first and last digits 1 ... 5. Now we add the digits in pairs from the left a digit step at a time. So 1+3=4; 3+2=5: 2+4=6; 4+3=7; 3+5=8. Write these new digits between 1 and 5 and we get 1456785=132435*11. But we had no carryovers here. Now try 864753*11. Write down the first and last digits: 8 ... 3. 8+6=14, so we cross out the 8 and replace it with 8+1=9, giving us 94 ... 3. Next pair: 6+4=10. Again we go over 10 so we cross out 4 and make it 5. Now we have 950 ... 3. 4+7=11, so we have 9511 ... 3. 7+5=12, giving us 95122 ... 3; 5+3=8, giving us the final result 9512283.  Divisibility by 11. We add alternate digits and then we add the digits we missed. Subtract one sum from the other and if the result is zero the original number was divisible by 11. Example: 1456785. 1 5 7 5 make up one set of alternate digits and the other set is 4 6 8. 1+5+7=13. We drop the ten and keep 3 in mind to add to 5 to give us 8. Now 4 6 8: 4+6=10, drop the ten and add 0 to 8 to give us 8 (or ignore the zero). 8-8=0 so 11 divides into 1456785. Now 9512283: set 1 is 9 1 2 3 and set 2 is 5 2 8; 9+1=0 (when we drop the ten); 2+3=5; set 1 result is 5; 5+2+8=5 after dropping the ten, and 5-5=0 so 9512283 is divisible by 11. Nines remainder for checking arithmetic. We can check the result of addition, subtraction, multiplication and (carefully) division. Using Method 2 above we can reduce operands to a single digit. Take the following piece of arithmetic: 17*56-19*45+27*84. We'll assume we have carried out this sum and arrived at an answer 2365. We reduce each number to a single digit using Method 2: 8*2-1*9+9*3. 9's have no effect so we can replace 9's by 0's: 8*2 is all that remains. 8*2=16 and 1+6=7. This tells us that the result must reduce to 7 when we apply Method 2: 2+3+6=11; 1+1=2 and 2+5=7. So, although we can't be sure we have the right answer we certainly don't have the wrong answer because we arrived at the number 7 for the operands and the result. For division we simply use the fact that a/b=c+r where c is the quotient and r is the remainder. We can write this as a=b*c+r and then apply Method 2, as long as we have an actual remainder and not a decimal or fraction. Divisibility by 3. This is similar to Method 2. We reduce a number to a single digit. If this digit is 3, 6 or 9 (in other words, divisible by 3) then the whole number is divisible by 3. Divisibility by 6. This is similar to Method 6 but we also need the last digit of the original number to be even (0, 2, 4, 6 or 8). Divisibility by 4. If 4 divides into the last two digits of a number then the whole number is divisible by 4. Using 4 or 2 times 2 instead of 25 for multiplication and division. 469/25=469*4/100=1876/100=18.76. 538*25=538*100/4=134.5*100=13450. We could also double twice: 469*2=938, 938*2=1876, then divide by 100 (shift the decimal point two places to the left). And we can divide by 2 twice: 538/2=269, 269/2=134.5 then multiply by 100 (shift the decimal point two places left or add zeroes). Divisibility by 8. If 8 divides into the last three digits of a number then the whole number is divisible by 8. Using 8 or 2 times 2 times 2 instead of 125 for multiplication and division. Similar to Method 9, using 125=1000/8. Using addition instead of subtraction. 457-178. Complement 178: 821 and add: 457+821=1278, now reduce the thousands digit by 1 and add it to the units: 278+1=279; 457-178=279. Example: 1792-897. First match the length of 897 to 1792 be prefixing a zero: 0897; complement this: 9102. 1792+9102=1894. Reduce the thousands digit by 1 and add to the result: 894+1=895. Example: 14703-2849. 2849 becomes 02849, then complements to 97150. 14703+97150=111853; reduce the ten-thousands digit by 1 and and add to the result: 11854. Squaring numbers ending in 5. Example: 75^2. Start by writing the last two digits, which are always 25. Take the 7 and multiply by 1 more than 7, which is 8, so we get 56. Place this before the 25: 5625 is the square of 75. The square of 25 is ...25, preceded by 2*3=6, so we get 625. All numbers ending in 0 or 5 are exactly divisible by 5 (see also Method 1). All numbers ending in zero are exactly divisible by 10. All numbers ending in 00, 25, 50 or 75 are divisible by 25. Divisibility by 7. Example: is 2401 divisible by 7? Starting from the left with a pair of digits we multiply the first digit by 3 and add the second to it: 24: 3*2+4=10; now we repeat the process because we have 2 digits: 3*1+0=3. We take this single digit and the one following 24, which is a zero: 3*3+0=9. When we get a single digit 7, 8 or 9 we simply subtract 7 from it: in this case we had 9 so 9-7=2 and the single digit is now 2. Finally in this example we bring in the last digit: 3*2+1=7, but 7 is reduced to 0. This tells us the remainder after dividing 2401 by 7 is zero, so 2401 is divisible by 7. Another example: 1378. 3*1+3=6; 3*6=18 before adding the next digit, 7 (we can reduce this to a single digit first): 3*1+8=3*1+1=4; now add the 7: 4+7=4+0=4;  3*4=12; 3*1+2+8=5+1=6, so 6 is the remainder after dividing 1378 by 7.  See also my solution to: http://www.mathhomeworkanswers.org/72132/addition-using-vedic-maths?show=72132#q72132
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how do you round 49,052 to the nearest thousand?

The 9 is in the thousands place. Look one digit to the right, in the hundreds place. If that digit is 5 or greater, then add 1 to the 9. If that digit is 4 or less, then leave the 9 alone. Replace everything to the right of the 9 with 0s. 0 in the hundreds place. 0 is 4 or less, so leave the 9 alone. Answer:  49,000   If the problem had been to round 49,552 to the nearest thousand, then the 5 in the hundreds place would have made the 9 go up by 1 and the answer would have been 50,000.
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1.8984375 rounded to the nearest dollar

If the question is "what is $1.8984375 rounded to the nearest dollar?" then: We're rounding to the ones place.  There's a 1 in the ones place.  The next digit to the right (8) is greater than or equal to 5, so the 1 rounds up. Answer:  $2 . If we wanted to be super-exact about it, 1.8984375 rounded to the nearest dollar is $0 because 1.8984375 is not a measure of money.  Your teacher is probably looking for the $2 version above though.
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How do I find thr greatest place value in a number for a fourth graders math assignment?

The greatest place value in a number is the place value of the digit on the far left. Example: 123.456 The digit on the far left is 1. The 1 is in the hundreds place. The greatest place value in 123.456 is the hundreds place. . Note:  Special Condition:  The greatest place value in a number is the place value of the non-zero digit on the far left. We don't usually write things with 0's on the left (00037 vs. 37), but if we do, the greatest place value in a number is the place value of the digit on the far left, ignoring any 0's to the left of the leftmost non-0 digit. Example: 0007.89 Ignore the 0's on the left 7.89 7 is on the far left 7 is in the ones place The ones place is the greatest place value in 0007.89 . Example: 00012300.456 We only remove the 0's on the far left 12300.456 The far left digit is 1 The 1 is in the ten thousands place The ten thousands place is the greatest place value in 00012300.456
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What is the greatest place value in a number, how do you find it?

The greatest place value in a number is the place value of the digit on the far left. Example: 123.456 The digit on the far left is 1. The 1 is in the hundreds place. The greatest place value in 123.456 is the hundreds place. . Note:  Special Condition:  The greatest place value in a number is the place value of the non-zero digit on the far left. We don't usually write things with 0's on the left (00037 vs. 37), but if we do, the greatest place value in a number is the place value of the digit on the far left, ignoring any 0's to the left of the leftmost non-0 digit. Example: 0007.89 Ignore the 0's on the left 7.89 7 is on the far left 7 is in the ones place The ones place is the greatest place value in 0007.89 . Example: 00012300.456 We only remove the 0's on the far left 12300.456 The far left digit is 1 The 1 is in the ten thousands place The ten thousands place is the greatest place value in 00012300.456
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Need eight digit combination using 1,2,9,0,1,0,0,0. How many possible combinations?

If by combinations you mean as in a safe, it's permutations of those digits you want isn't it? If all the numbers were different, the answer would be 8 factorial (8!), which is 8*7*6*5*4*3*2*1 = 40,320. However, you've got some duplicate digits. So we have to reduce this number by the number of permutations we've "lost". That's 4! (24) for the four zeroes and 2! (2) for the two 1's. That leaves us with 840 when we divide by 48. The same answer would apply of course for any digits similarly duplicated. If asked by the police, I shall of course deny all knowledge!
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what is 3.984 rounded to 1 decimal place

Since you need to round 3.984 to one decimal place, then that will be to the nearest tenth which 9. But the # at the right of 9 is 8, so it will be added by 1... 9 + 1 = 10 which is a 2-digit #, the 1 should be added now to 3, and what remains in the tenth digit is 0. 3.984 = 4.0 (final answer)
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