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1004.497 nearest whole number

Find the nearest whole number of 1004.497

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What is 1004.8 to the nearest whole number - Answers.com


What is 1004.8 to the nearest whole number? ... so the nearest whole number is 2. ... The above does not say what happens to 0.497, for example.
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What is the nearest whole number for 0.497 - answers.com


It is 0.
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How to Round a Number - WebMath


How to Round a Number. Rounding a number is when you take a number and "bump it up" or "bump it down" to a nearby and "cleaner" number.
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CLARK COUNTY SCHOOL DISTRICT LICENSED EMPLOYEE SALARY ...


J 58,459 16,369 6,764 848 497 29 82,966 ... J 69,265 19,394 6,764 1,004 589 35 97,051 ... All figures rounding to the nearest whole number. Title:
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Decimal Rounding | Math For You


Whole Number Comparison; Whole Number Addition; ... Round 0.762 to the nearest tenth ... Tenth (10th) Round 0.821 to the nearest hundredth ... Hundredth (100th)
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FORM SECOND QUARTER - 2000 85 KANSAS SCHEDULE 1 - IFTA Fuel ...


PAGE 1 Round amounts in Columns B through F to nearest whole gallon and mile ... NAME AS SHOWN ON FORM 85 IFTA License Number Tax Period ETHANOL April 1, 2000 ...
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Monthly operating rept for Oct 1995 for Indian Point Unit 2 ...


12. Number. Of Hours Reactor ... 11, 497 27 967 12' 953 28. 967 ... Compute to the nearest whole megawatt. ,. . (9/77) UNIT SHUTDOWNS AND POWER REDUCTIONS
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Rounding to the Nearest Thousandth Calculator | Calculator ...


Rounding to the Nearest Thousandth Calculator rounds given decimals and positive integer or negative integers to the nearest ... nearest whole number calculator:
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Suggested Questions And Answer :


Round off 9704.8372 to the: nearest thousand,nearest whole number , two decimal places , three significant figures

Nearest thousand is 10000, which is nearer than 9000. 9704.8372 is 704.8372 away from 9000; but only 295.1628 away from 10000. The guide is usually the digit in the hundred position: if it's 5 or more then we go up to the next thousand (9 to 10 in this case), otherwise we just take the thousand digit as it is (9 in this case). 7, the hundred digit, is bigger than 5, so we take 10 thousand. Nearest whole number is 9705, which is nearer than 9704, because the next digit 8 is bigger than 5 and the 4 in the ones position goes up to 5. 2 dec places is 9704.84, which is nearer than 9704.83. 3 sig figures is 9700. The 3 figures are 9, 7 and 0, because the zero is occupying a place separating it from the next digit 4, so it is significant. The 4 would make it 4 sig figures. The size of the number has to be preserved because, for example, 970 would not suggest the right number of thousands. Leading zeroes in whole numbers or mixed numbers are always discounted because they do not have any effect on the magnitude of the number.
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estimate to the nearest whole number 3.68/8.44

"Solution: Given 3.68/8.44 =>368/844 => 0.4360" Whole Number Definition Whole number is collection of positive numbers and zero. Whole number also called as integer. The whole number is represented as {0, 1, 2, 3, 4, 5, 6, 7, 8, 9 ….}. The opposite of whole numbers are negative numbers. The representation of negative number is {-1, -2, -3, -4, -5, -6, -7, -8, -9..}
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how do i convert this function: 2^n-1 so that it only returns whole number solutions.

Make it into the sum of a geometric progression where the first term (n=0) is 1 and a sub n+1=2a sub n. The sum to n terms is 2^n-1: 1, 2, 4, 8, 16. The sum of the first 5 terms is 31. The next sum would be 63=2^6-1. Sn=∑2^i for i=0 to n-1=2^n-1.
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what is three tenths rounded to the nearest whole number?

round(0.3)=0 ..........
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what is three tenths rounded to the nearest whole number?

round(0.3)=0 ............
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Interpolate the data set (1, 150), (3, 175), (4, 185), (6, 200), (8, 300) to estimate the amount of money Gracie may earn if she displays her items for 7 hours

Since 7 is halfway between 6 and 8, Gracie should earn an amount about halfway between 200 and 300, that is, 250 (triangular interpolation). This is the simplest interpolation, but see later. Interpolating for 2 and 5 we get 162.5 and 192.5, that is, respectively, a difference of 12.5 (162.5-150 or 175-162.5) and 7.5 (192.5-185 or 200-192.5), while 250 is a difference of 50 from 200 and 300. So the interpolated figures fluctuate. For a more sophisticated approach, we need to take the whole dataset and look for a formula that best fits. One way to do this is to fit a polynomial F(x)=ax^4+bx^3+cx^2+dx+e into the five given points. This polynomial has 5 unknown coefficients, so with 5 simultaneous equations we should be able to find them. The process can be simplified slightly by taking the lowest "x" coord and using that as the zero starting point. In this case the lowest coord is 1 (hour) so we subtract 1 from the first coord of each pair to get: (0,150), (2,175), etc. F(0)=e=150. So we have the constant 150. The next step is to subtract 150 from each of the other "y" coords so we arrive at the following set of equations: (1) F(2)=16a+8b+4c+2d=25 (2) F(3)=81a+27b+9c+3d=35 (3) F(5)=625a+125b+25c+5d=50 (4) F(7)=2401a+343b+49c+7d=150 and we already have F(0)=150=e. We can now eliminate d from (1) and (2): 2F(3)-3F(2): (162-48)a+(54-24)b+(18-12)c=70-75=-5; (5) 114a+30b+6c=-5. and we can eliminate d from (3) and (4): 5F(7)-7F(5): (12005-4375)a+(1715-875)b+(245-175)c=750-350; 7630a+840b+70c=400 which simplifies to (6) 763a+84b+7c=40 or 109a+12b+c=40/7 We can eliminate c between (5) and (6): 6(6)-(5): (654-114)a+(72-30)b=240/7+5=275/7; (7) 540a+42b=275/7 or 90a+7b=275/42. So b=(275/42-90a)/7. From (6) we have: 109a+12b+c=109a+12(275/42-90a)/7+c=40/7, so c=40/7-109a-12(275/42-90a)/7; c=40/7-550/7+(1080/7-109)a=-510/7+317a/7=(317a-510)/7. We now have b and c in terms of a. We can continue to find d in terms of a. From (1) d=(25-16a-8b-4c)/2=25-16a-8(275/42-90a)/7-4(317a-510)/7; d=25-1100/147+2040/7+(-16+720/7-1268/7)a= (3675-1100+42840)/147+(-112+720-1268)a/7; d=45415/147-660a/7. We have b, c and d in terms of a, so we can find a by substituting into an equation containing all four coefficients (but not (1), because we used it to find d). Let's pick (2) and hope we get a sensible result! 81a+27(275/42-90a)/7+9(317a-510)/7+3(45415/147-660a/7)=35. From this a=5143/2772=1.855. Therefore b=-22.919, c=78.510, d=-67.687, e=150. And F(x)=1.855x^4-22.919x^3+78.510x^2-67.687x+150. This results need to be checked before we use F to find an interpolated value. Unfortunately, this polynomial approach produces inconsistent results, and needs to be discarded. Lagrange's method seems the obvious choice, even if it is tedious to do. We have 5 x values which we'll symbolise as x0, x1, x2, x3, x4 and 5 function values f0, f1, f2, f3, f4. If the function we're looking for is f(x) then: f(x)=(x-x1)(x-x2)(x-x3)(x-x4)f0/((x0-x1)(x0-x2)(x0-x3)(x0-x4))+         (x-x0)(x-x2)(x-x3)(x-x4)f1/((x1-x0)(x1-x2)(x1-x3)(x1-x4))+         (x-x0)(x-x1)(x-x3)(x-x4)f2/((x2-x0)(x2-x1)(x2-x3)(x2-x4))+... x0=1, x1=3, x2=4, x3=6, x4=8; f0=150, f1=175, f2=185, f3=200, f4=300. We want x=7, so f(7) is given by: 4.3.1.-1.150/(-2.-3.-5.-7)+6.3.1.-1.175/(2.-1.-3.-5)+ 6.4.1.-1.185/(3.1.-2.-4)+6.4.3.-1.200/(5.3.2.-2)+ 6.4.3.1.300/(7.5.4.2) This comes to: -60/7+105-185+240+540/7=1600/7=228.57 (229 to the nearest whole number) compared with 250 from the simple interpolation. 
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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 many different distributions can the manager make if every employee receives at least one voucher?

There must be 100 vouchers because each is worth RM5 and the total value is 500 ringgits or RM500. (i) Each employee receives at least 1 voucher, so that means there are 95 vouchers left to distribute. We can write each distribution as {A,B,C,D,E}: starting with {0,0,0,0,95}, then {0,0,0,1,94}, {0,0,0,2,93}, ..., {0,0,0,95,0}, {0,0,1,0,94}, ..., {0,0,95,0,0}, ..., ..., {95,0,0,0,0}. So when A=B=C=0, {D,E} range from {0,95}, {1,94}, ..., to {95,0}, 96 ways. When A=B=0 and C=1, {D,E} range from {0,94} to {94,0}, 95 ways. Finally when C=95 so {C,D,E}={95,0,0} we will have covered 96+95+...+1=96*97/2=4656 ways. (The sum of the whole numbers from 1 to n is given by S=n(n+1)/2.) That was for B=0; when B=1, {C,D,E} ranges from {0,0,94} to {94,0,0} to cover 95*96/2=4560 ways. When B=2 it's 94*95/2=4465 ways. So for A=0 we have 4656+4560+4465+...+3+1 = 96^2+94^2+...+2^2 = 4(48^2+47^2+46^2+...+2^2+1^2)=4*48*49*97/6=152096 (the sum of the squares of the whole numbers from 1 to n is given by S=n(n+1)(2n+1)/6. Also, the sum of whole numbers between 1 and n taken in pairs gives us: n(n+1)/2+(n-1)n/2 for each pair. This is n^2/2+n/2+n^2/2-n/2=n^2. For the next pair we get (n-2)^2 and so on.) That was just for A=0! For A=1 we have {1,0,0,0,94} to {1,94,0,0,0}. This will give us 4560+4465+...+10+6+3+1 = 95^2+93^2+...+5^2+3^2+1. There is a formula for this sum. It is S=(n+1)(2n+1)(2n+3)/3, where 2n+1=95, so n=47. So for A=1, the number of ways is 48*95*97/3=147440. For A=2 we have {2,0,0,0,93} to {2,93,0,0,0} which produces 94^2+92^2+...+4^2+2^2 = 4(47^2+46^2+...+1) = 4*47*48*95/6 = 142880. So we alternate between two formulae as A continues to go from 3 to 95.  More to follow...
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what is 300170.7631 rounded to the nearest integer?

300170.7631 rounded to the nearest integer the 7 to the right of the decimal place is greater than or equal to 5, so the 0 in the ones place rounds up to 1 Answer:  300171
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round 1 1/9 to the nearest wholw number

1 1/9 is 10/9, because we multiply the whole number 1 by the denominator (bottom) of the fraction and add the result to the numerator (top) of the fraction part making it 10, making the improper fraction 10 over 9. That's a division, 10 divided by 9. 9 goes into 10 once (the whole number 1) remainder 1, which is 1/9 or one ninth. The nearest whole number is therefore 1. Rounding to the nearest whole number just means: is 1 1/9 nearer to 1 or 2? If the remainder when 9 was divided into 10 had been more than half of 9, we would have rounded to whole number 2. How many ninths would cause us to round to 2? The number 4 is less than half of 9, so we would still be nearer to 1 than 2; but if we had 5 or more (5, 6, 7 or 8) we would round to 2. 
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