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if x=round(26.3,10)-1 then x= how much and how explain?
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Answer / ramprakash
ROUND FUNCTION
rounds the first argument to the nearest multiple of the
second argument, or to
the nearest integer when the second argument is omitted.
So the answer for x will be 29...
| Is This Answer Correct ? | 16 Yes | 1 No |
Answer / radhiv kumar sure
X=ROUND(26.3,10)-1;
AFTER ROUNDING THE VALUE
X=30-1;
X=29;
| Is This Answer Correct ? | 11 Yes | 1 No |
Answer / samba
Hello Its Me.. u r wrong. the answer is 29.
The ROUND function rounds the first argument to a value that is very close to a multiple of the second argument. The result might not be an exact multiple of the second argument.
| Is This Answer Correct ? | 2 Yes | 0 No |
Answer / its me
The answer is 25 and not 29.
Can any one challenge on this?
It first rounds of the first argument to the nearest value.
The nearest value is 26 and then deducts 1 from that to be
25.
| Is This Answer Correct ? | 4 Yes | 12 No |
what is _error_?
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/* This is example of age caluculate wihtout to display perfect days and years in output window */ data age; retain dob "12jun2003"d now "24may2011"d; age1=now-dob; age=(now-dob)/365.25; years=int(age); days1=round((age-years)*365.25); months=month(now)-1; if days1 gt 30 and months in(12,10,8,6,4,2)then do; month1=days1/30.4375; month=int(days1/30.4375); if day(now)=1 then days=1; else days=round((month1-month)*30.4375)+1; drop days1 month1 month; end; else if days1 gt 30 and months in (1,3,5,7,9,11)then do; month1=days1/30.4375; month=int(days1/30.4375); if day(now)=1 then days=1; else days=round((month1-month) *30.4375); drop days1 month1 month; end; drop age age1; proc print data=age; format dob now date.; run;
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