Let a = 693, b = 567 and c = 441
Now first we will find HCF of 693 and 567 by using Euclid’s division algorithm as under
693 = 567 x 1 + 126
567 = 126 x 4 + 63
126 = 63 x 2 + 0
Hence, HCF of 693 and 567 is 63
Now we will find HCF of third number i.e., 441 with 63 So by Euclid’s division alogorithm for 441 and 63
441 = 63 x 7+0
=> HCF of 441 and 63 is 63.
Hence, HCF of 441, 567 and 693 is 63.
Answer:
b
Step-by-step explanation:
if you subtract 5x then you will get y=-10-5x
bearing in mind that the an hour has 15 + 15 + 15 + 15 = 60 minutes, so 15 minutes in 1/4 of an hour, thus 45 minutes is 3/4 of an hour.
now, from 11PM, if we add the 5 hours first, we'll be at 4AM, pass midnight of course.
now let's add the minutes, 32 and then 45, that gives us 77 minutes.
so the time will be 4AM plus 77 minutes, since 60 minutes is 1 hr, so 4AM plus 1 hr and 17 minutes, that'd be 5:17AM.
X + x + 36 = 90
2x = 54
x = 27 Answer
Answer:
The standard deviation of the sample mean differences is _5.23_
Step-by-step explanation:
We have a sample of a population A and a sample of a population B.
For the sample of population A, the standard deviation 
is
The sample size 
is:
.
For the sample of population B, the standard deviation 
is
The sample size 
is:
.
Then the standard deviation for the difference of means has the following form:
Finally
