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.
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$0.42 that is what your answer would be
Answer:
A. No, the student is not right. The central limit theorem says nothing about the histogram of the sample values. It deals only with the distribution of the sample means.
Step-by-step explanation:
No, the student is not right. The central limit theorem says nothing about the histogram of the sample values. It deals only with the distribution of the sample means. The central limit theorem says that if we take a large sample (i.e., a sample of size n > 30) of any distribution with finite mean 
and standard deviation 
, then, the sample average is approximately normally distributed with mean and variance 
.
Alright, so since 20/100=1/5, and 1/5 of 45=9, 20% off of 45=9 and 45-9=36 dollar sale. Since 30=30/100, we can divide 95 by 100 and multiply that by 30 to get 0.95*30=28.5. 95-28.5=66.5
Next, we need to find 8% of 36+66.5=102.5. 8=8/100, so we divide 102.5 by 100 and multiply it by 8, getting 1.025*8=8.2
