To solve for the value of the residues A, we can use the
formula:
<span>A^2 = 1 (mod
prime)</span>
This is only true for prime numbers. The given number
which is 317 is a prime number therefore the values of the residues A are:
A = + 1, - 1
Since I believe the problem specifically states for the list
of positive integers only and less than 317, a value of A = - 1 is therefore
not valid. However, a value of – 1 in this case would simply be equal to:
317 – 1 = 316
<span>Therefore the residue values A are 1 and 316.</span>
Here is the graphed part. Hope this helped!
Answer:The first issue one most notice is the words “at least” We are trying to find the probability of at least 2 girls.
The five possible outcomes for girls are 0,1,2,3,4. The odds of 1 girl out of 4 is .25 and the odds of 1 boy out of 4 is .25 (same as the odds of 3 out of 4 girls). Therefore the odds of 1 OR 3 girls must be .5 because 1 girl and 3 girls each has a .25 probability. If the probability of (1 OR 3 girls) equals .5, then the probability of 2 girls must be a different number.
The probability of 2 or more girls, is the sum of the probability of 4 girls (.06125)(—-.5 to the 4th power—— ), plus the probability of 3 girls (.25)——(the same as the probability of 1 boy)—- plus the probability of 2 girls. Since we know the probability of zero boys is .0625 (again, .5 to the 4th power) and the probability of 1 boy is .25 (the same as the probability of 3 girls )———then the probability of 2 girls is ((1 minus (the sum of the probability of 0 OR 1 boys) plus the (sum of the probability of 3 or 4 girls)), or 1-((.0625+.25)+(.0625+.25)), or .375. We had to derive the probability of two from the other known probabilities. Therefore .375+.25+.0625=.6875 is the probability of both AT LEAST 2 girls and also NO MORE than 2 boys. Notice this adds up to 1.375 because the probability of the central number 2 (i.e., .375) appears on both sides.
Answer:
3 follows 4 from 5 backwards
I think that it is around 47 degrees.