高一米唐 人 2022-03-22
You can still ask an expert for help

Expert Community at Your Service

  • Live experts 24/7
  • Questions are typically answered in as fast as 30 minutes
  • Personalized clear answers
Learn more

Solve your problem for the price of one coffee

  • Available 24/7
  • Math expert for every subject
  • Pay only if we can solve it
Ask Question

Expert Community at Your Service

  • Live experts 24/7
  • Questions are typically answered in as fast as 30 minutes
  • Personalized clear answers
Learn more

Relevant Questions

asked 2022-03-28

 

 

asked 2022-05-23
asked 2022-04-03
asked 2022-04-25

How to solve a cyclic quintic in radicals?
Galois theory tells us that
z111z1=z10+z9+z8+z7+z6+z5+z4+z3+z2+z+1 can be solved in radicals because its group is solvable. Actually performing the calculation is beyond me, though - here what I have got so far:
Let the roots be ζ1,ζ2,,ζ10, following Gauss we can split the problem into solving quintics and quadratics by looking at subgroups of the roots. Since 2 is a generator of the group [2,4,8,5,10,9,7,3,6,1] we can partition into the five subgroups of conjugate pairs [2,9],[4,7],[8,3],[5,6],[10,1].
A0=x1+x2+x3+x4+x5A1=x1+ζx2+ζ2x3+ζ3x4+ζ4x5A2=x1+ζ2x2+ζ4x3+ζx4+ζ3x5A3=x1+ζ3x2+ζx3+ζ4x4+ζ2x5A4=x1+ζ4x2+ζ3x3+ζ2x4+ζx5
Once one has A0,,A4 one easily gets x1,,x5. It's easy to find A0. The point is that τ takes Aj to ζjAj and so takes Aj5 to Aj5. Thus Aj5 can be written down in terms of rationals (if that's your starting field) and powers of ζ. Alas, here is where the algebra becomes difficult. The coefficients of powers of ζ in A15 are complicated. They can be expressed in terms of a root of a "resolvent polynomial" which will have a rational root as the equation is cyclic. Once one has done this, you have A1 as a fifth root of a certain explicit complex number. Then one can express the other Aj in terms of A1. The details are not very pleasant, but Dummit skilfully navigates through the complexities, and produces formulas which are not as complicated as they might be. Alas, I don't have the time nor the energy to provide more details.

asked 2022-03-31

Find the length of the confidence interval given the following data 

 

S=3 n=275 confidence level 95 %

asked 2022-04-25

Simple Linear Regression - Difference between predicting and estimating?
Here is what my notes say about estimation and prediction:
Estimating the conditional mean
"We need to estimate the conditional mean β0+β1x0 at a value x0, so we use Y0^=β0^+β1^x0 as a natural estimator." here we get
Y0^~N(β0+β1x0,σ2h00)  where  h00=1n+(x0x)2(n1)sx2
with a confidence interval for E(Y0)=β0+β1x0 is
(b0^+b1^x0csh00,b0^+b1^x0+csh00)
where c=tn2,1α2 Where these results are found by looking at the shape of the distribution and at E(Y0^) and var(Y0^)
Predicting observations
"We want to predict the observation Y0=β0+β1x0+ϵ0 at a value x0"
E(Y0^Y0)=0  and  var(Y0^Y0)=σ2(1+h00)
Hence a prediction interval is of the form
(b0^+b1^x0csh00+1,b0^+b1^x0+csh00+1)

asked 2022-04-28

Sally has caught covid but doesn’t know it yet. She is testing herself with rapid antigen kits which have an 80% probability of returning a positive result for an infected person. For the purpose of this question you can assume that the results of repeated tests are independent. 

a) If sally uses 3 test kits what is the probability that at least one will return a positive result? 

b) In 3 tests, what is the expected number of positive results?

c) Sally has gotten her hands on more effective tests, these ones have a 90% probability of returning a positive result for an infected person. If she tested herself
twice with the new tests, how many positive results would she expect to see?

Solve your problem for the price of one coffee

  • Available 24/7
  • Math expert for every subject
  • Pay only if we can solve it
Ask Question