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Write the limit as n goes to infinity of the summation from k equals 1 of the product of the 10th power of the quantity 5 plus 2
1 answer
1,874 views
John is building a fence that is 6 feet high and 18 feet wide to put around his garden. He has two sizes of boards that he can
1 answer
481 views
The squares of a 3×3 grid are filled with non-negative integers such that the sum of each row and the sum of each column is 7.
0 answers
445 views
Trahen is building a fence that is 6 feet high and 18 feet wide to put around his garden. He has two sizes of boards that he can
1 answer
525 views
For every positive integer n, consider all monic polynomials f(x) with integer coefficients, such that for some real number a
x(f
2 answers
584 views
According to the legend, Zeus, the king of ancient Greek gods, had the power to create thunderbolts in his hands and throw them
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405 views
A helium ion is at rest in a laboratory when it is put in an electric field of E=2 N/C. An infrared light, of wavelength 2000
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483 views
A boy is spinning on a chair and holding a paper bag so that it is horizontal, and perpendicular to the axis of the rotation.
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376 views
Exact solutions for many-body problems are rarely encountered in physics. The following problem deals with a non-trivial motion
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451 views
What is the probability that after 3 hours of randomly picking balls every minute from a box containing one yellow ball and one
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354 views
A truck of mass 3000 kg is harnessed to a trailer (Trailer A) of mass 1000 kg. The trailer is hooked to ANOTHER trailer (Trailer
1 answer
864 views
Length of Hospital Stays: The average length of “short” hospital stays for men is slightly longer thatn that for women: 5.2
1 answer
1,983 views
Answers (3)
@cle we are trying to calculate the sum of (1-p)^(x-1)=1+(1-p)^2+(1-p)^4+....+(1-p)^2k=1+sum(1-p)^2k for k=1,2,..... Thus (using the sum of geometric series formula we get => 1+(1-p)^2/(1-(1-p)^2)=1+(1-p)^2/(2p-p^2)=1/(2p-p^2)
@Hamzeh I think the right answer is: 20*(3^6-2*3*(2^6)-3)=1335/5832=0/2289
5) If K=1 then the probability is (1/3)*(3/4)+(2/3)*(1/4) If K=2 then (1/3)*(1/4)*(3/4)+(2/3)*(3/4)*(1/4) If K=3 then (1/3)*(1/4)^2(3/4)+(2/3)*(3/4)^2*(1/4) For K=n then (1/3)*(1/4)^(n-1)(3/4)+(2/3)*(3/4)^(n-1)*(1/4) We are looking for the expected value
