anonymous
  • anonymous
see attachment
Mathematics
schrodinger
  • schrodinger
See more answers at brainly.com
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this
and thousands of other questions

anonymous
  • anonymous
1 Attachment
anonymous
  • anonymous
3 are brown and so evidently 6 are not brown use the binomial \[\binom{9}{3}(.21)^3(.79)^6\]
anonymous
  • anonymous
i think you can use the formula (n x) . p^x . q ^(n-x)

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
what about the second one?
anonymous
  • anonymous
sorry third one
kropot72
  • kropot72
For the third one you use the fact that the binomial distribution can be approximated by the normal distribution when n is large and p is neither small nor near 1.\[\mu=np=175\times 0.21\] \[\sigma=\sqrt{np(1-p)}=\sqrt{175\times 0.21\times 0.79}\] Then you find the required probability by means of a standard normal distribution table.
anonymous
  • anonymous
oh, fourth one?
kropot72
  • kropot72
Basically you use the same method for the fourth one as for the third, but change the values of n and p accordingly.
kropot72
  • kropot72
You need to find one value of z-score for the third one and two values of z-score for the fourth.
anonymous
  • anonymous
because there two values 17% and 24%

Looking for something else?

Not the answer you are looking for? Search for more explanations.