The number of boys is a random variable that is binomial distributed. Recall that mean and standard deviation for a binomial distribution are
and
, respectively, where n = number of trials and p = probability of success.
In this case, n = 51. Since the gender selection method has no effect of boys and girls, the probability for a boy to be born is .5. So p = .5. Then the mean is
and the standard deviation is
This is obviously a problem for system of two equations.
Let
Z=number of zebras (4 legged)
P=number peacocks (2 legged).
We know that
P=8-Z (total of 8 animals)
and
4Z+2P=4Z+2(8-Z)=28
expand and solve for Z
4Z+16-2Z=28
2Z=12
Z=6 There are 6 zebras.
Another way to solve, without pen and paper:
if all are zebras, there are 4*8=32 legs, or 4 legs too many.
Each exchange with a peacock reduces 2 legs, so we exchange 2 zebras with 2 peacocks to get 6 zebras.
Answer: 4 also fish
Step-by-step explanation:
Answer: 3(2x-1) (3x-4)
Step-by-step explanation:
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