Notice that
11/12 = 1/6 + 3/4
so that
tan(11π/12) = tan(π/6 + 3π/4)
Then recalling that
sin(x + y) = sin(x) cos(y) + cos(x) sin(y)
cos(x + y) = cos(x) cos(y) - sin(x) sin(y)
⇒ tan(x + y) = (tan(x) + tan(y))/(1 - tan(x) tan(y))
it follows that
tan(11π/12) = (tan(π/6) + tan(3π/4))/(1 - tan(π/6) tan(3π/4))
tan(11π/12) = (1/√3 - 1)/(1 + 1/√3)
tan(11π/12) = (1 - √3)/(√3 + 1)
tan(11π/12) = - (√3 - 1)²/((√3 + 1) (√3 - 1))
tan(11π/12) = - (4 - 2√3)/2
tan(11π/12) = - (2 - √3) … … … [A]
Answer:
no they don't
hope this helps
have a good day :)
Step-by-step explanation:
It would be.703 i think hope this helps
Answer:
0.3891 = 38.91% probability that only one is a second
Step-by-step explanation:
For each globet, there are only two possible outcoes. Either they have cosmetic flaws, or they do not. The probability of a goblet having a cosmetic flaw is independent of other globets. So we use the binomial probability distribution to solve this question.
Binomial probability distribution
The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

In which
is the number of different combinations of x objects from a set of n elements, given by the following formula.

And p is the probability of X happening.
17% of its goblets have cosmetic flaws and must be classified as "seconds."
This means that 
Among seven randomly selected goblets, how likely is it that only one is a second
This is P(X = 1) when n = 7. So


0.3891 = 38.91% probability that only one is a second