Hope this helps you. Thank you....
The answer:
let be A(x) =<span>2√ 3 cos(x)csc(x)+4cos(x)-3csc(x)-2 √ 3
this function can be represented as </span><span>the product of the factors
proof
</span>2√ 3 cos(x)csc(x)+4cos(x)-3csc(x)-2 √ 3 =
2√ 3 cos(x)csc(x)+4cos(x)csc(x) / csc(x) - 3csc(x)- 2 √ 3 csc(x) / csc(x)
this method doesn't change nothing inside the function A(x)
so we have
[ 2√ 3 cos(x) +4cos(x) / csc(x) - 3 - 2 √ 3 / csc(x) ] . csc(x) this is a product of two factors,
[ 2√ 3 cos(x) +4cos(x) / csc(x) - 3 - 2 √ 3 / csc(x) ] and csc(x)
for more explanation
A(x) =[ 2√ 3 cos(x) - 3 + (4cos(x) - 2 √ 3 ) / csc(x) ] . csc(x)
A/b = c/d |cross multiply
ad = bc |divide both sides by a
d = bc/a
There are 5 blue pens total, 3 are from the first box and 2 are from the second. The probability of pulling a blue pen from the first box and not the second is 3/5.
Conditional probability:
P ( B \ A ) = P ( A and B ) / P ( A )
Probability of an event B given by that A has already occurred.
P ( A and B ) = 0.051
P ( A ) = 0.32
P ( B \ A ) = 0.051 / 0.32 = 0.159375 ≈ 0.159
Answer:
B ) 0.159
