I'm guessing you'd combine them. And the end equation would be
81x6y
Answer:
you want 4 correct and 16 incorrect
there are 20 questions
each question has four answers, so
P(right answer) = 1/4
P(wrong answer) = 3/4
----
Since you want 4 correct of 20 we have a combination of 20C4
This is a binomial problem where p = 1/4, q = 3/4 and we get
(20 "choose" 4)*(probability correct)^(number correct)*(probability incorrect)^(number incorrect)
putting numbers in we get
(20c4)*(1/4)^4*(3/4)^16
This gives us
~ .189685
Step-by-step explanation:
There is no solution. Here is the explanation with both of the methods :)
78.00 = 59.99+0.10x
10.01=0.10x
x= 180.1 so she can send 180 texts plus the 100 free ones for a total of 280 texts
double check: 180 x 0.10 = 18.00 + 59.99 = 77.99
<span>(1 + cos² 3θ) / (sin² 3θ) = 2 csc² 3θ - 1
Starting with the left: Note that cos²θ + </span><span>sin²θ = 1.
In the same way: </span><span>cos²3θ + <span>sin²3θ = 1
</span></span>Therefore cos²3θ = 1 - <span>sin²3θ
</span> From the top: (1 + cos² 3θ) = 1 + 1 - sin²3θ = 2 - <span>sin²3θ
</span>
(1 + cos² 3θ) / (sin² 3θ) = (<span>2 - sin²3θ) / (sin² 3θ) = 2/</span><span>sin² 3θ - </span><span>sin²3θ/</span>sin²3θ
= 2/<span>sin² 3θ - 1; But 1/</span><span>sinθ = csc</span><span>θ, Similarly </span>1/sin3θ = csc3θ
= 2 *(1/sin<span>3θ)² - 1</span>
= 2csc²3θ - 1. Therefore LHS = RHS. QED.
