My best guess would be 60 but im not 100% sure
Our angle teta is:
teta = 3pi/4
since that is larger than pi/2 but less than pi that means that our angle lies in II quadrant (x negative y positive)
sin(3pi/4) = √2/2
cosine and tangent of that angle must be negative because of position of the angle.
cos(3pi/4) = -√2/2
tan(3pi/4) = -1
Yes that statement is true :)
Solution:
we are given that ![y=cos x](https://tex.z-dn.net/?f=y%3Dcos%20x)
Here we are going to plot two curves one for cosx and the othere also of cosx but after making a phase shift of ![\pi/2](https://tex.z-dn.net/?f=%5Cpi%2F2)
When we do a phase shift by an angle of theta , in that case actaully we add angle negative theta .
For example when we shift the phase of sinx by
we get ![sin(x-\pi/2).](https://tex.z-dn.net/?f=sin%28x-%5Cpi%2F2%29.)
Hence we are going to plot the curve of
Hence the correct option is B.
Answer:
It is known that if Z is a binomial random variable with parameters n and p and in addition, W is a binomial random variable, independent of X, with parameters m and p, it is assumed that the variable R = Z + W, is a binomial random variable with parameters (n + m and p. In this case, m = 1. Therefore, R is a binomial random variable with parameters (n + 1) and p.
Step-by-step explanation: