just some more q's (1 Viewer)

[Pace_T]

just some q's if you wouldn't mind answering them for me, thanks

Q1. Urn A contains 6 white and 4 black balls. Urn B contains 2 white and 2 black balls. From urn A two balls are selected at random and placed in urn B. From urn B two balls are then selected at random. What is the probability that exactly one of these two balls is white.

A: Ok I got the answer to this one:
(6C1*4C1*3C1*3C1)(/10C2*6C2 ) + (6C2*4CO*4C1*2C1)(/10C2*6C2) +(6C0*4C2*2C1*4C2)(/10C2*6C2)


but the question is, how come the bottom is 10C2*6C2, and not 10C1*9C1*6C1*5C1. I thought for this question it would have to be not replaced back. Am I only able to use the 10C1*9C1*6C1*5C1 only when they SPECIFICALLY say that it isn't replaced? I am confused

Q2. Solve the equation: 3cos2x - 2sin2x = 1.
I'm assuming we use the Acos(x-a) expansion, but what to do about the 2x in the cos and sin?

Q3. Solve the equation: 5cos@ + 12cos@ - 13 = 0

Q4. Solve the equation: 4cos@ + 3sin@ = 1

Thank you for your time.

 

[KFunk]

For Q3. Solve the equation: 5cos@ + 12cos@ - 13 = 0

I think you may have mistyped it but that's just
17cosθ = 13
cosθ = 13/17 and solve as per usual. Is there a square, a sinθ or a 2θ in there somewhere?

For Q4. Solve the equation: 4cos@ + 3sin@ = 1 use:

acosθ + bsinθ = rsin(θ - α )
where r = √(a² + b²) and tanα = b/a, and where α is in the first quadrant.

 

[KFunk]

Q2. Solve the equation: 3cos2x - 2sin2x = 1.
I'm assuming we use the Acos(x-a) expansion, but what to do about the 2x in the cos and sin?

You're right, you use the Acos(x-a) transformation. You'll just end up with Acos(2x-a) instead so that you have to halve all the values you calculate (or if you have a range you'll be going through double the range).

 

[Pace_T]

thanks for ur help

i think i got the hang of it now
thanks!