Can you figure this out for me? (1 Viewer)

[lillaila]

1. I dun get this soln to this q.

Q: A game of poker uses a normal deck of cards. There are 4 suits each with 13 cards. If Joe is dealt 5 cards find prob.y of getting a FLUSH (all cards same suit).
A: 4(1/4)(12/51)(11/50)(10/49)(9/48)= 33/16660

2. I dunno how to solve this prob.

Q: Find domain and range for each.....

[sin^-1] is inv. sine or "arc sine"

a) y= [sin^-1](x^2)
b) y= x[cos^-1]x

A: a) D~ -1<_ x<_1
R~ 0<_y<_pi/2
b) D~ -1<_x<_1
R~ -pi<_y<pi/2

 

[lillaila]

A: b) R~ -pi<_y<pi/2

 

[lillaila]

weird ....smthng wrong wif the computer.

the answer should be " y is btwn -pi and pi on 2 or equal to pi"
in other words:
-pi<_ Y < pi on 2

 

[jogloran]

How do you work out the domain if the function is composite like that? Someone remind me?

 

[stag_j]

well the probability question looks ok to me...
for a particular suit you have 13/52 chance of picking one of the cards, then 12/51, 11/50, 10/49, 9/48 chance of picking cards of the same suit as the 2nd, 3rd, 4th, 5th card.
That can be done in 4 ways (4 suits)....

 

[...]

arc sine?!?!?!


woah, looks 4u to me

 

[stag_j]

for those inverse trig questions, the first one looks ok to me.
the second one i'm getting a different range... anyone else getting that too?
im in a hurry and havent figured it out exactly, but from my graph its looking to be between -pi and about 0.58 - can't be bothered figuring out exactly what the value is in terms of pi, altho because of the x out front it probably isnt a round number...
either way, i'm not getting the pi/2 that lillaila gave as the answer

 

[stag_j]

yeh i have no idea about that second one... i went to the point of guessing a a root of the derivative (maximum) and using newtons method which got pretty excessive and i got a different answer each time. but i definitely cant get an answer of pi/2

 

[LadyMoon]

ok

2a) simple one:

you know that for any sine inverse the variable eg: in Sin(^-1)@ (sine inverse of @) must lie between -1 and 1.
There for you put -1<@<1
(*
Less than or equal to, but you cant type that)
so instead of @, you substitue x^2, and you get -1<x^2<1, square root both sides,
(since you cant square root a negative with out doing 4 units....but remebering that a negative squared is positive)

-1<x<1 --is the domain, and then you simply subitute -1 into the orginal function to get the first value for the range and subistute 1 into getting the second value for the range, therefore the range is between those two values!


2b) y=x(cos(^-1)x)

since the 'x' in front of cos can take any value you ignore that when you are finding out the domain.
therefore you are only concerned with cos^(-1)x.

so to find the domain, in cos^(-1)@, @ must lie between 1 and -1, as like sine, so the domain becomes -1<x<1.

but remeber the maximum for cos^-1(x) is when x=0. (look at the graph)

and you subsitue x=-1 into y=x(cos(^-1)x), to get one value for the range, but the range the answer give i think are wrong, you cant have pi<y<-pi/2,

the range is actually pi<y<0
because when x=0 cos^(-1) is equal to pi/2, BUT
y=x(cos(^-1)x) when you subsitute x into that, 0 multiplied by pi/2 equals ZERO, not pi/2!!!


Hope i helped!

 

[stag_j]

ladymoon,
did you graph the function? if you do you should find that the upper limit of the range doesnt occur at x=0 or x=1 - in fact it occurs at approximately x=0.65