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Suppose that a small chest contains three drawers.

First drawer - contains two $1 bills,

Second drawer - contains two $100 bills,

Third drawer - contains one $1 bill and one $100 bill.

Suppose that first drawer is selected at random and then on of the bills inside the drawer is select at random.

A = the first drawer is selected

B = the second drawer is selected

C = the third drawer is selected

D = a $1 bill is selected

a) Suppose that you randomly select one drawer and then one bill from that drawer, the bill you obtain is a $1 bill. What is the probability that the second bull in this drawer is a $100 bill? In other words, find the probability P(C|D) because for the second null to be $100, it has to be the third drawer. Answer this question intuitively without making any calculation.

b) 
First, select a drawer by rolling a die once.

If either 1 or 2 occurs, the first drawer is selected ;

If either 3 or 4 occurs, the second drawer is selected ;

If either 5 or 6 occurs, the third drawer is selected.

Whenever C occurs, then select a bill by tossing a coin once.

If you obtain head, assume that you select a $1 bill ;

If you obtain tail, assume that you select a $100 bill.

Repeat this process 100 times. How many times in these 100 repetitions did the event D occur? What proportion of the time did C occur when D occurred? Use this proportion to estimate P(C|D). Does this estimate support your guess of P)C|D) in part (a) ?

(a) Show that for a fixed number x≠1, 3x² + 3x³ + ... +3xⱽ is a geometric series, find its sum in terms of x and v.

(b) The series Tᵥ(x) is given by

Tᵥ(x) = x + 4x² + 7x³ + ... + (3v-2)xⱽ , for v≠1.

By considering Tᵥ(x) - xTᵥ(x) and using the results from (a), show that

Tᵥ(x) = [ x + 2x² - (3v+1)xⱽ⁺¹ + (3v-2)xⱽ⁺² ] / [ (1-x )² ].

2)

Use the binomial expansions of ( √3 + 2 )⁶ & ( √3 - 2 )⁶ to evaluate ( √3 + 2 )⁶ + ( √3 - 2 )⁶. Hence, show that 2701 < ( √3 + 2 )⁶ < 2702.

:D

1) Suppose that x and y are positive integers such that

(x√y) + (y√x) - √(2014x) - √(2014y) + √(2014xy) = 2014

Find the value of xy.

2) Find all real numbers such that

x = √[ x - (1/x) ] + √[ 1 - ( 1/x ) ]

3) Find an integer x such that

[ 1 + (1/x) ]^(x+1) = ( 1 + 1/2014 )^(2014)

Do either one also can guys, I just want the answer as soon as possible, I will very appreciate for everything you guys that help me.

:D