Consider an urn initially containing n balls, numbered 1 through n, and suppose that balls will be randomly drawn from the urn, one by one, and without replacement (so that after n draws, it is empty'). Letting X be the number of successes that will occur, where a success is considered to occur on the ith draw if the ball obtained is numbered i or smaller, give the expected value of X, expressed as a fairly simple function of n. (E. g., if n = 5, and the balls are drawn in the order 3, 1, 5, 4, 2, then r = 3, because the 2nd, 4th, and 5th draws result in successes, but the 1st and 3rd draws don't.) (Hint: Express X as a sum of simple random variables which can only assume the values 0 and 1.)

definition: a circle is the  set of all points in a plane that are equidistant  from a given  point  called the center of the circle. we use the symbol ⊙ to represent a circle. the a line segment from the center of the circle to  any point  on the circle is a radius of the circle.

step-by-step explanation:

i googled it and this is what i got i don't know how it is. sorry.

geometric series converge the common ratio r has magnitude less than one.

which is equivalent to

for the example we need

|2-x| < 1

-1 < 2-x   and 2-x < 1

x < 3 and x > 1

we can write that

answer: 1 < x < 3

1a. common ratio r=250/25=10 > 1     diverges

1b.   r=50/100=1/2 converges

when these converge they converge to

where a is the first term.

answer: 200

1c. r=-1   so it's not the case that |r| < 1.     diverges

2a. x is the common ratio so we converge when |x|< 1

2b.   r=-10x / -2 = 5x.

we converge when

|5x| < 1

|x| < 1/5

2c.   r=3-2x

-1 < 3-2x and 3-2x < 1

-4 < -2x and 2 < 2x

x < 2 and 1 < x

1 < x < 2

2d. [/tex] r=-48x^3/12 = 4x^3[/tex]

|4x^3| < 1

|x^3| < \frac 1 4

|x| < 4^{-1/3}

abby has a greater debt becasue she owea 5 more then alan does