A committee has ten members. There are two members that currently serve as the board's chairman and vice chairman. Each member is equally likely to serve in any of the positions. Two members are randomly selected and assigned to be the new chairman and vice chairman. What is the probability of randomly selecting the two members who currently hold the positions of chairman and vice chairman and reassigning them to their current positions?

1/90 = 1.11%

Step-by-step explanation:

We have that the number of ways of total selections and assignments possible is a permutation.

We know that permutations are defined like this:

nPr = n! / (n-r)!

In our case n = 10 and r = 2, replacing:

10P2 = 10! / (10 - 2)! = 10! / 8!

10P2 = 90

In addition to this, there will only be one way to randomly select the two members currently holding the positions of President and Vice President and reassign them to their current positions. Thus,

Probability would come being the following:

P = 1/90 = 1.11%

the correct answer is

c. 4

step-by-step explanation:

the given polynomial is

the degree of the polynomial is the highest exponent which is 4.

the -4 is called the leading coefficient.

this polynomial has only one term so it is a monomial.

the correct answer is option c.

the simplified form is

step-by-step explanation:

we want to simplify the expression,

let us change the middle bar to a normal division sign to obtain,

we multiply the first fraction by the reciprocal of the second fraction to obtain,

we factor to obtain,

we cancel the common factors to get,

we simplify to get,

the correct answer is b