Harlie is a hotel manager. Their responsibilities include renting rooms for conferences. The hotel charges $250 per day plus $15 per person for the grand ballroom. Create a formula for the cost, C, in dollars, to rent the ballroom for any number of people for 1 day.
Hint: Use C for cost and n for the number of people. C =

How much should Charlie charge to rent the hall for 50 people? What about 100 people?

Rearrange your formula to express n in terms of C (rewrite your formula to solve for n)

How many people could attend a wedding reception if the planners have a budget of:
$4000

$2000

Formula: C= $15n + $250

50 People: $1000

100 People: $1750

Formula: n= (C- $250)/ 15

$4000 Budget: 250 Guests

$2000 Budget: 116.67 Guests

Step-by-step explanation:

Since the rental is $250 with an additional charge of $15 per person, we write this as C = $250 + $15n.

With the given equation, simply substitute n for the amount of guests.

C = $250 + $15(50) = $1000

C = $250 + $15(100) = $1750

To solve for n, everything would need to be moved to isolate n. This can be achieved by first subtracting $250 from each side: C- 250= 15n

Then divide both sides by 15, making the new equation: (C-250)/15 = n

Again, substitute C with allowed cost to find the amount of guests allowed.

n = ($4000- 250)/ 15 = 250 People

n = ($2000-250)/15 = 116.67 People

4.5%

step-by-step explanation:

i think it's eleven too, don't woosh me here, but according to the "experts" it's 2.

the proof starts from the peano postulates, which define the natural

numbers n. n is the smallest set satisfying these postulates:

  p1.   1 is in n.

  p2.   if x is in n, then its "successor" x' is in n.

  p3.   there is no x such that x' = 1.

  p4.   if x isn't 1, then there is a y in n such that y' = x.

  p5.   if s is a subset of n, 1 is in s, and the implication

      (x in s => x' in s) holds, then s = n.

then you have to define addition recursively:

  def: let a and b be in n. if b = 1, then define a + b = a'

      (using p1 and p2). if b isn't 1, then let c' = b, with c in n

      (using p4), and define a + b = (a + c)'.

then you have to define 2:

  def:   2 = 1'

2 is in n by p1, p2, and the definition of 2.

theorem:   1 + 1 = 2

proof: use the first part of the definition of + with a = b = 1.

      then 1 + 1 = 1' = 2   q.e.d.

note: there is an alternate formulation of the peano postulates which

replaces 1 with 0 in p1, p3, p4, and p5. then you have to change the

definition of addition to this:

  def: let a and b be in n. if b = 0, then define a + b = a.

      if b isn't 0, then let c' = b, with c in n, and define

      a + b = (a + c)'.

you also have to define 1 = 0', and 2 = 1'. then the proof of the

theorem above is a little different:

proof: use the second part of the definition of + first:

      1 + 1 = (1 + 0)'

      now use the first part of the definition of + on the sum in

      parentheses:   1 + 1 = (1)' = 1' = 2   q.e.d.