In its first year of operations, Roma Company reports the following. Earned revenues of $53,000 ($45,000 cash received from customers). Incurred expenses of $29,500 ($23,050 cash paid toward them). Prepaid $8,750 cash for costs that will not be expensed until next year.

Complete question :

In its first year of operations, Roma Company reports the following. Earned revenues of $53,000 ($45,000 cash received from customers). Incurred expenses of $29,500 ($23,050 cash paid toward them). Prepaid $8,750 cash for costs that will not be expensed until next year. calculate the first year's net income under both the cash basis and the accrual basis of accounting.

Kindly check explanation

Step-by-step explanation:

Given the following :

Earned revenue = $53,000

Cash received from customers = $45000

Incurred expenses = $29,500

Cash paid towards incurred expenses = $23,050

Cash which will not be expensed till next year = $8,750

Cash Accounting:

Cash received $45,000

Expenses ($23,050 + $8750) = $31,800

Net Income $(45,000 - 31,800) = $13,200

Accrual Accounting :

Revenue earned $53,000

Expenses incurred $29,500

Net income $(53,000 - 29,500) = $23500

i wish i could you, i have no idea how to answer this

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.