AFC = FC / Quantity printed
<span>So given she prints 1,000 posters: AFC = 250.00/1000 = $0.25 </span>
<span>Given she prints 2,000 posters: AFC = 250.00/2000 = $0.125 </span>
<span>Given she prints 10,000 posters: AFC = 250.00/2000 = $0.025 </span>
<span>ATC = TC / Quantity printed </span>
<span>where TC = FC + Variable C * Quantity printed </span>
<span>If she prints 1000: TC = 250 + 2000*1000 = 2,000,250 </span>
<span>ATC = 2,000,250/1000 = 2000.25 </span>
<span>If she prints 2000: TC = 250 + 1600*2000 = 3,200,250 </span>
<span>ATC = 3,200,250/2000 = 1600.125 </span>
<span>If she prints 10000: TC = 250 + 1600*2000 + 1000*8000 ($1000 for each additional poster after 2000) = 11,200,250 </span>
<span>ATC = 11,200,250/10000 = 1120.025</span>
Answer:
8.34/1
Step-by-step explanation:
Answer:
3. b
2. c
Step-by-step explanation:
The Range (Statistics) The Range is the difference between the lowest and highest values.
The mean is the sum divided by the count.
Answer:
See below.
Step-by-step explanation:
There is an infinite n umber of systems of equations that has (1, 4) as its solution. Are you given choices? Try x = 1 and y = 4 in each equation of the choices. The set of two equations that are true when those values of x and y are used is the answer.
Answer:
A: an = 6n - 9
Step-by-step explanation:
third term in an arithmetic sequence is 9
a3 = a + 2d
a + 2d = 9
the fifth term is 21
a5 = a + 4d
a + 4d = 21
a + 2d = 9 (1)
a + 4d = 21 (2)
Subtract (1) from (2) to eliminate a
4d - 2d = 21 - 9
2d = 12
d = 12/2
d = 6
Substitute d = 6 into (1)
a + 2d = 9 (1)
a + 2(6) = 9
a + 12 = 9
a = 9 - 12
a = -3
nth term of this sequence = a + (n - 1)d
= -3 + (n - 1)6
= -3 + 6n - 6
= 6n - 9
an = 6n - 9
