Answer:
11.25
Step-by-step explanation:
7500 times 0.03 times 5 = 1125
1125 divided 100 = 11.25
simple interest formula = Amount times Rate times Time
Divide 100
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:
See answer below
Step-by-step explanation:
The statement ‘x is an element of Y \X’ means, by definition of set difference, that "x is and element of Y and x is not an element of X", WIth the propositions given, we can rewrite this as "p∧¬q". Let us prove the identities given using the definitions of intersection, union, difference and complement. We will prove them by showing that the sets in both sides of the equation have the same elements.
i) x∈AnB if and only (if and only if means that both implications hold) x∈A and x∈B if and only if x∈A and x∉B^c (because B^c is the set of all elements that do not belong to X) if and only if x∈A\B^c. Then, if x∈AnB then x∈A\B^c, and if x∈A\B^c then x∈AnB. Thus both sets are equal.
ii) (I will abbreviate "if and only if" as "iff")
x∈A∪(B\A) iff x∈A or x∈B\A iff x∈A or x∈B and x∉A iff x∈A or x∈B (this is because if x∈B and x∈A then x∈A, so no elements are lost when we forget about the condition x∉A) iff x∈A∪B.
iii) x∈A\(B U C) iff x∈A and x∉B∪C iff x∈A and x∉B and x∉C (if x∈B or x∈C then x∈B∪C thus we cannot have any of those two options). iff x∈A and x∉B and x∈A and x∉C iff x∈(A\B) and x∈(A\B) iff x∈ (A\B) n (A\C).
iv) x∈A\(B ∩ C) iff x∈A and x∉B∩C iff x∈A and x∉B or x∉C (if x∈B and x∈C then x∈B∩C thus one of these two must be false) iff x∈A and x∉B or x∈A and x∉C iff x∈(A\B) or x∈(A\B) iff x∈ (A\B) ∪ (A\C).
Answer:
x=9
Step-by-step explanation:
hope this helps i think it's right 60=9x - 21 you add 21 to both sides and get 81 = 9x then divide 81 by 9 and get x=9