From what I understood in the problem, the total budget that covers all types of media is only $1,000 per month. For the allocation, each type of media would get at least 25% of the budget. If we infer on this information, there should only be 4 types of media, at least. This is because four 25% portions would equal to 100%. If it exceeds 25% for each of the four types, it would be over the $1000 budget. With that being said, it is also possible that there will be 3 or 2 types of media. Nevertheless, let's just stick to the least assumption of 25% for each of the 4 types.
If local newspaper advertising is one of the four types, then:
$1000(25%) = $250
It would get $250 from the overall budget.
Answer and Explanation:
The Preparation of direct material budget is shown below:-
Direct Material budget
Particulars Amount
Units to be produced $90,000 Y
Material per unit 2
Total pounds needed for
production M $180,000 2Y
Add: Desired ending Direct
Material Inventory 20% $36,000 (.2 × 2Y = .4Y)
Total Material requirement $216,000 (2.4Y
)
Less: beginning Raw material
Inventory $9,000 (.1Y)
Material to be purchased
Account $207,000 (2.3Y)
Cost per pound C $5
Total cost of direct Material
Purchases A $1,035,000
2Y + .4Y - .1Y = $207,000
Y = $207,000 ÷ 2.3 $90,000
Answer:
The correct answer is D
Explanation:
Solid minerals contained in the land
(Coal, iron, ore, gold or silver)
Hope this helps! (づ ̄3 ̄)づ╭❤~
Answer:
1.6 hour
Explanation:
Given
Rate of Arrival =30 per hour
Rate of Processing = 25 per hour
Open Time = 8am
Close Time = 4pm
How long the last package has to wait before it is processed is calculated by;
Duration = ∆Time/∆Rate
∆Time = 4pm - 8am
∆Time = 8 hours
∆Rate = Rate of Arrival - Rate of Processing
∆Rate = 30 - 25
∆Rate = 5 per hour
Duration = 8 hours ÷ 5 per hour
Duration = 1.6 hours
Answer:
74.64%
Explanation:
Average sales (μ) = 50 hot dogs
Standard deviation (σ) = 7 hot dogs
In a normal distribution, the z-score for any given number of hot dogs sold, X, is determined by:

For X = 45 hot dogs:

For X = 65 hot dogs:

A z-score of -0.7143 falls in the 23.75th percentile of a normal distribution while a z-score of 2.1429 falls in the 98.39th percentile.
Therefore, the probability that he vendor will sell between 45 and 65 hot dogs is:
