Answer: 289 units
Step-by-step explanation:
Given the following :
Inventory (I) = 180
Lead time (L) = 7 days
Review time (T) = 2 weeks = 14 days
Demand (D) = 20
Standard deviation (σ) = 5
Zscore for 95% probability = 1.645
Units to be ordered :
D(T + L) + z(σT+L)
(σT+L) = √(T + L)σ²
= √(14 + 7)5²
= √(21)25
= 22.9
D(T + L) + z(σT+L) - I
20(14 + 7) + 1.645(22.9 + 7) - I
= 420 + 49.1855 - 180
= 289.1855
= 289 quantities
So here is how we are going to answer the problem above:
Given that f is the cost of French fries and b is the cost of Hamburger.
So: 5f + 6b = 4.55
5f = 4.55 - 6b (multiply both sides by 6) and we get
30f = 27.3 - 36b
Next,
6f + 10b = 7.00 (multiply both sides by 5) and we get
30f + 50b = 35 (Now substitute the value of 30f)
27.3 - 36b + 50b = 35
14b = 35 - 27.3
14b = 7.7 (divide both sides by 14)
b = 0.55
Therefore, the cost per hamburger is $0.55
Now go back to the cost of fries:
30f = 27.3 - 36b
30f = 27.3 - 36(0.55)
30f = 27.3 - 19.8
30f = 7.5 (divide both sides by 30)
f = 0.25
Therefore, the cost per french fries is $0.25
Hope this helps.
<em><u>Hope</u></em><em><u> </u></em><em><u>this</u></em><em><u> </u></em><em><u>will</u></em><em><u> </u></em><em><u>help</u></em><em><u> </u></em><em><u>u</u></em><em><u>.</u></em><em><u>.</u></em><em><u>.</u></em><em><u>.</u></em><em><u>.</u></em>
<em><u>Have</u></em><em><u> </u></em><em><u>a</u></em><em><u> </u></em><em><u>nice</u></em><em><u> </u></em><em><u>day</u></em><em><u>.</u></em><em><u>.</u></em><em><u>.</u></em><em><u>.</u></em><em><u>.</u></em><em><u>.</u></em>
Hello!
you have to find what one inch is equal to
To find this you do 50/3
50/3 = 16.6666
So one inch is equal to 16.6666 miles
Multiply this by 5
16.6666 * 5 = 83.3333
the answer is 83.3333
Hope this helps!
The question is somehow incomplete but the answer is it in
the inferential stage of probability-based inference. It is in
complex networks of codependent variables is an lively theme in statistical
research, encouraged by such varied presentations as predicting, pedigree examination
and troubleshooting.
