Answer:
11∗2+13∗4+15∗6+⋯+199∗100=?
=(1−12)+(13−14)+(15−16)+⋯+(199−1100)
=(1+13+15+17+⋯+199)−(12+14+16+⋯+1100)
=(1+13+15+17+⋯+199)−12(1+12+13+⋯+150)
=∑k=15012k−1−∑k=15012k
Continue further.
Answer:
(B)93
Explanation:
Since we are using a fixed-order-interval model,
The Amount to Order=Expected Demand During protection Interval+Safety Stock-Amount at Hand
Where:
d=weekly demand
OI=Order Interval
LT=Lead Time
z=Standard Deviation of Desired Service Level
=Standard Deviation of weekly Demand
A= Amount at Hand
![[30(3 + 1)] + [1.964*4*\sqrt{3 + 1} - 43](https://tex.z-dn.net/?f=%20%5B30%283%20%2B%201%29%5D%20%2B%20%5B1.964%2A4%2A%5Csqrt%7B3%20%2B%201%7D%20-%2043)
The answer is 10.
6^2/2(3)+4 is the formula
6^2=36
36/2(3)+4
2(3)=6
36/6+4
36/6
6+4
10
I did PEMDAS. P is for Parentheses. E is for exponents. M is for multiplication. D is for Division. A is for adding. S is for subtracting.
Hi!
It would be B, because when we do $4.25-$2.60=$1.65. When we divide that be 0.25, that equals 6.6, but you can't have a percent of a coin, so we round down.
Hope this helped!
Your Answer:
what type of question is this?
