A)<span> How much would it cost to ship a package weighing 3.2 pounds? Explain how you arrived at your answer.
Answer:Shipping cost of 3.2 package is $4.13.
</span>We can't use the cost assigned to 3 lbs because the package is over 3 lbs but it is not over 4 lbs so we use the cost of 4lbs. (check the attachment for price)
b)<span>What type of graph is needed to represent the Media Mail shipping prices as a function of the weight of the books shipped? Explain your thinking.
Answer:Line graph </span>is needed to represent the Media Mail shipping prices as a function of the weight of the books shipped.
In line graph, x values represent the weight in pounds and the y values represent the cost.
c)<span>Graph the Media Mail shipping prices as a function of the weight of the books shipped.
Answer;When we g</span>raph the Media Mail shipping prices as a function of the weight of the books shipped, it will be;
f(x) = 2.69 + 0.48(x-1)
<span>Since: v =sqrt(3)/2 s^2h
6779 liters x 0.0353cu ft/1 liter= 239.299 cu ft
but by proportion s/h = 10/25
s = 10/25 h
and v = sqrt(3)/2 (10/25 h)^2 h
239.299 = 0.139 h^3
h = (239.299/0.139)^(1/3) = 11.985 ft</span>
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
Answer:
MArginal productivity: 
We can interpret this as he will reduce his time an <em>additional </em>0.0002 seconds for every <em>additional </em>yard he trains.
Step-by-step explanation:
The marginal productivy is the instant rate of change in the result for an increase in one unit of a factor.
In this case, the productivity is the time he last in the 100-yard. The factor is the amount of yards he train per week.
The marginal productivity can be expressed as:

where dt is the variation in time and dL is the variation in training yards.
We can not derive the function because it is not defined, but we can approximate with the last two points given:

Then we can interpret this as he will reduce his time an <em>additional </em>0.0002 seconds for every <em>additional </em>yard he trains.
This is an approximation that is valid in the interval of 60,000 to 70,000 yards of training.