What is the best way to manage oil transport on

aleksley [76]

I can't see the page it is glitches out try reopening it

7 0

3 years ago

Weekly demand for a particular item averages 30 units, with a standard deviation of 4. This item is managed with a fixed-order-i

aivan3 [116]

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

$=d(OI+LT)+z\sigma_{d}\sqrt{OI+LT}-A$

Where:

d=weekly demand

OI=Order Interval

LT=Lead Time

z=Standard Deviation of Desired Service Level

$\sigma_{d}$ =Standard Deviation of weekly Demand

A= Amount at Hand

$=d(OI+LT)+z\sigma_{d}\sqrt{OI+LT}-A$

$[30(3 + 1)] + [1.964*4*\sqrt{3 + 1} - 43$

$= 120 + 15.712 - 43 = 92.7$

4 0

3 years ago

The area of triangle ABC is _____. A 60 B65 C130 D150

Romashka [77]

Is there a picture that goes with the question ?

7 0

3 years ago

If the cost of a pound of nails increased from $2.29 to $2.48. what is the percent of the increase to the whole number percent?

masya89 [10]

100.8 percent

hope it helps

7 0

3 years ago

Read 2 more answers

The top and bottom margins of a poster are each $3$ cm and the side margins are each $2$ cm. If the area of printed material on

babymother [125]

Answer:

Therefore the dimension of the poster is 12 cm by 8 cm.

Step-by-step explanation:

Let the length of the poster be x and the width be y.

Given that the area of the poster is 96 cm².

∴xy =96

$\Rightarrow y= \frac{96}{x}$

The sides margins each are 2 cm and the top and bottom margins of the poster are each 3 cm.

The length of printing space is =(x- 2.3) cm

= (x-6) cm

The width of the printing space is =(y-2.2) cm

=( y-4 )cm

The area of the printing space is A=(x-6)(y-4) cm²

∴A=(x-6)(y-4)

$\Rightarrow A=(x-6)(\frac{96}{x}-4)$ [ Putting $y= \frac{96}{x}$ ]

$\Rightarrow A=120-\frac{576}{x}-4x$

Differentiating with respect to x

$A'= \frac{576}{x^2}-4$

Again differentiating with respect to x

$A''=-\frac{1152}{x^3}$

To find the minimum area, we set A'=0

$\therefore \frac{576}{x^2}-4=0$

$\Rightarrow \frac{576}{x^2}=4$

$\Rightarrow x^2=\frac{576}{4}$

$\Rightarrow x^2 =144$

$\Rightarrow x=\pm 12$

Dimension can't be negative.

Therefore x=12

If x=12, the value of A''>0,then at x=12, the area of the poster will be minimum.

If x=12, the value of A''<0,then at x=12, the area of the poster will be minimum.

$\therefore A''|_{x=12}=-\frac{1152}{12^3}$

Therefore at x= 12 cm the area of the poster will be maximum.

The width of the poster is $y=\frac{96}{12}$ = 8 cm

Therefore the dimension of the poster is 12 cm by 8 cm.

3 0

3 years ago