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4 years ago

Production and sales estimates for June are as follows:

Business

1 answer:

anastassius [24]4 years ago

5 0

Answer:

Production= 13,000

Explanation:

Giving the following information:

Estimated inventory (units), June 1 18,500

Desired inventory (units), June 30 19,000

Expected sales volume (units):

Area X 3,000

Area Y 4,000

Area Z 5,500

Total= 12,500

To calculate the production for the period, we need to use the following formula:

Production= sales + desired ending inventory - beginning inventory

Production= 12,500 + 19,000 - 18,500

Production= 13,000

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Sarah is a financial planner and wants to sell fixed-income investments. Which license does she need to procure?

dimaraw [331]

The answer is series 7 which is for investment agents <span>who want to sell </span>fixed-income<span> investment products such as bonds, stocks, and packaged products.</span>

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3 years ago

Read 2 more answers

An oil refinery is located on the north bank of a straight river that is 3km wide. A pipeline is to be constructed from the refi

Mariulka [41]

Answer:

$6,598,076.21

Explanation:

<h2>THE KEY IS TO FIND OUT THE COST FUNCTION, the calculations are very easy!!!</h2><h2></h2><h3>In order to find the cost function, take a look at the drawing attached. </h3>

We can see the river (sort of) that is 3 km wide and the storage tanks on the other side of the river 8 km apart.

<h3 />

Laying pipes under (across) the river costs 1,000,000 the km & laying pipes over land costs 500,000 per km.

<h3 /><h3>So basically the cost function is 1,000,000 multiplied by something plus 500,000 multiplied by another something.</h3><h3 />

The distance across the river can be found by using Pythagoras Theorem. A side is 3 km the other is unknown, so we call it X. And it is equal to:

$\sqrt{3^{2} +x^{2}}=\\\sqrt{9 +x^{2}}$

And we multiply it by 1,000,000; the cost of laying pipe under the river, the we get:

$1000000\sqrt{9+x^{2}$

The distance over the land is (8-x), as we can see in the drawing. So we multiply it by its cost, 500,000. And we get 500,000(8-x).

So the cost function f(x) would be:

$f(x)=1000000\sqrt{9+x^{2}} + 500000(8-x)$

<h2>From here, we just have to differentiate and the derivative found must be equal to zero in order to minimize cost. </h2><h3>The value of x when the derivative is zero is plugged in the original function to get the cost.</h3><h3 /><h2>LET'S DO THIS</h2>

$f(x)=1000000\sqrt{9+x^{2}} + 500000(8-x)\\f(x)=1000000(9+x^{2})^{1/2}+4000000-500000x\\f'(x)=\frac{1}{2} 1000000(9+x^{2})^{-1/2}(2x)-500000\\\\f'(x)=\frac{1000000x}{\sqrt{9+x^2}} - 500000$

$f'(x)=\frac{1000000x}{\sqrt{9+x^2}} - 500000=0\\\frac{1000000x}{\sqrt{9+x^2}} = 500000\\\frac{2x}{\sqrt{9+x^2}} = 1\\2x={\sqrt{9+x^2}}\\4x^2=9+x^2\\3x^2=9\\x^2=3\\x=\sqrt{3} \\$

And we plug square root of 3 in the original cost function ad we get

$f(\sqrt{3} )=1000000\sqrt{9+x^{2}} + 500000(8-x)\\f(\sqrt{3})=1000000\sqrt{9+(\sqrt{3} )^{2}} + 500000(8-(\sqrt{3}))\\f(\sqrt{3})=1000000\sqrt{9+3} + 500000(8-(\sqrt{3}))\\f(\sqrt{3})=1000000\sqrt{12}+500000(6.27)\\f(\sqrt{3})=1000000(3.46)+500000(6.27)\\f(\sqrt{3})=3464101.62+3133974.60\\f(\sqrt{3})=6598076.21\\$

<h2>so the minimal cost is $6,598,076.21</h2><h2 /><h3 />

6 0

4 years ago

Money serves as a good store of value except when an economy experiences a period of _____.

Marat540 [252]

Money serves as a good store of value except when an economy experiences a period of INFLATION.

3 0

3 years ago

Incident managers begin planning for the demobilization process when:

vampirchik [111]

Incident managers begin planning for the demobilization process when B. INCIDENT ACTIVITIES SHIFT FROM RESPONSE TO RECOVERY.

Demobilization occurs when the incident objectives have been met.

<span>Demobilization is the orderly, safe, and efficient return of all resources used in the incident to its original status and location. It must be done as soon as possible to facilitate accountability of the resources.</span>

5 0

3 years ago

Mill Company began operations on January 1, 20X1, and recognized income from construction-type contracts under different methods

Vinvika [58]

Answer:

Deferred tax asset balance on December 31, 20X3 = $115,500

Explanation:

The computation of the amount of deferred income taxes should Mill report is shown below:

<u>Year Tax purpose Book purpose Difference Deferred tax book </u>

20X1 $400,000 $0 $400,000 $84,000

20X2 $625,000 $375,000 $250,000 $52,500

20X3 $750,000 $850,000 ($100,000) ($21,000)

Deferred tax asset balance on December 31, 20X3 = $115,500

5 0

3 years ago