Which of the following factor is a cause for team failure?

John occasionally borrows the car of his friend, Sophie. Sophie has a PAP with liability limits of 100/300/50. John also has a P

Tamiku [17]

Answer:

Sophie's policy will pay up to its maximum amount of $100,000 and John's policy should pay the rest ($200,000).

Explanation:

Personal Auto Policies (PAP) provide coverage in case of injury to the insured or other third parties involved in a car accident. PAP policies provide coverage even when you occasional borrow a car. If you regularly drive someone else's car you need extended non owned liability coverage endorsement.

3 0

3 years ago

Lewis Company had the following transactions involving notes payable.

Fiesta28 [93]

Answer and Explanation:

The journal entries are shown below

1. Cash Dr $50,500

To Note payable $50,500

(Being the amount borrowed is recorded)

2. Cash Dr $61,200

To Note payable $61,200

(Being the amount borrowed is recorded)

3. Interest expense $2,020

To Interest payable $2,020

(Being the interest expense is recorded)

The computation is shown below:

= $50,500 × 8% × 6 months ÷ 12 months

= $2,020

4. Interest expense $612

To Interest payable $612

(Being the interest expense is recorded)

The computation is shown below:

= $61,200 × 6% × 2 months ÷ 12 months

= $612

5. Note payable $61,200

Interest expense $306

Interest payable $612

To Cash $62,118 ($61,200 + $918)

(Being the principal and the interest is recorded)

= $61,200 × 6% × 3 months ÷ 12 months

= $2,020

5. Note payable $50,500

Interest expense $1,010

Interest payable $2,020

To Cash $62,118 ($50,500 + $3,030)

(Being the principal and the interest is recorded)

= $50,500 × 8% × 9 months ÷ 12 months

= $3,030

8 0

3 years ago

What term refers to the business model of selling goods online through an Internet store site?

balu736 [363]

It would be B.) E-commerce this term is used to describe buying and selling products online.

4 0

3 years ago

the Bailey Brothers want to issue 20-year , zero coupon bonds that yield 9% .what price should it charge for these bonds if the

prohojiy [21]

Answer:

the amount charged is $178.43

Explanation:

The computation of the price charged is shown below:

As we know that

Future value = Present value × (1 + rate)^number of years

So,

Present value = Future value ÷ (1 + rate)^no of years

= $1,000 ÷ (1 + 0.09)^20

= $1,000 ÷ 1.09^20

= $178.43

Hence, the amount charged is $178.43

6 0

3 years ago

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

3 years ago