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lisabon 2012 [21]
3 years ago
14

What kind of PPE should be worn operating a pit​

Business
1 answer:
earnstyle [38]3 years ago
5 0

Answer:

A mask, gloves, gown, face shield.

Explanation:

Try covering up as much as possible, but these are the four most essential protective coverings you should wear while operating a pit. Hope this helps!

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Using the fixed-time-period inventory model, and given an average daily demand of 75 units, 10 days between inventory reviews, 2
viva [34]

Answer:

a. 863

Explanation:

Calculation for the order quantity

Order quantity = 75 x (10 + 2) + (1.64 x 8) - 50

Order quantity = (75 x 12) + (1.64 x 8) - 50

Order quantity= 900 + 13.12 - 50

Order quantity= 863.12

Order quantity = 863

Therefore the Order quantity will be 863

8 0
3 years ago
Consider the following marginal cost function. a. Find the additional cost incurred in dollars when production is increased from
weqwewe [10]

Answer:   (a) $197,500

(b) $ 189,500

Explanation:

Given : The marginal cost function : C′​(x)=4000−0.4x

To find the cost function, we need to integrate the above function with respect to x.

Now, the additional cost incurred in dollars when production is increased from 100 units to 150 units will be:-

\int^{150}_{100}\ C'(x)\ dx\\\\=\int^{150}_{100} (4000-0.4x)\ dx\\\\=[4000x-\dfrac{0.4x^2}{2}]^{150}_{100}\\\\=[4000(150)-\dfrac{0.4(150)^2}{2}-4000(100)+\dfrac{0.4(100)^2}{2}]\\\\=[600000-4500-400000+2000]\\\\=197500

Hence, the additional cost incurred in dollars when production is increased from 100 units to 150 units= $197,500

Similarly,  the additional cost incurred in dollars when production is increased from 500 units to 550 units :-

\int^{550}_{500}\ C'(x)\ dx\\\\=\int^{550}_{500} (4000-0.4x)\ dx\\\\=[4000x-\dfrac{0.4x^2}{2}]^{550}_{500}\\\\=[4000(550)-\dfrac{0.4(550)^2}{2}-4000(500)+\dfrac{0.4(500)^2}{2}]\\\\=[2200000-60500-2000000+50000]\\\\=189,500

Hence, the additional cost incurred in dollars when production is increased from 500 units to 550 units = $ 189,500

4 0
3 years ago
After learning more about implied warranties and disclaimers, would you ever buy an item sold "as is"? Imagine a car salesman wh
atroni [7]

Answer:

one should go to buy a car for $8000

Explanation:

given data

car = $8,000

price down = $6,500

solution

As here Implied Warranty is the sale contract environment oral or written that provides some assurance that the products sold are suitable for trade and purpose. It arises from the operation of the law.

  • Disclaimer is a statement that order are used to prevent the creation of a warranty or contract.
  • After learning about the implied warranty and disclaimer, I was not going through the items sold.
  • For someone who does not offer special consumer protection, they should go to buy a car for $8000.
7 0
3 years ago
Which element of the marketing mix is most relevant to the activity "creating value"? select one:
defon
The answer is "price".

The marketing mix alludes to the arrangement of activities, or strategies, that an organization uses to advance its image or item in the market.There are 4 p's included in the marketing mix, that are; promotion (communicating value), product (creating value), price (capturing value), and place (delivering value).
7 0
3 years ago
Using the continuous compounding equation, if someone invested $5,000 at an interest rate of 3.5%, and someone else invested $5,
UNO [17]

Answer:

Therefore after 16.26 unit of time, both accounts have same balance.

The both account have $8,834.43.

Explanation:

Formula for continuous compounding :

P(t)=P_0e^{rt}

P(t)=  value after t time

P_0= Initial principal

r= rate of interest annually

t=length of time.

Given that, someone invested $5,000 at an interest 3.5% and another one  invested $5,250 at an interest 3.2% .

Let after t year the both accounts have same balance.

For the first case,

P= $5,000, r=3.5%=0.035

P(t)=5000e^{0.035t}

For the second case,

P= $5,250, r=3.5%=0.032

P(t)=5250e^{0.032t}

According to the problem,

5000e^{0.035t}=5250e^{0.032t}

\Rightarrow \frac{e^{0.035t}}{e^{0.032t}}=\frac{5250}{5000}

\Rightarrow e^{0.035t-0.032t}=\frac{21}{20}

\Rightarrow e^{0.003t}=\frac{21}{20}

Taking ln both sides

\Rightarrow lne^{0.003t}=ln(\frac{21}{20})

\Rightarrow 0.003t}=ln(\frac{21}{20})

\Rightarrow t}=\frac{ln(\frac{21}{20})}{0.003}

\Rightarrow t= 16.26

Therefore after 16.26 unit of time, both accounts have same balance.

The account balance on that time is

P(16.26)=5000e^{0.035\times 16.26}

              =$8,834.43

The both account have $8,834.43.

7 0
3 years ago
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