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kolbaska11 [484]
2 years ago
13

Wayne Inc., a health insurance company, pays clerks an incentive based on the average amount of work completed per hour. Wayne p

ays $10 for processing 20 invoices per hour. An employee who processes 30 invoices would earn $15 per hour. Hence, Wayne pays the same rate per invoice no matter how many invoices an employee processes per hour. Which type of incentive pay does this scenario illustrate?
Mathematics
1 answer:
yulyashka [42]2 years ago
5 0

Answer: Straight piecework plan

Step-by-step explanation:

The Straight piecework plan is a type of incentive system in which the rate per unit output is constant and fixed at a specific value. And the total earnings of a worker is calculated by multiplying the value of its total output by the rate per unit output.

In the case above the rate per unit invoice processed is fixed, so employees are paid based on the number of invoice processed per hour which makes it a Straight piecework plan system.

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At the start of 2014, Mike's car is worth 12000. the car depreciates by 30 percent every year. how much is his car worth in 2017
Anarel [89]

Answer:

$4,116

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Worth of Mike's car at the start of 2014 = $12,000

If the car is said to depreciates every year by 30% = 30/100 = 0.3

The worth of the car at the start of 2017 is what we are to determine.

This means that the car depreciated by 30% (0.3) for 3 years since 2014 (2017 - 2014 = 3 yrs)

The worth at the start of 2017 would be calculated as follows:

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Jane wishes to bake an apple pie for dessert. The baking instructions say that she should bake the pie in an oven at a constant
Viktor [21]

Answer:

Therefore k= \frac{ln2 }{18}, A=184

Step-by-step explanation:

Given function is

T(t)=230 -e^{-kt}

where T(t) is the temperature in °C and t is time in minute and A and k are constants.

She noticed that after 18 minutes the temperature of the pie is 138°C

Putting T(t) =138°C and t= 18 minutes

138=230 -Ae^{-k\times 18}

\Rightarrow  -Ae^{-18k}=138-230

\Rightarrow  Ae^{-18k}=92 .....(1)

Again after 36 minutes it is 184°C

Putting T(t) =184°C and t= 36 minutes

184=230-Ae^{-k\times 36}

\Rightarrow Ae^{-36k}=230-184

\Rightarrow Ae^{-36k}=46.......(2)

Dividing (2) by (1)

\frac{Ae^{-36k}}{Ae^{-18k}}=\frac{46}{92}

\Rightarrow e^{-18k}=\frac{46}{92}

Taking ln both sides

ln e^{-18k}=ln\frac{46}{92}

\Rightarrow -18k =ln (\frac12)

\Rightarrow -18k= ln1-ln2

\Rightarrow k= \frac{ln2 }{18}

Putting the value k in equation (1)

Ae^{-18\frac{ln2}{18}}=92

\Rightarrow A e^{ln2^{-1}}=92

\Rightarrow A.2^{-1}=92

\Rightarrow \frac{A}{2}=92

\Rightarrow A= 92 \times 2

⇒A= 184.

Therefore k= \frac{ln2 }{18}, A=184

7 0
2 years ago
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