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mestny [16]
2 years ago
4

It cost a contractor $13,750 to manufacture their first unit. The company expects to experience a 93% unit learning curve. Estim

ate the cost of the 14th unit. (Round intermediate calculations to 4 decimal places)
Mathematics
1 answer:
Pepsi [2]2 years ago
8 0

Answer:

$275

Step-by-step explanation:

Price of first unit = $ 13,750

Unit Learning curve = 93%

T find the cost of 14th unit

Unit learning curve = 93%

So decrease in price in every  unit = 7 %

Decrease in Price of one unit = 7 % of $13750

                                                 = \frac{7}{100} * 13750

                                                 =$ 962.5

Price after one unit = 13750 - 962.5 = $ 12787.5

Decrease in price of 14 units = 14* 962.5

                                                =$13475

Estimated price of 14th unit = $ 13750 - $13475

                                             = $275


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zubka84 [21]

Answer:

<em>24 square yards</em>

Step-by-step explanation:

Find the diagram attached.

The area of the diagram = Area of rectangle + Area of triangle

Area of rectangle = 3 * 4

Area of rectangle = 12 square yards

Area of triangle = 1/2 * base * height

Area of triangle = 1/2 * 6 * 4

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7 0
2 years ago
The dye dilution method is used to measure cardiac output with 3 mg of dye. The dye concentrations, in mg/L, are modeled by c(t)
Lemur [1.5K]

Answer:

Cardiac output:F=0.055 L\s

Step-by-step explanation:

Given : The dye dilution method is used to measure cardiac output with 3 mg of dye.

To Find : Find the cardiac output.

Solution:

Formula of cardiac output:F=\frac{A}{\int\limits^T_0 {c(t)} \, dt} ---1

A = 3 mg

\int\limits^T_0 {c(t)} \, dt =\int\limits^{10}_0 {20te^{-0.06t}} \, dt

Do, integration by parts

[\int{20te^{-0.6t}} \, dt]^{10}_0=[20t\int{e^{-0.6t} \,dt}-\int[\frac{d[20t]}{dt}\int {e^{-0.6t} \, dt]dt]^{10}_0

[\int{20te^{-0.6t}} \, dt]^{10}_0=[\frac{-20te^{-0.6t}}{0.6}+\frac{20}{0.6}\int {e^{-0.6t} \,dt]^{10}_0

[\int{20te^{-0.6t}} \, dt]^{10}_0=[\frac{-20te^{-0.6t}}{0.6}+\frac{20e^{-0.6t}}{(0.6)^2}]^{10}_{0}

[\int{20te^{-0.6t}} \, dt]^{10}_0=[\frac{-200e^{-6}}{0.6}+\frac{20e^{-6}}{(0.6)^2}]+\frac{20}{(0.60^2}

[\int{20te^{-0.6t}} \, dt]^{10}_0=\frac{20(1-e^{-6}}{(0.6)^2}-\frac{200e^{-6}}{0.6}

[\int{20te^{-0.6t}} \, dt]^{10}_0\sim {54.49}

Substitute the value in 1

Cardiac output:F=\frac{3}{54.49}

Cardiac output:F=0.055 L\s

Hence Cardiac output:F=0.055 L\s

4 0
2 years ago
Ali took five Math tests during the semester and the mean of his test score was 85. If his mean after the first three was 83, Wh
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Answer:

88

Step-by-step explanation:

Mean of five test = 85

Sum of five tests = 85*5 = 425

Mean of first three test =83

Sum of first three test = 83 * 3 = 249

Sum of 4th and 5th test scores = 425 - 249

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How long is bob's journey from aberystwyth to shrewsbury
steposvetlana [31]

Answer: bob will take a while

Step-by-step explanation:

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