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2 years ago

HOW TO CALCULATE MARGINAL RATE

Engineering

1 answer:

Basile [38]2 years ago

8 0

Answer:

Divide the difference in tax by the amount of income from the investment, and you'll get the economic marginal tax rate from investing. Most people refer to marginal tax rates as being identical to tax brackets.

hope this helps

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Explanation:

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When using fall arrest, free fall must be kept at or below how many feet

FromTheMoon [43]

6 feet
Explanation: A few extra feet of free fall can significantly increase the arresting force on the employee, possibly to the point of causing injury. Because of this, the free fall distance should be kept at a minimum, and, as required by the standard, in no case greater than 6 feet (1.8 m).

5 0

2 years ago

A bakery wants to determine how many trays of doughnuts it should prepare each day. Demand is normal with a mean of 15 trays and

shutvik [7]

The number of trays that should be prepared if the owner wants a service level of at least 95% is; 7 trays

<h3>How to utilize z-score statistics?</h3>

We are given;

Mean; μ = 15

Standard Deviation; σ = 5

We are told that the distribution of demand score is a bell shaped distribution that is a normal distribution.

Formula for z-score is;

z = (x' - μ)/σ

We want to find the value of x such that the probability is 0.95;

P(X > x) = P(z > (x - 15)/5) = 0.95

⇒ 1 - P(z ≤ (x - 15)/5) = 0.95

Thus;

P(z ≤ (x - 15)/5) = 1 - 0.95

P(z ≤ (x - 15)/5) = 0.05

The value of z from the z-table of 0.05 is -1.645

Thus;

(x - 15)/5 = -1.645

x ≈ 7

Complete Question is;

A bakery wants to determine how many trays of doughnuts it should prepare each day. Demand is normal with a mean of 15 trays and standard deviation of 5 trays. If the owner wants a service level of at least 95%, how many trays should he prepare (rounded to the nearest whole tray)? Assume doughnuts have no salvage value after the day is complete. 6 5 4 7 unable to determine with the above information.

Read more about Z-score at; brainly.com/question/25638875

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4 0

1 year ago

5. A typical paper clip weighs 0.59 g and consists of BCC iron. Calculate (a) the number of

marta [7]

Answer:

(a) $3.185*10^{21} cells$

(b) $6.37*10^{21} atoms$

Explanation:

(a)

Volume, V of unit cell

$V=(2.866*10^{-8})^{3}=2.354*10^{-23}$

Number of unit cells, N

$N=\frac {W_{mat}}{V\rho_{mat}}$ Where $W_{mat}$ is weight of material and $\rho_{mat}$ is density of material

$N=\frac{0.59}{7.87*(2.354*10^{-23}}=3.185*10^{21} cells$

(b)

Number of atoms in paper clip

This is a product of number of unit cells and number of atoms per cell

Since iron has 2 atoms per cell

Number of atoms of iron= $3.185*10^{21} cells*2 atoms/cell=6.37*10^{21} atoms$

8 0

2 years ago

The snowmobile has a weight of 250 lb, centered at G1, while the rider has a weight of 150 lb, centered at G2. Ifh=3ft, determin

KATRIN_1 [288]

Answer:

See explaination

Explanation:

Please kindly check attachment for the step by step solution of the given problem.

5 0

3 years ago

A binary geothermal power plant uses geothermal water at 160°C as the heat source. The cycle operates on the simple Rankine cycl

bogdanovich [222]

A binary geothermal power operates on the simple Rankine cycle with isobutane as the working fluid. The isentropic efficiency of the turbine, the net power output, and the thermal efficiency of the cycle are to be determined

Assumptions :

1. Steady operating conditions exist.

2. Kinetic and potential energy changes are negligible.

Properties: The specific heat of geothermal water ( $c_{geo}$ [) is taken to be 4.18 kJ/kg.ºC.

Analysis (a) We need properties of isobutane, we can obtain the properties from EES.

a. Turbine

P $P_{3} = 3.25mPa = (3.25*1000) kPa\\= 3250kPa\\from the EES TABLE\\h_{3} = 761.54 kJ/kg\\s_{3} = 2.5457 kJ/kg\\P_{4} = 410kPa\\\\s_{4} = s_{3} \\h_{4s} = 470.40kJ/kg\\\\T_{4} = 179.5^{0} C\\\\h_{4} = 689.74 kJ/KG\\\\ The isentropic efficiency, n_{T} = \frac{h_{3}-h_{4} }{h_{3}- h_{4s} }$

= $=\frac{761.54-689.74}{761.54-670.40} \\=\frac{71.8}{91.14} \\=0.788$

b. Pump

$h_{1} = h_{f} @ 410kPa = 273.01kJ/kg\\v_{1} = v_{f} @ 410kPa = 0.001842 m^{3}/kgw_{p,in} = \frac{v_{1}(P_{2}-P_{1}) }{n_{p} } \\\\= \frac{0.01842(3250-410)}{0.9} \\\\ =5.81kJ/kg\\h_{2} =h_{1} + w_{p,in}\\ = 273.01+5.81\\ = 278.82 kJ/kg\\\\w_{T,out} = m^{.} (h_{3} -h_{4} )\\=(305.6)(761.54-689.74)\\=305.6(71.8)\\=21,942kW\\\\$

$W^{.} _ {P,in} = m^{.} (h_{2} -h_{1}) \\=m^{.} w_{p,in \\=305.6(5.81)\\\\=1,777kW\\W^{.} _{net} = W^{.} _{T, out} - W^{.} _{P,in} \\= 21,942-1,777\\=20,166 kW\\\\HEAT EXCHANGER\\\\Q_{in} = m^{.} _{geo} c_{geo} (T_{in-T_{out} } )\\=555.9(4.18)(160-90)\\=162.656kW\\$

c. $The thermal efficiency of the cycle n_{th} =\frac{W^{.} _{net} }{Q^{._{in} } } \\\\= \frac{20,166}{162,656} \\=0.124\\=12.4%$

7 0

3 years ago

Read 2 more answers