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Paha777 [63]
3 years ago
5

Eagletrons are all-electric automobiles produced by Mogul Motors, Inc. One of the concerns of Mogul Motors is that the Eagletron

s be capable of achieving appropriate maximum speeds. To monitor this, Mogul executives take samples of ten Eagletrons at a time. For each sample, they determine the average maximum speed and the range of the maximum speeds within the sample. They repeat this with 35 samples to obtain 35 sample means and 35 ranges. They find that the average sample mean is 103.50 miles per hour, and the average range is 4.00 miles per hour. Using these results, the executives decide to establish an R-chart. They would like this chart to be established so that when it shows that the range of a sample is not within the control limits, there is only approximately a 0.0027 probability that this is due to natural variation. The control limits for the chart based on the above requirement for the given information are: UCL Subscript Upper R overbar = nothing miles per hour (round the reponse to three decimal places). amison Kovach Supply Company manufactures paper clips and other office products. Although inexpensive, paper clips have provided the firm with a high margin of profitability. Sample size is 200 . Results are given for the last 10 samples.                                                                                                                         Sample 1 2 3 4 5 6 7 8 9 10 Defectives 4 8 3 5 6 5 4 5 2 11 The type of control chart that is best to monitor this process is ▼ c minus chart x overbar minus chart p minus chart. Eagletrons are all-electric automobiles produced by Mogul Motors, Inc. One of the concerns of Mogul Motors is that the Eagletrons be capable of achieving appropriate maximum speeds. To monitor this, Mogul executives take samples of ten Eagletrons at a time. For each sample, they determine the average maximum speed and the range of the maximum speeds within the sample. They repeat this with 35 samples to obtain 35 sample means and 35 ranges. They find that the average sample mean is 103.50 miles per hour, and the average range is 4.00 miles per hour. Using these results, the executives decide to establish an R-chart. They would like this chart to be established so that when it shows that the range of a sample is not within the control limits, there is only approximately a 0.0027 probability that this is due to natural variation. The control limits for the chart based on the above requirement for the given information are: UCL Subscript Upper R overbar = nothing miles per hour (round the reponse to three decimal places). amison Kovach Supply Company manufactures paper clips and other office products. Although inexpensive, paper clips have provided the firm with a high margin of profitability. Sample size is 200 . Results are given for the last 10 samples.                                                                                                                          Sample 1 2 3 4 5 6 7 8 9 10 Defectives 4 8 3 5 6 5 4 5 2 11 The type of control chart that is best to monitor this process is ▼ c minus chart x overbar minus chart p minus chart.
Engineering
1 answer:
Ronch [10]3 years ago
3 0

Answer:

Mogul executives takes samples of n=8 eagletrons at a time.

They found that the average is 3.25 mph.

For n = 8,

D4 =1.864(table value)

D3=0.136(table value)

The upper and lower control limits are found as

UCL =1.864 ×3.25= 6.058

LCL = 0.136 ×3.25= 0.442

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If the contact surface between the 20-kg block and the ground is smooth, determine the power of force F when t = 4 s. Initially,
grigory [225]

Answer:

The power of force F is 115.2 W

Explanation:

Use following formula

Power  = F x V

F_{H} = F cos0

F_{H} = (30) x 4/5

F_{H} = 24N

Now Calculate V using following formula

V = V_{0} + at

V_{0} = 0

a = F_{H} / m

a = 24N / 20 kg

a = 1.2m / S^{2}

no place value in the formula of V

V = 0 + (1.2)(4)

V = 4.8 m/s

So,

Power = F_{H} x V

Power = 24 x 4.8

Power = 115.2 W

3 0
3 years ago
(a) Determine the dose (in mg/kg-day) for a bioaccumulative chemical with BCF = 103 that is found in water at a concentration of
solmaris [256]

Answer:

0.064 mg/kg/day

6.25% from water, 93.75% from fish

Explanation:

Density of water is 1 kg/L, so the concentration of the chemical in the water is 0.1 mg/kg.

The BCF = 10³, so the concentration of the chemical in the fish is:

10³ = x / (0.1 mg/kg)

x = 100 mg/kg

For 2 L of water and 30 g of fish:

2 kg × 0.1 mg/kg = 0.2 mg

0.030 kg × 100 mg/kg = 3 mg

The total daily intake is 3.2 mg.  Divided by the woman's mass of 50 kg, the dosage is:

(3.2 mg/day) / (50 kg) = 0.064 mg/kg/day

b) The percent from the water is:

0.2 mg / 3.2 mg = 6.25%

And the percent from the fish is:

3 mg / 3.2 mg = 93.75%

3 0
3 years ago
A 150-lbm astronaut took his bathroom scale (aspring scale) and a beam scale (compares masses) to themoon where the local gravit
Nonamiya [84]

Answer:

a) W = 25.5 lbf

b) W = 150 lbf

Explanation:

Given data:

Mass of astronaut = 150 lbm

local gravity = 5.48 ft/s^2

a) weight on spring scale

it can be calculated by measuring force against local gravitational force which is equal to weight of body

W = mg

W = (150 \times 5.48)\times \frac{1 lbm}{32.32 lbm. ft/s^2} = 25.5 lbf

b) As we know that beam scale calculated mass only therefore no change in mass due to variation in gravity

thus W= 150 lbf

7 0
4 years ago
A transformer has 300,000 windings in its primary coil and uses 12,000V AC input. (4 points) How many windings would be needed t
viva [34]

Answer:

  2750

Explanation:

The number of windings and the voltage are proportional.

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Let n represent the number of windings to produce 110 Vac. Then the proportion is ...

  n/110 = 300,000/12,000

  n = 110(300/12) = 2750 . . . . multiply by 110

2750 windings would be needed to produce 110 Vac at the output.

7 0
2 years ago
A 11.5 nC charge is at x = 0cm and a -1.2 nC charge is at x = 3 cm ..At what position or positions on the x-axis is the electric
diamong [38]

Answer:

Explanation:

Given

q_1=11.5\ nC charge is placed at x=0\ cm

another charge of q_2=-1.2\ nC is at x=3\ cm

We know that Electric field due to positive charge is away  from it and Electric field due to negative charge is towards it.

so net electric field is zero somewhere beyond negatively charged particle

Electric Field due to q_2 at some distance r from it

E_2=\frac{kq_2}{r^2}

Now Electric Field due to q_1 is

E_1=\frac{kq_1}{(3+r)^2}

Now E_1+E_2=0

\frac{k\times 11.5}{(r+3)^2}\frac{k\times (-1.2)}{r^2}=0

\frac{3+r}{r}=(\frac{11.5}{1.2})^{0.5}

\frac{3+r}{r}=3.095

thus r=1.43\ cm

Thus Electric field is zero at some distance r=1.43 cm right of q_2

3 0
3 years ago
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