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3 years ago

18

Engineering

1 answer:

svp [43]3 years ago

5 0

Answer:

серйозних порушень точності,

∵∴⊕∴∵

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Pls follow me in brainly

Sergio [31]

Answer:

sure

Add on:

5 0

3 years ago

Read 2 more answers

Which of the following would NOT fall into one of the WHMIS 1988 classes?

sweet [91]

A. Compressed Gas Yup

8 0

3 years ago

Read 2 more answers

Technician A says that as long as the dog (clutch) teeth of the synchronizer are rounded, the synchronizer can be reused. Techni

Sidana [21]

Neither of the technicians are correct.

<h3>Who is a technician?</h3>

A technician simply means an individual who looks after the technical equipments in a company.

In this case, the teeth of the synchronizer are not rounded, the synchronizer can be reused. This shows that he is incorrect likewise technician B.

Learn more about technicians on:

brainly.com/question/1548867

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6 0

2 years ago

The time to failure for a gasket follows the Weibull distribution with ß = 2.0 and a characteristic life of 300 days. What is th

Aleks04 [339]

Answer:

64.11% for 200 days.

t=67.74 days for R=95%.

t=97.2 days for R=90%.

Explanation:

Given that

β=2

Characteristics life(scale parameter α)=300 days

We know that Reliability function for Weibull distribution is given as follows

$R(t)=e^{-\left(\dfrac{t}{\alpha}\right)^\beta}$

Given that t= 200 days

$R(200)=e^{-\left(\dfrac{200}{300}\right)^2}$

R(200)=0.6411

So the reliability at 200 days 64.11%.

When R=95 %

$0.95=e^{-\left(\dfrac{t}{300}\right)^2}$

by solving above equation t=67.74 days

When R=90 %

$0.90=e^{-\left(\dfrac{t}{300}\right)^2}$

by solving above equation t=97.2 days

7 0

3 years ago

7.4 A pretimed four-timing-stage signal has critical lane group flow rates for the first three timing stages of 200, 187, and 21

Irina18 [472]

Answer:

16 seconds

Explanation:

Given:

C = 60

L = 4 seconds each = 4*4 =16

In this problem, the first 3 timing stages are given as:

200, 187, and 210 veh/h.

We are to find the estimated effective green time of the fourth timing stage. The formula for the estimated effective green time is:

$g = (\frac{v}{s}) (\frac{C}{X})$

Let's first find the fourth stage critical lane group ratio $\frac{v}{s}$ , using the formula:

$C = \frac{1.5L +5}{1 - ( \frac{200}{1800} + \frac{187}{1800} + \frac{210}{1800}) + ( \frac{v}{s})}$

$60 = \frac{1.5*16 + 5}{1 - ( \frac{200}{1800} + \frac{187}{1800} + \frac{210}{1800}) + ( \frac{v}{s})}$

$60 = \frac{24+5}{1 - (0.332 + ( \frac{v}{s}))}$

Solving for $(\frac{v}{s})$ , we have:

$(\frac{v}{s}) = 0.185$

Let's also calculate the volume capacity ratio X,

$X = (\frac{200}{1800} + \frac{187}{1800} + \frac{210}{1800} + 0.185)(\frac{60}{60-16}$

X = 0.704

For the the estimated effective green time of the fourth timing stage, we have:

$g_4 = (\frac{v}{s}) (\frac{C}{X})$

Substituting figures in the equation, we now have:

$g_4 = (0.185) (\frac{60}{0.704})$

$g_4 = 15.78 seconds$

15.78 ≈ 16 seconds

The estimated effective green time of the fourth timing stage is 16 seconds

8 0

3 years ago