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grandymaker [24]
3 years ago
15

In addition to passing an ASE certification test, automotive technicians must have __________ year(s) of on the job training or

__________ year(s) of on the job training and a two-year degree in automotive repair to qualify for certification.
Engineering
1 answer:
Lunna [17]3 years ago
8 0
Two years or one year
You might be interested in
True/False
agasfer [191]

Answer: This is true.

Hope this helps!  :)

Explanation:

4 0
3 years ago
Design a sequential circuit with two D flip-flops A and B, and one input x_in.
DENIUS [597]

Answer:

View Image

Explanation:

The question is basically asking you to build a 2-bit asynchronous counter.

What the counter does is it increase it's value by 01₂ every clock pulse. So at 0₂, nothing happens, but at 1₂ it'll count up by 1. It then reset to 00₂ when it overflows.

The design for it is pretty much universal so I kinda did this from memory.

a.) A count-up counter (from 00-11) is simply made by connecting Q' to D, and the output of the previous DFF to the clock of the next one.

b.) A count-down counter (from 11-00) is simply made by using the same circuit as the count-up counter, but you connect Q' to the clock instead of Q.

7 0
4 years ago
A thick steel sheet of area 150 in.2 is exposed to air near the ocean. After a one-year period it was found to experience a weig
Bingel [31]

Answer:

Rate of corrosion = 24.95 mpy

Rate of corrosion = 0.63 mm/yr

Explanation:

given data

steel sheet area = 150 in²

weight loss = 485 g

density of steel = 7.9 g/cm³

time taken = 1 year

to find out

rate of corrosion in (a) mpy and (b) mm/yr

solution

we get here the rate of corrosion that is express as

rate of corrosion = (k × W) ÷ (D × A × T)     ..................1

here k is constant  and w is total weight lost and t is time taken for loss and A is surface area and D is density of steel

so put her value in equation 1 we get

Rate of corrosion = \frac{534*485*1000}{7.9*150*24hr*365day/yr}

Rate of corrosion = 24.95 mpy

and

Rate of corrosion = \frac{87.6*485*1000}{7.9*150*2.54^2*24hr*365days/yr}

Rate of corrosion = 0.63 mm/yr

5 0
3 years ago
By increasing the cross-sectional area of the restriction, one can significantly increase the flow velocity. a) True b) False
natka813 [3]

Answer:

b)false

Explanation:

As we know that

Volume flow rate Q

Q = A x V

For constant volume flow rate,if velocity will increase then automatically area will decrease and vice versa.

Generally nozzle are used to increase the velocity and diffuser are used to decrease the exit velocity of flow.

So by increasing the cross sectional area of the restriction ,the velocity of the flow will decrease.

8 0
3 years ago
A company, studying the relation between job satisfaction and length of service of employees, classified employees into three le
Wewaii [24]

Answer:

Below see details

Explanation:

A) It is attached. Please see the picture

B) First to calculate the overall mean,  

μ=65∗25/75+80∗25/75+95∗25/75  

μ=65∗25/75+80∗25/75+95∗25/75 = 80

Next to calculate E(MSTR) = σ2+(1/r−1) ∑ni(μi−μ)^2 = 5634

And E(MSE) = σ^2= 9

C) Yes, it is substantially large than E(MSE) in this case.

D) If we sampled 25 employees from each group, we are likely to get a F statistics to indicate differences of job satisfactions among three types of length of service of employees.

3 0
3 years ago
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