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liberstina [14]
3 years ago
15

Technician A says that the original equipment manufacturer (OEM) offer scan tools to service their specific line of vehicles. Te

chnician B says that scan tools supplied by aftermarket tool and equipment providers usually provide greater functionality
a. Technician A
b. Technician B
c. Both A and B
d. Neither A nor B
Engineering
2 answers:
Anastasy [175]3 years ago
8 0

Answer:

a. Technician A

Explanation:

Aftermarket parts are replacement parts that are not made by the original equipment manufacturer. They can not guarantee as much functionaluty when compared to OEMs. Original Equipment Manufacturer (OEM) usually their own proprietary scan tools which offer better and functionality to service their special line of vehicles. So Technician A alone is correct.

aleksklad [387]3 years ago
4 0

Answer:

C. Both A and B

Explanation:

Three types of scan tools are available to the technician:

1. Generic On-Board Diagnostics II (OBD-II) scan tool (many brands)

2. Aftermarket scan tool with enhanced coverage (Snap-On, OTC, others)

3. OEM scan tool (developed and approved by an OEM, used in dealerships)

Not too long ago, an OEM scan tool was often the exception rather than the rule at the average automotive repair facility. Although dealership technicians have normally used manufacturer-specific OEM scan tools, such equipment was not commonly found in an independent automotive repair shop.

Traditionally, most independent shops have relied on aftermarket scan tools that were designed to work with a variety of Asian, domestic, and/or European vehicles. While this remains true for many independent shops, more and more independent shops are using personal computer-based original equipment manufacturer (OEM) scan tools, particularly on late-model vehicles, as access to such tools improves and in some cases costs of the tool has gone down.

Many of today’s vehicles have several dozen modules, each with their own data stream and diagnostic trouble codes (DTCs). Hybrid and electric vehicles can have especially complex self-diagnostic systems. Some aftermarket scan tools manufacturers may be challenged to keep up with the demands of scan data in new vehicles. The technician can expect to work with both OEM and aftermarket scan tools during his or her career.

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You are to design two CONCEPTUALLY different synchronous state machines (Mealy and Moore) that perform the task described below.
allochka39001 [22]
Answer:








Explanation:









I hope this helps!
3 0
3 years ago
The cube measures 3.0-ft on all sides and has a density of 3.1 slugs/ft3. How much does it weigh?
kodGreya [7K]

Answer:

W = 2695.14 lb

Explanation:

given,

side of cube = 3 ft

density of the cube = 3.1 slugs/ft³

we know,

density = \dfrac{mass}{volume}

mass = density x volume

volume = 3³ = 27 ft³

mass =  3.1  x 27

    m = 83.7 slugs.

weight calculation

converting mass from slug to pound

weight of 1 slug is equal to 32.2 lb

now,

weight of the cube is equal to

  W = 83.7 slugs x 32.2 lb/slug

  W = 2695.14 lb

hence, weight is equal to W = 2695.14 lb

4 0
3 years ago
Which of these are not referenced in an assembly?
Otrada [13]
E. Parts they don’t resemble
7 0
3 years ago
: Câu nào dưới đây thể hiện sự thiếu tự chủ?
sukhopar [10]

Answer:

thành thật mà nói bởi vì cách những chiếc lá đang chuyển và cách mặt trời chiếu sáng.

3 0
2 years ago
A student takes 60 voltages readings across a resistor and finds a mean voltage of 2.501V with a sample standard deviation of 0.
Lena [83]

Answer:

There are 2 expected readings greater than 2.70 V

Solution:

As per the question:

Total no. of readings, n = 60 V

Mean of the voltage, \mu = 2.501 V

standard deviation, \sigma = 0.113 V

Now, to find the no. of readings greater than 2.70 V, we find:

The probability of the readings less than 2.70 V, P(X\leq 2.70):

z = \frac{x - \mu}{\sigma} = \frac{2.70 - 2.501}{0.113} = 1.761

Now, from the Probability table of standard normal distribution:

P(z\leq 1.761) = 0.9608

Now,

P(X\geq 2.70) = 1 - P(X\leq 2.70) = 1 - 0.9608 = 0.0392 = 3.92%

Now, for the expected no. of readings greater than 2.70 V:

P(X\geq 2.70) = \frac{No.\ of\ readings\ expected\ to\ be\ greater\ than\ 2.70\ V}{Total\ no.\ of\ readings}

No. of readings expected to be greater than 2.70 V = P(X\geq 2.70)\times Total\ no.\ of\ readings

No. of readings expected to be greater than 2.70 V = 0.0392\times 60 = 2.352 ≈ 2

7 0
3 years ago
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