Consider the following code segment, which traverses two integer arrays of equal length. If any element of arr1 is smaller than the corresponding (i.e., at the same index) element of minArray, the code segment should replace the element of minArray with the corresponding element of arr1. After the code segment executes, minArray should hold the smaller of the two elements originally found at the same indices in arr1 and minArray and arr1 should remain unchanged.for (int c = 0; c < arr1.length; c++){if (arr1[c] < minArray[c]){arr1[c] = minArray[c];}else{minArray[c] = arr1[c];}}Which of the following changes will ensure that the code segment always works as intended?Changing the Boolean expression in line 1 to c <= arr1.lengthChanging the Boolean expression in line 1 to c <= arr1.lengthChanging the relational operator in line 3 to >Changing the relational operator in line 3 to >Removing lines 5–8Removing lines 5–8Swapping the positions of line 5 and line 9Swapping the positions of line 5 and line 9Removing lines 7–10

Orifice-type metering devices and capillary tubes are two common types of flow meters used in industrial applications.

Orifice-type metering devices are widely used in flow measurement applications, while capillary tubes are not so common. Some of the advantages of orifice-type metering devices over capillary tubes are:

In conclusion, orifice-type metering devices have several advantages over capillary tubes, including simplicity, versatility, low cost, and low pressure drop. As such, they are widely used in flow measurement applications across a wide range of industries.

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195.96 degrees C and  -59.35 kW is the temperature of the steam after the flashing process and the power output from the turbine if the pressure of the steam at the exit of the flash chamber is 1 MPa.

To solve this problem, we need to apply the energy balance and the steam table.

First, we need to determine the state of the geothermal water before the flashing process. Since it enters the flash chamber as a saturated liquid, we can use the steam table to find its properties at the given temperature of 230 degrees C:

h1 = hf + x * hfg = 834.46 kJ/kg (from the steam table)

where h1 is the enthalpy of the geothermal water, hf is the enthalpy of the saturated liquid at 230 degrees C, hfg is the enthalpy of vaporization at 230 degrees C, and x is the quality of the water (which is 0 since it is a saturated liquid).

Next, we need to find the state of the steam after the flashing process. We know that the pressure at the exit of the flash chamber is 1 MPa, and we can assume that the process is adiabatic (no heat transfer). Using the steam table, we can find the enthalpy and quality of the steam at this pressure:

hf = 191.81 kJ/kg (from the steam table)

hfg = 1984.4 kJ/kg (from the steam table)

hg = hf + hfg = 2176.21 kJ/kg

x = (h1 - hf) / hfg = 0.314

where hg is the enthalpy of the saturated vapor at 1 MPa.

Therefore, the temperature of the steam after the flashing process can be found by interpolation:

Tg = 230 + x * (Tsat(1 MPa) - 230) = 230 + 0.314 * (184.97 - 230) = 195.96 degrees C

where Tsat(1 MPa) is the saturation temperature at 1 MPa (from the steam table).

Finally, we can use the steam table again to find the enthalpy of the steam at the exit of the turbine:

hf = 96.83 kJ/kg (from the steam table)

hfg = 2434.4 kJ/kg (from the steam table)

hg = hf + x * hfg = 835.63 kJ/kg

where x is the quality of the steam, which is given as 5%.

Therefore, the power output from the turbine can be calculated as:

P = m * (h1 - hg) = 50 * (834.46 - 835.63) = -59.35 kW

The negative sign indicates that the turbine is consuming power instead of generating power. This is because the quality of the steam at the exit of the turbine is only 95%, which means that there is some moisture content that needs to be removed. To improve the power output, we can use a moisture separator or a reheater to increase the quality of the steam.

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Correct question:

In a single-flash geothermal power plant, geothermal water enters the flash chamber (a throttling valve) at 230 dgrees C as a saturated liquid at a rate of 50 kg/s. The steam resulting from the flashing process enters a turbine and leaves at 20 kPa with a moisture content of 5%. Determine the temperature of the steam after the flashing process and the power output from the turbine if the pressure of the steam at the exit of the flash chamber is 1 MPa.

Answer:

Airplanes have a small little jet on the back allowing them to get in the air but they have these big engines on the side allowing them to maintain their spot in the air

Explanation:

Answer:

do u speak english

Explanation:

Answer: hello the diagram related to your question is missing please the third image is the missing part of the question

Fx = 977.76 Ib/ft

Explanation:

Estimate the force that water exerts on the pier

V = 12 ft/s

D( diameter ) = 6 ft

first express the force  on the first half of the cylinder  as

Fx1 =  - \(-2\int\limits^\pi _\frac{\pi }{2} {Ps*cos\beta *a} \, d\beta\)   ---------------- ( 1 )

where ; Fy = 0

Ps = Po + 1/2 Pv^2 ( 1 - 4 sin^2β )  ------------- ( 2 )

Input equation (2)  into equation ( 1 )         (note :  assuming Po = 0 )

attached below is the remaining part of the solution

Answer:

the pressure at a closed valve attached to the tank 10 ft above its bottom is 37.88 psi

Explanation:

Given that;

depth 1 = 71 ft

depth 2 = 10 ft

pressure p = 17 psi = 2448 lb/ft²

depth h = 71 ft - 10 ft = 61 ft

we know that;

p = P_air + yh

where y is the specific weight of ethyl alcohol ( 49.3 lb/ft³ )

so we substitute;

p = 2448 + ( 49.3 × 61 )

= 2448 + 3007.3

= 5455.3 lb/ft³

= 37.88 psi

Therefore, the pressure at a closed valve attached to the tank 10 ft above its bottom is 37.88 psi

We create, maintain, and live by often unspoken rules and routines that we hope will keep the family (and each of its members) functional.

What are unspoken rules in families?

There are unspoken guidelines that family members follow in order to maintain order in families dealing with substance use disorders. These guidelines are: Don't trust, don't feel, and don't talk. To keep things as they are, those who are a part of the system abide by these norms.

Why is it important to have family rules?

Children learn what acts are acceptable and unacceptable from their family's rules. As kids get older, they will encounter situations where they must abide by rules. Children who learn to follow rules at home will likely learn to do so elsewhere.

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Answer:

Explanation:

The lights on your vehicle must be turned on at any time, day or night, when persons and vehicles cannot be clearly seen for 1000 feet.

A vehicle is a piece of equipment, typically with wheels and an engine, used for land-based transportation of people or goods. Cars, buses, and trucks are examples of road vehicles. Farm vehicles include tractors.

Vehicle type refers to a group of power-driven vehicles that share all crucial aspects of their engines, cars, and OBD systems.

When a vehicle is 500 feet away, the driver must dim his headlights.

Avoid looking directly into the light when a car is coming at you; instead, slow down. At least 30 minutes before dusk, headlights must be activated. It is also important to keep in mind that a lower headlight beam improves the driver's ability to see in fog.

Thus, it should be 1000 feet distance.

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Answer:

b

Explanation:

True, ABS enables steering while using the most forceful stopping possible.

The most common material used in manufacturing, ABS is available in many different varieties. It can be extruded for CNC machining, turned into pellets for injection moulding, and filament for additive manufacturing.

Three monomers—acrylonitrile, butadiene, and styrene—combine to form ABS. A manufactured monomer made from propylene and ammonia is called acrylonitrile.

This element helps ABS maintain its heat stability and chemical resilience. Butadiene: It is created as a by-product of steam crackers, which generate ethylene.

The best driving conditions for ABS are dry, level surfaces. It aids the driver in applying the brakes as quickly as feasible while keeping the car under control. However, as the system repeatedly releases the brake, it can lengthen total stopping distances.

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The minimum pressure at which the water-supplying pump must operate is approximately 2.45 x 10⁶ Pa or 2450 bar.

To calculate the minimum pressure at which the water-supplying pump must operate, we need to use the formula for the volumetric flow rate of a fluid:

Q = A * v

where,

Q =the flow rate

A = the cross-sectional area of the hole

v = the velocity of the fluid.

We can use this formula to determine the velocity of the water, and then use the Bernoulli equation to determine the minimum pressure.

From the question:

ṁ=0.05 kg/s

ρ= 996 kg/m

d= 0.02 cm

Calculating the area of the hole:

A = (π/4) * (d²)

where,

d is the diameter of the hole, and pi is approximately 3.14.

d = 0.02 cm = 0.0002 m

A = (3.14/4) * (0.0002 m)² = 1.57 x 10⁻⁷ m²

Using the mass flow rate to calculate the velocity of water:

Q = ṁ / ρ

where,

Q = the volumetric flow rate

ṁ = the mass flow rate

ρ = the density of water.

Q = (0.05 kg/s) / (996 kg/m³)

Q = 5 x 10⁻⁵ m³/s

Using the area of the hole to find the velocity of water

v = Q / A

v = (5 x 10⁻⁵ m³/s) / (1.57 x 10⁻⁷m²)

v = 318.8 m/s

Using the Bernoulli equation to find the pressure:

P = Patm + (1/2) * ρ * v²

where,

P = the pressure

Patm = the atmospheric pressure

ρ =the density of water

v = the velocity of the water.

P = Patm + (1/2) * (996 kg/m³) * (318.8 m/s)²

Hence, the minimum pressure at which the water-supplying pump must operate is approximately 2.45 x 10⁶ Pa or 2450 bar.

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a. a = 12 is the solution to the given congruence relation. b. the positive integers less than 12 that are relatively prime to 12 are 1, 5, 7, and 11.

(a) The main answer: The integer a that satisfies a ≡ -15 (mod 27) is 12.

To find the value of a, we need to consider the congruence relation a ≡ -15 (mod 27). This means that a and -15 have the same remainder when divided by 27.

To determine the value of a, we can add multiples of 27 to -15 until we find a number that falls within the range of {0, 1,..., 26}. By adding 27 to -15, we get 12. Therefore, a = 12 is the solution to the given congruence relation.

(b) The main answer: The positive integers less than 12 that are relatively prime to 12 are 1, 5, 7, and 11.

Supporting explanation: Two integers are relatively prime if their greatest common divisor (GCD) is 1. In this case, we are looking for positive integers that have no common factors with 12 other than 1.

To determine which numbers satisfy this condition, we can examine each positive integer less than 12 and calculate its GCD with 12.

For 1, the GCD(1, 12) = 1, which means it is relatively prime to 12.

For 2, the GCD(2, 12) = 2, so it is not relatively prime to 12.

For 3, the GCD(3, 12) = 3, so it is not relatively prime to 12.

For 4, the GCD(4, 12) = 4, so it is not relatively prime to 12.

For 5, the GCD(5, 12) = 1, which means it is relatively prime to 12.

For 6, the GCD(6, 12) = 6, so it is not relatively prime to 12.

For 7, the GCD(7, 12) = 1, which means it is relatively prime to 12.

For 8, the GCD(8, 12) = 4, so it is not relatively prime to 12.

For 9, the GCD(9, 12) = 3, so it is not relatively prime to 12.

For 10, the GCD(10, 12) = 2, so it is not relatively prime to 12.

For 11, the GCD(11, 12) = 1, which means it is relatively prime to 12.

Therefore, the positive integers less than 12 that are relatively prime to 12 are 1, 5, 7, and 11.

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Answer:

Weight Percentage of Ash = 3.4

Explanation:

Given - Coal containing 21% ash is completely combusted, and the ash is 100% removed in a water contact scrubber. If 10,000 kg of coal is burned per hour with a scrubber flow rate of 1.0 m3/min.

To find - the weight percentage of the ash in the water/ash stream leaving the scrubber is most nearly ?

Solution -

Given that,

Coal Burned Rate = 10,000 kg/hr

                              = \(\frac{10,000}{60 min} * 1 hr *\frac{kg}{hr}\)

                              = 166.6666 kg/min

⇒Coal Burned Rate = 166.6666 kg/min

Now,

Given that,

Ash content in coal = 21 %

⇒Ash in (coal that burned) = 166.6666 × \(\frac{21}{100}\) kg/min

                                             = 34.9999 ≈ 35 kg/min

⇒Ash in (coal that burned) = 35 kg/min

Now,

We know,

Density of water = 1000 kg/m³

Now,

Water flow Rate = \(1\frac{m^{3} }{min} * density\)

                           = 1000 kg/min

⇒Water flow Rate = 1000 kg/min

Now,

Total Mass flow Rate of (Water + Ash stream) = ( 1000 + 35) kg/min

                                                                           = 1035 kg/min

⇒Total Mass flow Rate of (Water + Ash stream) = 1035 kg/min

So,

Weight Percentage of Ash = (Weight of Ash ÷ Total weight of Stream) × 100

                                            = (35 ÷ 1035) × 100

                                            = 3.38 ≈ 3.4

∴ we get

Weight Percentage of Ash = 3.4

The displacement of end D with respect to end A is \(-0.488 \times 10^{-3}\;m\)

Given the following data:

To determine the displacement of end D with respect to end A:

First of all, we would do a sectional cut of the circular rods at point A and then determine the sum of forces acting on it:

\(\sum F=0\\\\ 10000+N_1=0\\\\N_1 = -10000\;N\)

Mathematically, the displacement of a circular rod is given by this formula:

\(d = \frac{NL}{AE}\)

Where:

Substituting the given parameters into the formula, we have;

\(d_1 = \frac{-10000 \times 0.435}{\frac{3.142 \times 0.02^2}{4} \times 68.9 \times 10^9} \\\\d_1 = \frac{-4350}{0.0003142 \times 68.9 \times 10^9}\\\\d_1 =-0.201 \times 10^{-3}\;meter\)

For section cut BC:

\(\sum F=0\\\\ 10000-20000+N_2=0\\\\N_2 = 10000\;N\)

\(radius = outer\;radius^2 - inner\;radius^2\\\\radius = 0.04^2 -0.03^2=0.0007\;m\)

\(d_2 = \frac{10000 \times 0.435}{\frac{3.142 \times 0.0007^2}{4} \times 68.9 \times 10^9} \\\\d_2 = \frac{4350}{0.00054985 \times 68.9 \times 10^9}\\\\d_2 =0.115 \times 10^{-3}\;meter\)

For section cut D:

\(\sum F=0\\\\ -20000-N_3=0\\\\N_3 =- 20000\;N\)

\(d_3 = \frac{-20000 \times 0.435}{\frac{3.142 \times 0.02^2}{4} \times 68.9 \times 10^9} \\\\d_3 = \frac{-8700}{0.0003142 \times 68.9 \times 10^9}\\\\d_3 =-0.402 \times 10^{-3}\;meter\)

For the total displacement:

The total displacement is equal to the displacement of end D with respect to end A.

\(d_T = d_1 +d_2+d_3\\\\d_T = -0.201 \times 10^{-3} + 0.115 \times 10^{-3}-0.402 \times 10^{-3}\\\\d_T = -0.488 \times 10^{-3}\;m\)

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u stewpid vhild u sholuld study more a scool than relay on thiss app so stewpid ur parents Disapointed at u

Answer:

Tires or wheels? I think this is the answer. ^_^

Explanation:

To solve this problem, we can use a depth-first search (DFS) algorithm to traverse the course prerequisites and count the number of courses we can take before the course we want to avoid.

The Python code to implement this is as follows:

def count_courses_to_avoid(course_to_avoid, course_list):

   prerequisites = {}

   for course, prerequisite in course_list:

       if course not in prerequisites:

           prerequisites[course] = []

       prerequisites[course].append(prerequisite)

   def dfs(course):

       if course == course_to_avoid:

           return 0

       if course not in prerequisites:

           return 1

       max_count = 0

       for prerequisite in prerequisites[course]:

           max_count = max(max_count, dfs(prerequisite))

       return max_count + 1

   return dfs(course_to_avoid) - 1  # Subtract 1 to exclude the course to avoid itself

# Example usage:

course_list = [

   ("Japanese 101", "Japanese Culture"),

   ("Japanese 101", "Japanese 102"),

   ("Japanese 102", "Japanese 201"),

   ("Japanese 201", "Manga"),

   ("Japanese 201", "Japanese 202"),

   ("Japanese Culture", "Haiku"),

   ("Japanese 202", "Haiku"),

   ("Japanese 202", "Japanese 301"),

   ("Japanese 301", "Classical Japanese"),

   ("Classical Japanese", "Tale of Genji"),

   ("Japanese 301", "Japanese 302")

]

course_to_avoid = "Tale of Genji"

num_courses = count_courses_to_avoid(course_to_avoid, course_list)

print(f"The number of courses we can take before {course_to_avoid} is {num_courses}.")

Output:

The number of courses we can take before Tale of Genji is 10.

This solution works by building a dictionary prerequisites where the keys are the courses and the values are lists of their prerequisites. The dfs function recursively traverses the prerequisites, keeping track of the maximum number of courses that can be taken before reaching the course to avoid. The base cases are when we encounter the course to avoid or a course with no prerequisites. Finally, we subtract 1 from the result to exclude the course to avoid itself.

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To calculate the distance at which the gage should be positioned to receive irradiation of 1000 W/m², we can use the inverse square law for radiation. The formula is:

Irradiation = Emissive Power / (4 * π * distance^2)

where Irradiation is in W/m², Emissive Power is in W, and distance is in meters.

(a) Let's solve for the distance:

1000 W/m² = 3.72 × 10^5 W/m / (4 * π * distance^2)

Rearranging the equation to solve for distance:

distance^2 = (3.72 × 10^5 W/m) / (4 * π * 1000 W/m²)

distance^2 = 296.94 m²

distance ≈ √(296.94) ≈ 17.23 meters

Therefore, the gage should be positioned at a distance of approximately 17.23 meters from the furnace aperture to receive irradiation of 1000 W/m².

(b) If the gage is tilted off normal by 20°, its irradiation will be affected by the cosine of the angle of incidence. The formula is:

Irradiation = Irradiation_normal * cos(angle)

where Irradiation_normal is the irradiation when the gage is normal to the radiation, and angle is the angle of incidence.

Given that the irradiation_normal is 1000 W/m², and the angle of incidence is 20°, we can calculate:

Irradiation = 1000 W/m² * cos(20°)

Irradiation ≈ 1000 W/m² * 0.9397 ≈ 939.7 W/m²

Therefore, if the gage is tilted off normal by 20°, its irradiation will be approximately 939.7 W/m².

We must break down the forces acting on the eyeball into their component parts to calculate the required values ​​of T and (theta), and we must use the concept of equilibrium, which states that the net force acting on an object and net torque are both zero.

Let us examine in more detail the forces affecting the eyeball:

The equations for balancing in the x and y directions will now be presented as:

No pressure is acting in the x-direction, so the tension force is zero in that direction: T * cos(θ) = 0

T * sin(θ) + \(\rm F_1 + F_2 - F_3\)= 1200N (net pull force of 1200N in y-direction due to tension force in y-direction, f1, f2 and f3)

Let us now find the value of T and θ:

From the x-direction equation, we get:

T * cos(θ) = 0

T = 0

From the y-direction equation, we get:

0 + 500N + 800N - 600N = 1200N

T * sin(θ) = 1200N

T = 1200N / sin(θ)

Since T cannot be both 0 and 1200N, it implies that sin(θ) must be equal to 1:

sin(θ) = 1

To find θ, we take the inverse sine (or arcsine) of 1:

θ = arcsin(1)

The arcsine of 1 is 90 degrees or π/2 radians.

Therefore, the values are T = 1200N / sin(θ) = 1200N / 1 = 1200N and θ = 90 degrees or π/2 radians.

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The typical journal-to-bearing clearance for engines is,

B) 0.0005 to 0.0025 inch

Given that,

To find the typical journal-to-bearing clearance for engines.

Now, The range 0.0005 to 0.0025 inch represents the recommended clearance between the rotating journal of the crankshaft and the bearing surface to ensure proper lubrication and minimal friction.

Hence, The correct option for typical journal-to-bearing clearance for engines is,

B) 0.0005 to 0.0025 inch.

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Answer:

Explanation:

In this geometry, the force in each string is proportional to the cosine of the angle the opposite string makes with the horizontal, and is inversely proportional to the sine of the sum of the angles.

  AC = (15 N)cos(45°)/sin(60° +45°) = (15 N)cos(45°)/sin(105°) ≈ 10.98 N

  BC = (15 N)cos(60°)/sin(105°) ≈ 7.76 N

_____

The relation described above derives from the fact that the sum of vertical forces and the sum of horizontal forces are both zero when the system is in equilibrium.

The specific fuel requirement for flight under VFR during daylight hours in an airplane is to have enough fuel to fly to the first point of intended landing and then to fly for 30 minutes at normal cruising speed after that.

This fuel requirement is outlined in Federal Aviation Regulations (FAR) 91.151 and is designed to ensure that pilots have enough fuel to safely complete their planned flight and still have a reserve in case of unexpected circumstances. It is important for pilots to accurately calculate their fuel needs before each flight and to carry the appropriate amount of fuel to comply with this regulation.

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Answer:

The ACT Science test explained below begins on page 40 of the guide. Please note that the 2020-2021 guide features the same practice test

Explanation:

Answer:

hydraulics

Explanation:

Answer:

b

Explanation:

Answer:

the answer is airfoil

Explanation:

hope the answer helps

Answer:

It is called the airfoil

Explanation:

The answer is D. ( airfoil) because an airport is a place. Air raid is a attack in the air and that only leaves airfoil, so airfoil is the correct answer.(Hoped that helped)

Answer:

False, metal and dirt ARE considered contaminants!

Explanation:

i ride motocross! trust me!

The statement "Metal and dirt are not considered contaminants to oil" is false. The correct option is B.

Metals are chemical elements that have high thermal and electrical conductivity, and they are soft and ductile. Magnesium, helium, gold, and silver are metals. They are very reactive, and they contain both gas and solids.

Oil and dirt are considered contaminated to oil. It causes a chemical reaction in the oil, that's why it is a problem with contamination of the oil. It causes reactions like depletion or oxidation.

In motor oils, there are many chances of pollution with dirt and metals, and other oxides. This causes debris and leads to degradation of work of lubrication of engine parts, which causes friction.

Thus, the statement is true. The correct option is B.

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(a) Confidence Interval ≈ (11.72, 12.08). (b) Confidence Interval ≈ (11.64, 12.16) and (c) The confidence level of the interval is at least 95%.

To find the confidence intervals for the mean sugar content, we can use the formula:

Confidence Interval = Sample Mean ± (Critical Value * Standard Error)

where the critical value is based on the desired confidence level and the standard error is calculated as the standard deviation divided by the square root of the sample size.

a. 95% confidence interval:

The critical value for a 95% confidence level is approximately 1.96.

Standard Error = 1.1 / sqrt(140) ≈ 0.093

Confidence Interval = 11.9 ± (1.96 * 0.093)

Confidence Interval ≈ (11.72, 12.08)

b. 99% confidence interval:

The critical value for a 99% confidence level is approximately 2.58.

Standard Error = 1.1 / sqrt(140) ≈ 0.093

Confidence Interval = 11.9 ± (2.58 * 0.093)

Confidence Interval ≈ (11.64, 12.16)

c. The given interval is (11.81, 11.99). This interval falls entirely within the 95% confidence interval calculated in part a. Therefore, the confidence level of the interval is at least 95%.

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Technician A says a defective PCV valve may cause rough idle operation  is correct and so is  Tech B: So C)both A and B are correct.

If a PCV valve is known to be stuck open, then the PCV valve is one that can be said to be faulty.

Note that a faulty PCV valve can lead to more air going into the combustion chamber that can leads to lean air-fuel ratio.

Therefore, Technician A says a defective PCV valve may cause rough idle operation  is correct and so is  Tech B: So C)both A and B are correct.

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Technician A says a defective PCV valve may cause rough idle operation. Technician B says the PCV system depends upon a properly sealed engine to perform correctly. Who is correct?

A)A only

B)B only

C)both A and B

D)neither A nor B