why is it important to create a class such as admin or overhead when you set up class tracking

The elevation and stationing of the high point, PVC, and PVT of the given vertical curve are;

High Point Station = 342+60

PVT Station = 340+00

PVC Station = 340+00

Elevation of High Point = 1350.0 ft

Elevation of PVT = 1325.0 ft

Elevation of PVC = 1345.0 ft

A 520-ft-long equal-tangent crest vertical curve connects tangents that intersect at station 340 + 00 and elevation 1325 ft. The initial grade is +4.0% and the final grade is 2.5%.

Determine the elevation and stationing of the high point, PVC, and PVT. A vertical curve is used to connect two tangents that have differing slope grades, and it is the arc of a parabola. The parabolic arc's vertex is referred to as the high point of the curve.

PVC (point of vertical curvature) is the point where the parabolic arc starts and the tangent line ends, and PVT (point of vertical tangency) is the point where the parabolic arc ends and the tangent line begins.

Therefore, by considering the given data, we can calculate the elevation and stationing of the high point, PVC, and PVT of the vertical curve in the following way;

Stationing of the curve = 340+00. High point of the curve = L/2 + 340+00 = 520/2 + 340+00 = 342+60PVC = 340+00. Elevation of PVC = Elevation of point of intersection + (Initial grade / 100) x Length of the curve 1325 + (4/100) x 520 = 1345 ft.

Elevation of PVT = Elevation of point of intersection + (Final grade / 100) x Length of the curve 1325 + (2.5/100) x 520 = 1350 ft.

Therefore, the elevation and stationing of the high point, PVC, and PVT of the given vertical curve are;

High Point Station = 342+60

PVT Station = 340+00

PVC Station = 340+00

Elevation of High Point = 1350.0 ft

Elevation of PVT = 1325.0 ft

Elevation of PVC = 1345.0 ft

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Answer: The correct way to achieve proper drive pinion gear bearing preload is to torque the pinion nut to the manufacturer’s specifications and then check for preload.

Explanation: Typical starting points for torque specifications are 55 lb-ft for 3/8-inch bolts/nuts, 75 lb-in for 7/16-inch bolt/nuts, and 125 lb-in for 1/2-inch nuts/bolts. Once the nut is properly torqued, check for preload by tapping the end of the pinion gear with a punch to ensure the races are fully seated, shake it to evaluate for excessive play.

Pinion bearings are bearings that support the weight of the drive shaft and transfer power from the differential to the wheels. Preload is the tension that is put on a bearing. In the case of pinion bearings, preload takes place in the form of a collapsible spacer that stays between the bearings. The required amount of tension is obtained by adjusting the space manually.

Other ways to achieve proper drive pinion gear bearing preload include using a crush sleeve or shims. Crush sleeves are used in place of spacers and are designed to be crushed when the pinion nut is tightened. This crushes the sleeve and creates tension on the bearings. Shims can also be used to adjust the space between bearings.

Common problems with drive pinion gear bearing preload include over-tightening or under-tightening of the pinion nut, which can lead to damage or failure of the bearings. It is important to follow manufacturer specifications when torquing the pinion nut and checking for preload.

Hope this helps, and have a great day!

Answer:

A pointer is spun on a fair wheel of chance having its periphery labeled Trom 0 to 100. (a) Whhat is the sample space for this experiment? (b)What is the probability that the pointer will stop between 20 and 35? (c) What is the probability that the wheel will stop on 58?

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Explanation:

thats all you said

Answer:

hii my name is RAGHAV what is your name

Explanation:

this question is which chapter

Using the formula of comfort condition it is possible to determine the safe operating speed on the highway.

In physics, we define angular deviation as the angle resulting from the deflection of light when reflected from a prism. A prism has the ability to separate white light into several other colors, which are reflected by the object.

The comfort condition formula is given by: \(L=2(\frac{NV^{3} }{C} )^{\frac{1}{2} }\)

Where, L is the length of the vertical curve, N is the grade, V is the safe speed operation and assuming the standart value for C (0,6 m/s3) it is possible to calculate:

\(C = 0,6 m/s^{3} = 1,968 ft/s^{3}\\ N= N1 - N2 = 7,5-0 = 7,5% = 0,075\)

Changing the values in the formula:

\(L=2(\frac{NV^{3} }{C}){\frac{1}{2} } \\300 = 2(\frac{0,075.V^{3} }{1,968})^{\frac{1}{2} } \\V = 83,89 ft/s\)

So, the safe operating speed on the highway is 83,89 ft/s.

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The characteristic impedance, the phase constant, the velocity on the line, and the input impedance are respectively;

A) Z_o = 40.82 Ω

B) β = 11.543 rad/m

C) v_p = 2.72 × 10^(8) m/s

D) Z_in = 115.91 Ω

We are given;

Inductance; L = 0.15 μH/m = 0.15 × 10^(-6) H/m

Capacitance; C = 90 pf/m = 90 × 10^(-12) f/m

Frequency; f = 500 MHz = 500 × 10^(6) Hz

Load impedance; Z_l = 80 Ω

Length of transmission line; l = 80cm = 0.8m

A) Formula for characteristic impedance is;

Z_o = √(L/C)

Thus;

Z_o = √((0.15 × 10^(-6))/(90 × 10^(-12)))

Z_o = 40.82 Ω

B) Formula for the phase constant is;

β = ω√(LC)

Where;

ω = 2πf

Thus;

β = (2π × 500 × 10^(6))√(0.15 × 10^(-6) × 90 × 10^(-12))

β = 11.543 rad/m

C) Formula for phase velocity is;

v_p = ω/β

v_p = (2π × 500 × 10^(6))/11.543

v_p = 2.72 × 10^(8) m/s

D) Formula for the input impedance is;

Z_in = Z_o[(Z_l + Z_o*tanβl)/(Z_o + Z_l*tanβl)]

Z_in = 40.82[(80 + 40.82*tan(11.543*0.8))/(40.82 + 80*tan(11.543*0.8))]

Z_in = 40.82(72.1335/25.4031)

Z_in = 115.91 Ω

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Answer:

u need to plan it out

Explanation:

u need to plan it out

Answer:

use the turn 1 degrees option and put a repeat loop on it

Explanation:

u can add sound in ur loop

Answer:

  B) leads the voltage

Explanation:

One way to think about it is that the current causes charge to be accumulated on the capacitor, changing its voltage. The current must be non-zero before the voltage can change. Hence current leads voltage.

Answer:

33.56 ft^3/sec.in

Explanation:

Duration = 6 hours

drainage area = 185 mi^2

constant baseflow = 550 cfs

Derive the unit hydrograph using the inverse procedure

first step : calculate for the volume of direct runoff hydrograph using the details in table 2 attached below

Vdrh = sum of drh *  duration

        = 29700 * 6 hours ( 216000 secs )

        = 641,520,000 ft^3.

next step : Calculate the volume of runoff in equivalent depth

Vdrh / Area = 641,520,000  / 185 mi^2

                    = 1.49 in

Finally derive the unit hydrograph

Unit of hydrograph = drh /  volume of runoff in equivalent depth

                                = 50 ft^3 / 1.49 in  =  33.56 ft^3/sec.in

Answer:

The resulting pressure of the gas when its volume decreases is 300 kN/m².

Explanation:

Given;

initial volume of the gas, V₁ = 80 L

number of moles of the gas, n = 0.5 moles

initial pressure of the gas, P₁ = 150 kN/m² = 150 kPa

Determine the constant temperature of the gas using ideal gas equation;

PV = nRT

where;

R is ideal gas constant = 8.315 L.kPa/K.mol

T is the constant temperature

\(T = \frac{P_1V_1}{nR} \\\\T = \frac{150.kPa \ \times \ 80 .L}{0.5 .mol \ \times \ 8.315(L.kPa/mol.K)} \\\\T = 2,886.35 \ K\)

When the gas is compressed to half of its volume;

new volume of the gas, V₂ = ¹/₂ V₁

                                             = ¹/₂ x 80L = 40 L

The new pressure, P₂ is calculated as;

\(P_2V_2 = nRT\\\\P_2 = \frac{nRT}{V_2} \\\\P_2 = \frac{0.5 \times 8.315\times 2886.35}{40} \\\\P_2 = 300 \ kPa = 300 \ kN/m^2\)

Therefore, the resulting pressure of the gas when its volume decreases is 300 kN/m².

AASHTO assumes a deceleration value in calculating stopping sight distance (SSD) of B. 11.2 ft/s^2. This doesn't take any road defects into account.

Explanation:

for 40hz

since they are in series they will have the same amper

so R=30ohm

C=30F

for 50Hz

R=25

C=25

Active-active architecture provides _______ in the event one application fails. Select an answer: a. mirroring rb. edundancy c. auto-recovery d. resiliency

Active-active architecture provides redundancy in the event one application fails.

Active-active architecture provides redundancy in the event one application fails. This means that multiple instances of the same application or service are running simultaneously, allowing for continued operation even if one of the instances fails. This can improve the overall resiliency and reliability of the system.

In general, redundancy refers to the inclusion of extra components or resources in a system that serve as backups in case the primary components fail. This can include things like backup power supplies, extra data storage, or multiple copies of the same application or service running simultaneously.

The goal of redundancy is to improve the overall reliability and resiliency of a system by providing fail-safes that can take over in the event of a failure. This can help prevent outages or downtime and ensure that the system continues to function even in the face of unexpected problems.

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Answer:

  157.38° CCW from +x axis

  67.38° CCW from +y axis

  n = (12/13)i +(5/13)j

Explanation:

The reference angle is ...

  arctan(15/36) ≈ 22.62° . . . . CW from the -x axis

The signs of the component vectors indicate this is a 2nd-quadrant vector, so the angles of interest are ...

  157.38° CCW from +x axis

  67.38° CCW from +y axis

The unit vector will be ...

  n = cos(157.38°)i +sin(157.38°)j

 n = (12/13)i +(5/13)j

True . The names of all registers updated during a procedure can be listed using the USES operator and the PROC directive.

The value that the ESP register references to is copied from the stack before the POP instruction is executed, at which point the register is incremented. The instruction ((n XOR m) XOR m) generates PUSH and POP in the 8085 microprocessor for any two integers n and m: When the PUSH instruction is carried out by the 8085 microprocessor, the stack pointer register is decremented by two, and when the POP instruction is carried out, the stack pointer is increased by two.

Mathematical calculations are carried out using arithmetic operators. A value can be assigned to a property or variable using assignment operators. Numeric, date, system, time, and tex are all acceptable assignment operators.

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The egg-shaped surfaces machined onto a camshaft are called the Camshaft lobes.

As the camshaft rotates, lobes (referred to as cams) push against the valves to open them; the valves' springs close the valves back up. This is a crucial task that can significantly affect how well an engine performs at various speeds.

As long as the wear on the lobes is not extreme, camshafts can be rebuilt. Examples of a rebuildable and non-rebuildable camshaft are shown above.

The term "camshaft lobes" refers to the egg-shaped surfaces machined onto a camshaft.

Thus, answer is Camshaft lobes.

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Answer:

A drone controller works by sending a radio signal from the remote control to the drone, which tells the drone what to do. Radio signals are sent from the radio transmitter in the drone controller and received by the drone's receiver.

Explanation:

Answer:

what's the question?...........

The asymptotes of the open loop transfer are:

Horizontal: y = 0

Vertical: x = -10 and x = -100

The open loop transfer function is given as:

f(s) = 100(s + 1)/((s + 10)(s + 100))

Set the numerator of the function to 0.

So, we have:

f(s) = 0/((s + 10)(s + 100))

Evaluate

f(s) = 0

This means that, the vertical asymptote is y = 0

Set the denominator of the function to 0.

(s + 10)(s + 100)  0

Split

s + 10 = 0 and s + 100 = 0

Solve for s

s = -10 and s = -100

This means that, the horizontal asymptotes are s = -10 and s = -100

See attachment for the graph of the asymptotes.

Therefore, The asymptotes of the open loop transfer are:

Horizontal: y = 0

Vertical: x = -10 and x = -100

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Answer:

Answer is explained in the explanation section below.

Explanation:

Solution:

Note: This question is incomplete and lacks necessary data to solve. But I have found the similar question on the internet. So, I will be using the data from that question to solve this question for the sack of concept and understanding.

Data Given:

x = 27 , 44 , 32 , 47, 23 , 40, 34, 52

y = 30, 19,  24,  13 , 29,  19,  21,  14

It is given that,

∑x = 299

∑y = 167

∑\(x^{2}\) = 11887

∑\(y^{2}\) = 3773

We are asked to verify the above values manually in this question.

So,

1. ∑x = 299

Let's verify it:

∑x = 27 + 44 + 32 + 47 + 23 + 40 + 34 + 52

∑x = 299

Yes, it is equal to the given value. Hence, verified.

2. ∑y = 167

Let's verify it:

∑y = 30 + 19 +  24 + 13 + 29 + 19 +  21 +  14

∑y = 169

No, it is not equal to the given value.

3. ∑\(x^{2}\) = 11887

Let's verify it:

For this to find,  first we need to square all the value of x individually and then add them together to verify.

∑\(x^{2}\) = \(27^{2}\) + \(44^{2}\) + \(32^{2}\) + \(47^{2}\) + \(23^{2}\) + \(40^{2}\) + \(34^{2}\) + \(52^{2}\)

∑\(x^{2}\) = 11,887

Yes, it is equal to the given value. Hence, verified.

4. ∑\(y^{2}\) = 3773

Let's verify it:

Again, for this we need to find the squares of all the y values and then add them together to verify it.

∑\(y^{2}\) = \(30^{2}\) + \(19^{2}\) +  \(24^{2}\) + \(13^{2}\) + \(29^{2}\) + \(19^{2}\) +  \(21^{2}\) +  \(14^{2}\)

∑\(y^{2}\)  = 3,845

No, it is not equal to the given value.

Explanation:

Dempwolf created by John Augustus, Among the most prominent innovative solutions in Southern California Pennsylvania was established by Dempwolf with  brother Reinhardt or uncle's son Frederick entered the company of J.A. Dozens of structures in 10 states were engineered by Dempwolf.

In Python, the term you're looking for is "keyword."

So, the correct answer is B.

A keyword is a reserved word that is part of the Python language and cannot be used as a variable name. These words have specific meanings and functions within the language, which is why they are reserved. Using keywords as variable names can cause confusion and errors in your code.

Examples of keywords include "if," "else," "while," "def," and "import."

To avoid issues, it's essential to choose descriptive and unique variable names that do not conflict with Python's keywords

Hence,the answer of the question is B.

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Answer:

a) \(L = 1.5\)

\(L_q = 0.9\)

\(W = \dfrac{1 }{8 } \, hour\)

\(W_q = \dfrac{3}{40 } \, hour\)

\(P = \dfrac{3}{5 }\)

b) The new backacter should be recommended

c) The additional backacter should not be deployed

Explanation:

a) The required parameters are;

L = The number of customers available

\(L = \dfrac{\lambda }{\mu -\lambda }\)

μ = Service rate

\(L_q\) = The number of customers waiting in line

\(L_q = p\times L\)

W = The time spent waiting including being served

\(W = \dfrac{1 }{\mu -\lambda }\)

\(W_q\) = The time spent waiting in line

\(W_q = P \times W\)

P = The system utilization

\(P = \dfrac{\lambda }{\mu }\)

From the information given;

λ = 12 trucks/hour

μ = 3 min/truck = 60/3 truck/hour = 20 truck/hour

Plugging in the above values, we have;

\(L = \dfrac{12 }{20 -12 } = \dfrac{12 }{8 } = 1.5\)

\(P = \dfrac{12 }{20 } = \dfrac{3}{5 }\)

\(L_q = \dfrac{3}{5 } \times \dfrac{3}{2 } = \dfrac{9}{10 } = 0.9\)

\(W = \dfrac{1 }{20 -12 } = \dfrac{1 }{8 } \ hour\)

\(W_q = \dfrac{3}{5 } \times \dfrac{1}{8 } = \dfrac{3}{40 } \, hour\)

(b) The service rate with the new backacter = 1.5 minutes/truck which is thus;

μ = 60/1.5 trucks/hour = 40 trucks/hour

\(P = \dfrac{12 }{40 } = \dfrac{3}{10}\)

\(W = \dfrac{1 }{40 -12 } = \dfrac{1 }{38 } \, hour\)

\(W_q = \dfrac{3}{10 } \times \dfrac{1}{38 } = \dfrac{3}{380 } \, hour\)

λ = 12 trucks/hour

Total cost = \(mC_s + \lambda WC_w\)

m = 1

\(C_s\) = GH¢ = 1300

\(C_w\) = 400

Total cost with the old backacter is given as follows;

\(1 \times 1000 + 12 \times \dfrac{1}{8} \times 400 = \$ 1,600.00\)

Total cost with the new backacter is given as follows;

\(1 \times 1300 + 12 \times \dfrac{1}{38} \times 400 = \$ 1,426.32\)

The new backacter will reduce the total costs, therefore, the new backacter is recommended.

c)

Here μ = 3 min/ 2 trucks = 2×60/3 truck/hour = 40 truck/hour

\(\therefore W = \dfrac{1 }{40 -12 } = \dfrac{1 }{38 } \, hour\)

Total cost with the one backacter is given as follows;

\(1 \times 1000 + 12 \times \dfrac{1}{8} \times 400 = \$ 1,600.00\)

Total cost with two backacters is given as follows;

\(2 \times 1000 + 12 \times \dfrac{1}{38} \times 400 = \$ 2,126.32\)

The additional backacter will increase the total costs, therefore, it should not be deployed.

Answer:

  C.  3325 Ω - 3675 Ω

Explanation:

5% of 3500 Ω is ...

  0.05 × 3500 = 175

The lower limit is this amount less than the nominal value:

  3500 -175 = 3325

The upper limit is the nominal value plus the tolerance:

  3500 +175 = 3675

The lower and upper limits are 3325 Ω and 3675 Ω, respectively.

Answer:

if user_tickets == 7:

  award_points = 1

else:

  award_points = user_tickets

Explanation:

Not sure what language you are using. But this can be used for python. Also don't know if you are required to ask the user for any input

Answer:

Technology

Explanation:

It’s made to meet desire

It should be noted that Anything that is made to meet a need or desire is Technology.

Technology can be regarded as the practical application of science.

Technology can be seen in areas such as

However, Technology came up because of the necessity for it, people desire it.

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Answer:

\(\| \vec A \| = 6.163\,m\)

Explanation:

Sean A, B y C vectores coplanares tal que:

\(\vec A = (\| \vec A \|\cdot \cos \theta_{A},\| \vec A \|\cdot \sin \theta_{A})\), \(\vec B = (\| \vec B \|\cdot \cos \theta_{B},\| \vec B \|\cdot \sin \theta_{B})\) y \(\vec C = (\| \vec C \|\cdot \cos \theta_{C},\| \vec C \|\cdot \sin \theta_{C})\)

Donde \(\| \vec A \|\), \(\| \vec B \|\) y \(\| \vec C \|\) son las normas o magnitudes respectivas de los vectores A, B y C, mientras que \(\theta_{A}\), \(\theta_{B}\) y \(\theta_{C}\) son las direcciones respectivas de aquellos vectores, medidas en grados sexagesimales.

Por definición de producto escalar, se encuentra que:

\(\vec A \,\bullet\, \vec B = \|\vec A \| \| \vec B \| \cos \theta_{B}\cdot \cos \theta_{A} + \|\vec A \| \| \vec B \| \sin \theta_{B}\cdot \sin \theta_{A}\)

\(\vec B \,\bullet\, \vec C = \|\vec B \| \| \vec C \| \cos \theta_{B}\cdot \cos \theta_{C} + \|\vec B \| \| \vec C \| \sin \theta_{B}\cdot \sin \theta_{C}\)

Asimismo, se sabe que \(\| \vec B \| = 5\,m\), \(\theta_{B} = 60^{\circ}\), \(\vec A \,\bullet \,\vec B = 30\,m^{2}\), \(\vec B\, \bullet\, \vec C = 35\,m^{2}\), \(\|\vec A \| = \| \vec C \|\) y \(\theta_{C} = \theta_{A} + 25^{\circ}\). Entonces, las ecuaciones quedan simplificadas como siguen:

\(30\,m^{2} = 5\|\vec A \| \cdot (\cos 60^{\circ}\cdot \cos \theta_{A} + \sin 60^{\circ}\cdot \sin \theta_{A})\)

\(35\,m^{2} = 5\|\vec A \| \cdot [\cos 60^{\circ}\cdot \cos (\theta_{A}+25^{\circ}) + \sin 60^{\circ}\cdot \sin (\theta_{A}+25^{\circ})]\)

Es decir,

\(30\,m^{2} = \| \vec A \| \cdot (2.5\cdot \cos \theta_{A} + 4.330\cdot \sin \theta_{A})\)

\(35\,m^{2} = \| \vec A \| \cdot [2.5\cdot \cos (\theta_{A}+25^{\circ})+4.330\cdot \sin (\theta_{A}+25^{\circ}})]\)

Luego, se aplica las siguientes identidades trigonométricas para sumas de ángulos:

\(\cos (\theta_{A}+25^{\circ}) = \cos \theta_{A}\cdot \cos 25^{\circ} - \sin \theta_{A}\cdot \sin 25^{\circ}\)

\(\sin (\theta_{A}+25^{\circ}) = \sin \theta_{A}\cdot \cos 25^{\circ} + \cos \theta_{A} \cdot \sin 25^{\circ}\)

Es decir,

\(\cos (\theta_{A}+25^{\circ}) = 0.906\cdot \cos \theta_{A} - 0.423 \cdot \sin \theta_{A}\)

\(\sin (\theta_{A}+25^{\circ}) = 0.906\cdot \sin \theta_{A} + 0.423 \cdot \cos \theta_{A}\)

Las nuevas expresiones son las siguientes:

\(30\,m^{2} = \| \vec A \| \cdot (2.5\cdot \cos \theta_{A} + 4.330\cdot \sin \theta_{A})\)

\(35\,m^{2} = \| \vec A \| \cdot [2.5\cdot (0.906\cdot \cos \theta_{A} - 0.423 \cdot \sin \theta_{A})+4.330\cdot (0.906\cdot \sin \theta_{A} + 0.423 \cdot \cos \theta_{A})]\)

Ahora se simplifican las expresiones, se elimina la norma de \(\vec A\) y se desarrolla y simplifica la ecuación resultante:

\(30\,m^{2} = \| \vec A \| \cdot (2.5\cdot \cos \theta_{A} + 4.330\cdot \sin \theta_{A})\)

\(35\,m^{2} = \| \vec A \| \cdot (4.097\cdot \cos \theta_{A} +2.865\cdot \sin \theta_{A})\)

\(\frac{30\,m^{2}}{2.5\cdot \cos \theta_{A}+ 4.330\cdot \sin \theta_{A}} = \frac{35\,m^{2}}{4.097\cdot \cos \theta_{A} + 2.865\cdot \sin \theta_{A}}\)

\(30\cdot (4.097\cdot \cos \theta_{A} + 2.865\cdot \sin \theta_{A}) = 35\cdot (2.5\cdot \cos \theta_{A}+4.330\cdot \sin \theta_{A})\)

\(122.91\cdot \cos \theta_{A} + 85.95\cdot \sin \theta_{A} = 87.5\cdot \cos \theta_{A} + 151.55\cdot \sin \theta_{A}\)

\(35.41\cdot \cos \theta_{A} = 65.6\cdot \sin \theta_{A}\)

\(\tan \theta_{A} = \frac{35.41}{65.6}\)

\(\tan \theta_{A} = 0.540\)

Ahora se determina el ángulo de \(\vec A\):

\(\theta_{A} = \tan^{-1} \left(0.540\right)\)

La función tangente es positiva en el primer y tercer cuadrantes y tiene un periodicidad de 180 grados, entonces existen al menos dos soluciones del ángulo citado:

\(\theta_{A, 1} \approx 28.369^{\circ}\) y \(\theta_{A, 2} \approx 208.369^{\circ}\)

Ahora, la magnitud de \(\vec A\) es:

\(\| \vec A \| = \frac{35\,m^{2}}{4.097\cdot \cos 28.369^{\circ} + 2.865\cdot \sin 28.369^{\circ}}\)

\(\| \vec A \| = 6.163\,m\)

The least cost input combination of producing a given output on an isoquant/isocost graph is given by the tangency between the isoquant curve and the isocost line. Therefore, the correct answer is c.

In economics, an isoquant represents all possible combinations of inputs that can produce a specific level of output. An isocost line, on the other hand, represents all combinations of inputs that can be purchased at a given cost. The least cost input combination occurs when the isoquant curve and the isocost line are tangent to each other.

At this point of tangency, the firm achieves the desired output level at the lowest possible cost, as it maximizes the use of inputs while staying within the budget constraint represented by the isocost line. Therefore, the tangency between the isoquant curve and the isocost line determines the least cost input combination.

Option C holds true.

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When the process is in control but does not meet specification, it is referred to as a special cause error.

In statistical process control, a process is considered to be in control when it operates within the defined limits and shows only random variations. However, when a process is in control but does not meet the desired specifications, it indicates the presence of a special cause error.

Special cause errors are attributed to specific factors or events that cause the process to deviate from the expected outcome. These errors are typically unpredictable and require investigation and corrective action to bring the process back within the desired specifications.

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