The given question is incomplete. The complete question is as follows.

A river has a steady speed of 0.360 m/s. A student swims upstream a distance of 1.00 km and swims back to the starting point.

(a) If the student can swim at a speed of 1.29 m/s in still water, how long does the trip take?

(b) How much time is required in still water for the same length swim?

(c) Intuitively, why does the swim take longer when there is a current?

Explanation:

(a) Relative velocity of the student while going upstream is as follows.

(1.29 - 0.360) m/s

= 0.93 m/s

Now, relative velocity of the student while coming downstream is as follows.

(1.29 + 0.360) m/s

= 1.65 m/s

When the student is going upstream then time taken to cover 1 km distance is calculated as follows.

= 1075.26 sec

When the student is going downstream then time taken by it to cover 1 km distance is as follows.

= 606.06 sec

Hence, total trip time will be calculated as follows.

Total trip time = (1075.26 sec + 606.06 sec)

= 1681.32 sec

(b) Time taken in still water is calculated as follows.

= 1550.38 sec

(c) When there will be current then there will occur a decrease in relative velocity of the swimmer. Therefore, he will take more time.

A small metal ball is given a negative charge, then brought near (i.e., within a few millimeters) to end A of the rod. What happens to end A of the rod when the ball approaches it closely this first time?

Answer:

They will attract.

Explanation:

We know, in conductors electrons are free to move> in the case given both metal ball and rod are conductor.

Now, when conductor are brought to a charge all the opposite charges moves through the point of conductor which is close to the charge and attraction takes place.

In this case too, when metal ball is bought near rod, rod also develop opposite charge and attraction takes place.

Hence, this is the required solution.

Charge q1 = 29.5 μC is located at r1 = (3.4 i − 3.5 j) m. A second charge q2 = −44.4 μC is located at r2 = (10.5 i + 7.5 j) m. The force of the second charge on the first charge can be written using unit vectors as follows: F = Fx i + Fy j What is the value of Fx in newtons?

Answer:

-0.234 N

Explanation:

Electrostatic force is:

Substitute the values:

Negative sign indicates that the force is attractive in nature which is also true as the charges are opposite in nature.

Compute the moon's centripetal acceleration in its orbit around the earth. Recall that the moon orbits the earth every 28.0 ���� and that it is about 240000 ����� from the earth. What force causes this acceleration? Be sure to convert to �� �����.

Answer:

0.0026 m/s²

Explanation:

Centripetal acceleration is given as follows:

Substitute the values:

Finding the work done in pulling a stranded climber to safety. In an unfortunate accident, a rock climber finds herself stuck 17 m from the top of a 80 m rock face. Rescuers sent to help the climber lower a harness attached to a cable that will pull the climber to the top of the rock face. If the climber, secured in the harness, weighs 56 kg and the cable weighs 0.7kgm, how much work is done in raising the climber, harness, and cable to the top of the rock face? Note: the weight of the climber and cable is really a mass so you will need to multiply the mass in kg by the acceleration due to gravity (approximately 9.8ms2).

Answer:

-9446.22 J

Explanation:

Parameters given:

Mass of climber and harness = 56kg

Mass of cable = 0.7kg

Distance between climber and top of rock face = 17m

The work done in pulling the climber is given as:

W = F * d

F is the force applied on the rope. It is opposite the force of gravity pulling the climber, hence, it is given as:

F = -Fg

Fg = force of gravity

Fg = m * g

g = acceleration due to gravity.

The mass of the climber, harness and cable = 56 + 0.7 = 56.7kg

=> Fg = 56.7 * 9.8

Fg = 555.66 N

Therefore, the work done will be:

W = - 555.66 * 17

W = -9446.22 J

The negative value of work means that the work done is opposite the value of the force acting on the climber.

Laptop batteries are rated in W⋅hW⋅h, the product of the power (in WW) that the battery can provide and the time (in hh) that it can provide this power. For instance, a 50 W⋅hW⋅h battery can provide 50 WW for 1.0 hh, 25 WW for 2.0 hh, and so on. A 10.8 VV laptop battery is rated at 74 W⋅hW⋅h.

Answer:

6.85C

Explanation:

the charge and the current are related by

Q= I×t

where Q = charge in columbs

v = 10.8

P = IV

I = P/V = 74/10.8

I =6.85 A

the battery will supply a charge of (6.85×1) C

6.85C

in 1 hour

how long would it take for a radio wave sent from a space satellite circling mars to reach Earth? Assume that radio waves (a form of electromagnetic radiation) travel at the speed of light?

Answer:

3 Minutes 2 seconds to 22 minutes 16 seconds

Explanation:

Lets assume the Mars to be at the closest distance to Earth. This distance (D) = 54.6 Million km

The signal travels at the speed of light, so speed of the signal (V) = 300000 km/s

So, the time (T) taken by the radio wave to reach Earth from Mars will be,

Thus, T = 182 Sec = 3 Minutes 2 seconds.

The radio wave will take minimum 03 Minutes 02 seconds to reach Earth. Here is should be noted that the distance between the two planets keep on changing as they revolve around the Sun. There will come a point when Mars is farthest from Earth and the distance (D) will be 401 Million km. Then, the time will change to ,

T = 1336.67 sec = 22.27 Minutes.

So the maximum time will be 22 minutes 16 seconds.

Ice at -50 oF and 1.00 atmis to be melted in a continuous flow process with saturated steam at 1 bar. Find the minimum quantity of steam required per pound of ice.The heat capacity of ice is 0.45 Btu/lbmoF.

Answer:

The minimum quantity of steam required per pound of ice is -22.5 BTU/lbm

Explanation:

Q = mCT

Q/m = CT

Q/m is the quantity of steam required per pound of ice

C is the heat capacity of ice = 0.45 BTU/lbm°F

T is the temperature of ice = -50 °F

Q/m = 0.45 BTU/lbm°F × -50 °F = -22.5 BTU/lbm

A woman wearing an in-ear hearing aid listens to a television set at a normal volume of approximately60 dB. To hear it, she requires an amplification of30 dB, so the hearing aid supplies sound at90 dB to the ear canal, which we assume to be circular with a diameter of 6.4 mm . What is the output power of the hearing aid?

Express your answer to two significant figures and include the appropriate units.

Answer:

the output power of the hearing aid in two significant figures is 3.2 × 10⁻⁸ W

Explanation:

Output power of sound is given as;

P = I*A

where;

I is the sound intensity = 90dB = 1 × 10⁻³ W/m²

A is the area through which the transmits;

since the canal is circular in shape.

Given;

D as diameter = 6.4mm = 0.0064 m

Output power of the hearing aid = (1 × 10⁻³ W/m²)*(3.217 × 10⁻⁵ m²)

= 3.2 × 10⁻⁸ W

Therefore, the output power of the hearing aid in two significant figures is 3.2 × 10⁻⁸ W.

Two forces, F? 1 and F? 2, act at a point, as shown in the picture. (Figure 1) F? 1 has a magnitude of 9.20 N and is directed at an angle of ? = 65.0 ? above the negative xaxis in the second quadrant. F? 2 has a magnitudeof 5.80 N and is directed at an angle of ? = 53.9 ? below the negative x axis in the third quadrant. Part A

What is the x component Fx of the resultant force?

Express your answer in newtons.

Part B

What is the y component Fy of the resultant force?

Express your answer in newtons.

Part C

What is the magnitude F of the resultant force?

Express your answer in newtons.

Part D

What is the angle ? that the resultant force forms with the negative x axis? In this problem, assume that positive angles are measured clockwise from the negative x axis.

Express your answer in degrees.

Explanation:

A) x-component of forces is 7.297 N

For x-component FX;

FX = F1 cos 65 + F2 cos 53.9

FX = 9.20 cos 65 + 5.8 cos 53.9

NB both forces are positive if resolved along the x-axis.

FX = 3.88 + 3.417 = 7.297 N

B) 6-component of forces is -3.654 N

For y-component FY

FY = -F1 sin 65 + F2 sin 53. 9

FY = - 9.2 sin 65 + 5.8 sin 53.9

NB F1 is negative along y-axis

FY = - 8.34 + 4.686 = - 3.654 N

C) The magnitude of the resultant force is 8.16 N

Magnitude = (FX^2 + FY^2)^0.5

(7.297^2 + (-3.654)^2)^0.5 = 8.16N

D) Angle made with negative x-axis is 26.56°

Tan^ -1 of FY/FX gives the angle of the resultant force.

= tan^-1 of -3.654/7.297

=tan^-1 of -0.5

= -26.56°

I.e 26.55° from the negative x-axis in a clockwise direction.

If the fraction of vacant sites at a temperature T1 is F1 and the fraction of vacant sites at a temperature T2 is F2, determine the relationship between T1, T2, F1, F2, Qv and k where Qv is the activation energy to form a vacancy and k is Boltzmann’s constant

Answer:

F₁ / F₂ = exp (-Q_{v} / k) exp [(T₂-T₁) / T₁T₂]

Q_{v} = -k ΔT / T₁T₂ ln (F₁ / F₂)

Explanation:

The Boltzmann equation for vacancies is

n = n₀ exp (-Eₙ /kT)

Where n and n₀ are the number of current and maximum vacancies, respectively, Eₙ is the activation energy (), K the Boltzmann constant and T the absolute temperature

The fraction is

F = n / n₀ = exp (- Q_{v} / kT)

Let's apply this equation to our case

For temperature T = T₁

F₁ = exp (-Qv / k T₁)

For temperature T = T₂

F₂ = exp (-Qv / kT₂)

The ratio of the vacancy fraction is

F₁ / F₂ = exp (-Q_{v} / k T₁) / exp (- Q_{v} / k T₂)

F₁ / F₂ = exp (-Q_{v} / k) exp [(T₂-T₁) / T₁T₂]

In general, the activation energy is sought by clearing this equation

Q_{v} = -k ΔT / T₁T₂ ln (F₁ / F₂)

An atom of helium and one of argon are singly ionized - one electron is removed from each. The two ions are then accelerated from rest by the electric field between two plates with a potential difference of 150 V. What happens after the two ions accelerate from one plate to the other?

a) The helium ion has more kinetic energy.

b ) The argon ion has more kinetic energy.

c) Both ions have the same kinetic energy.

d) There is not enough information to say which ion has more kinetic energy.

Answer:Both ions have the same kinetic energy.

Explanation:

The kinetic energy of the ions depends on the accelerating potential. The greater the accelerating potential, the greater the kinetic energy of the ions and vice versa. Ions are usually accelerated in such a way that they all have the same kinetic energy. Since the accelerating potential of the He+ and Ar+ are the same, they will both possess the same kinetic energy.

A ball is thrown straight upward. How does the sign of the work done by gravity while the ball is traveling upward compare with the sign of the work done by gravity while the ball is traveling downward?A. Work done by gravity is negative while the ball is traveling upward and positive while the ball is traveling downward.B. Work done by gravity is positive while the ball is traveling both upward and downward.C. Work done by gravity is positive while the ball is traveling upward and negative while the ball is traveling downward.D. Work done by gravity is negative while the ball is traveling both upward and downward.

Answer:

A.

Explanation:

- According to the work energy theorem, the work done by external forces on a mass, must equal to the change in kinetic energy.
- When thrown upward, the change in kinetic energy is negative, due to it must have some initial velocity, which it will diminishing while going upward, as gravity is slowing down it.
- So, in this part of the trajectory, work done by gravity (only force acting on the ball if we neglect air resistance) must be negative.
- In the downward part of the trajectory, as the change in kinetic energy is positive (the ball is speeding up when falling), the work done by gravity must be positive too.
- So, the option A is the one that is true.

The heating coil of a hot water heater has a resistance of 17 Ω and operates at 300 V. How long a time is required to raise the temperature of 115 kg of water from 18°C to 75°C? (The specific heat for water = 10 3 cal/kg ⋅°C and 1.0 cal = 4.186 J).

Answer:

5183 sec.

Explanation:

- The power dissipated in the resistance, will be used to raise the temperature of the water completely.
- This power can be found applying Joule's law, which states the following:

- If the resistance of the heating coil can be assumed as constant with the temperature, we can find the current I applying Ohm's Law, as follows:

- Now, from (1) we can find P, as follows:

- The energy supplied by this power, is just the product of the power times the time during which the energy was delivered:

- This energy, will be supplied as heat to the mass of water, as stated by the following equation (assuming no heat losses out of the heater):

- where c= specific heat of water = 4186 J/ºC*kg, m= 115 kg, and ΔT= 57ºC.
- From (2) and (3), if left sides are equal each other, so do right sides:

- Replacing by the values, we can solve for Δt, as follows:

- The time required to raise the temperature of 115 kg of water, from 18ºC, to 75ºC, is 5183 sec (approximately 1hr 26').

What is the resistance (in Ω) of thirty-seven 330 Ω resistors connected in series? 12210 Correct: Your answer is correct. Ω (b) What is the resistance (in Ω) of thirty-seven 330 Ω resistors connected in parallel?

Answer:

(a) 12210 Ω

(b) 9.92 Ω

Explanation:

From Electricity,

(a)

When resistors are connected in series, the total resistance is the sum of the individual resistance.

From the question,

Rt = R1 + R2 + R3................................. R37 ....................... Equation 1

Where Rt = Total resistance of combined resistance, R1 = resistance of the first resistor, R2 = Resistance of the second resistor .......... and so on up to the 37th resistor.

Note: Since The resistor are identical.

Rt = n(R) ..................... Equation 2

Where n = number of resistor, R = resistance of each resistor.

Given: n = 37, R = 330 Ω

Substitute into equation 2

Rt = 37(330)

Rt = 12210 Ω

(b)

For parallel connection,

Rt = R/n...................... Equation 2

Given: R = 330 Ω, n = 37

Substitute into equation 2

Rt = 330/37

Rt = 8.92 Ω

Hence the resistance when connected in parallel = 9.92 Ω

Monochromatic light with a wavelength of 381 nm passes through a single slit and falls on a screen 85 cm away. If the distance of the first-order dark band is 0.43 cm from the center of the pattern, what is the width of the slit?

Answer:

7.55 * 10^(-5) m

Explanation:

Parameters given:

Wavelength of light, λ = 381 nm = 3.81 * 10^(-7) m

Distance between screen and slit, L = 85 cm = 0.85 m

Distance of the first-order dark band from the center of the pattern, x = 0.43cm = 0.0043 m

The relationship between λ, L, x and the width of the slit, w is given as:

λ/w = x/L

=> w = λL/x

w = (3.81 * 10^(-7) * 0.85)/(0.0043)

w = 7.55 * 10^(-5) m

Rubber rods charged by rubbing with cat fur repel each other. Glass rods charged by rubbing with silk repel each other. A rubber rod and a glass rod charged respectively as above attract each other. A possible explanation is that a. Any two rubber rods charged this way have opposite charges on them.b. Any two glass rods charged this way have opposite charges on them.c. A rubber rod and a glass rod charged this way have opposite charges on them.d. All rubber rods always have an excess of positive charge on them.e. All glass rods always have an excess of negative charge on them.

Answer:

C. A rubber rod and a glass rod charged this way have opposite charges on them.

Explanation:

When a rubber rod is rubbed against cat fur, it acquires a negative charge, it becomes negatively charged.

When you then try to bring two rubber rod's together, they repel because like charges repel.

Meanwhile, when you rub a glass rod against silk, it loses electrons to the silk material and becomes positively charged.

When you bring two positively charged glass rod's together, they repel, because like charges repel.

However, when you bring the rubber rod and a glass rod together, the attract each other because unlike/opposite charges attract.

A +25µC point charge is placed 6.0cm from an identical +25µC point charge. How much work would be required by an external force to move a +0.18µC test charge from a point midway between them to a point 1.0cm closer to either charge?

The work done needed by an external force is .3375 J

calculation of work required:Required work done = charge moved × potential difference at two points

Here

Potential at midpoint

= 2 x 9 x 10⁹ x 25 x 10⁻⁶ / 3 x 10⁻²

= 150 x 10⁵ V

Potential at the second point

= 9 x 10⁹ x 25 x 10⁻⁶ / 4 x 10⁻² + 9 x 10⁹ x 25 x 10⁻⁶ / 2 x 10⁻²

= 56.25 x 10⁵ + 112.5 x 10⁵

= 168.75 x 10⁵ V

Potential difference

= (168.75 - 150 ) x 10⁵ V

= 18.75 x 10⁵

So, Work done

= .18 x 10⁻⁶ x 18.75 x 10⁵ J

= 3.375 x 10⁻¹ J

= .3375 J

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Answer:

Explanation:

Required work done = charge moved x potential difference at two points

Potential at mid point

= 2 x k q / r

= 2 x 9 x 10⁹ x 25 x 10⁻⁶ / 3 x 10⁻²

= 150 x 10⁵ V

Potential at the second point

= 9 x 10⁹ x 25 x 10⁻⁶ / 4 x 10⁻² + 9 x 10⁹ x 25 x 10⁻⁶ / 2 x 10⁻²

= 56.25 x 10⁵ + 112.5 x 10⁵

= 168.75 x 10⁵ V

Potential difference

= (168.75 - 150 ) x 10⁵ V

= 18.75 x 10⁵

Work done

= .18 x 10⁻⁶ x 18.75 x 10⁵ J

= 3.375 x 10⁻¹ J

= .3375 J

You are standing on a train station platform as a train goes by close to you. As the train approaches, you hear the whistle sound at a frequency of f1 = 98 Hz. As the train recedes, you hear the whistle sound at a frequency of f2 = 76 Hz. Take the speed of sound in air to be v = 340 m/s(a) Find an equation for the speed of the sound source v., in this case it is the speed of the train. Express your answer in terms of f2. and v.

The dopler effect allows finding the result for the speed of the train as a function of the frequency of when it moves away and the speed of sound is:

- The expression for the speed of the train is:

- The velocity for this case is: vs = 43 m / s

The Dopler effect is the change in the frequency of the sound due to the relative movement of the source and the observer.

In this case the observer is standing on the platform whereby the movement is from the sound source.

Where f’ is the perceived frequency, f the emitted frequency, v the speed of sound and the speed of the source, the negative sign corresponds when they are approaching and the positive sign when they are moving away.

Indicate that when the train approaches the perceived frequency is f₁ = 98 Hz and when the train moves away the frequency is f₂ = 76 Hz, the speed of sound is v = 340 m / s.

We have a system of two equation with two unknowns, we can solve it.

let's resolve.

They indicate that the result is given as a function of f₂ and the speed of sound, let's substitute.

Let's calculate the speed of the train.

= 43 m / s

In conclusion using the dopler effect we can find the result for the speed of the train as a function of the frequency of when it moves away and the speed of sound is:

- The expression for the speed of the train is:

- The velocity for this case is: = 43 m / s

Learn more about the Doppler effect here: brainly.com/question/5871604

Answer: v = (33320 - 340f2)/( f2 + 98)

Explanation: this question refers to Doppler effect, hence,

f1 = { V / (V - v)} x f2

f1 = 98Hz, f2 = 76 Hz, V = 340m/s, v = ?, fs = fsourcs

For f1,

98 = { 340 / (340 + v)} x fs ...(i)

for f2,

f2 = { 340 / (340 - v)} x fs ... (ii)

Divide (ii) by (i)

98/f2 = [{ 340 / (340 + v)} x fs]÷ [ 340 / (340 - v)} x fs]

98/f2 = {340 / (340 + v)} x fs x (340 - v)} / 340 x fs

98/f2 = (340 + v)/ (340 - v)

Cross multiplying

98(340 - v) = f2 (340 + v)

33320 - 98v = 340f2 + vf2

Collecting like terms

vf2 +98v = 33320 - 340f2

v(f2 +98) = 33320 - 340f2

v = (33320 - 340f2)/( f2 + 98)

When a current flows in an aluminum wire of diameter 2.91 mm 2.91 mm , the drift speed of the conduction electrons is 0.000191 m/s 0.000191 m/s . How many electrons are flowing past a given point each second? The conduction electron density in aluminum is 6.00 × 10 28 m − 3 6.00×1028 m−3 .

There are electrons flowing past a given point each second.

Conduction current:The current I in a wire with cross-sectional area A is given by:

where n is the conduction electron density = 6×10²⁸m⁻³

is the drift velocity = 0.000191 m/s

A is the area = ( here d is the diameter = 0.000291m)

In terms of rate of flow of charge the current can be expressed as:

where N is the number of electrons passing through a point per unit time.

Equating both the expressions we get:

Learn more about drift velocity:

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Answer:

Number of electrons are flowing per second is 2.42 x 10¹⁹

Explanation:

The electric current flows through a wire is given by the relation :

....(1)

Here I is current, e is electronic charge, v is drift velocity of electrons and A is the Area of the wire.

But electric current is also define as rate of electrons passing through junction times their charge, i.e. ,

....(2)

Here N is the rate of electrons passing through junction.

From equation (1) and (2).

But area of wire,

Here d is diameter of wire.

So,

Substitute 2.91 x 10⁻³ m for d, 0.000191 m/s for v and 6 x 10²⁸ m⁻³ for n in the above equation.

N = 2.42 x 10¹⁹ s⁻¹

Points A, B, and C form the vertices of a triangle in a nonuniform electrostatic field. The electrostatic work done on a particle of charge q as the particle travels from A to B is WAB and that done on the particle as it travels from A to C is WAC=−WAB/3. a) How much electrostatic work is done on the particle as it travels from B to C?

Answer:

Explanation:

Let electric potential at A ,B and C be Va , Vb and Vc respectively.

Work done = charge x potential difference

Wab = q ( Va - Vb )

Wac = q ( Va - Vc )

Given

Wac = - Wab / 3

3Wac = - Wab

Now

Wbc = q ( Vb - Vc )

= q [ ( Va-Vc ) - ( Va - Vb )]

= Wac - Wab

= Wac + 3Wac

= 4Wac

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