which of the following code segments will move the robot from the center square along the same path as the code segment above?

Answers

Answer 1

count ← 0

REPEAT 4 TIMES

{

count count + 1

REPEAT count TIMES

{

MOVE_FOWARD ()

}

ROTATE_LEFT ()

}

Which of the following best describes the typical method by which messages are transmitted over the Internet?

The message is broken into bundles that are communicated in a predefined request. Every bundle should be gotten in the request it was sent for the message to be accurately reassembled by the beneficiary's gadget.

Which of the following best shows how keylogging can be used to get into a computer system without permission?

A user installs a program on their computer that records and transmits all user input unintentionally.

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Related Questions

If it is struck by a rigid block having a weight of 550 lblb and traveling at 2 ft/sft/s , determine the maximum stress in the cylinder. Neglect the mass of the cylinder. Express your answer to three significant figures and include appropriate units.

Answers

This question is incomplete, The missing image is uploaded along this answer below;

Answer:

the maximum stress in the cylinder is 3.23 ksi

 

Explanation:

Given the data in the question and the diagram below;

First we determine the initial Kinetic Energy;

T = [tex]\frac{1}{2}[/tex]mv²

we substitute

⇒ T = [tex]\frac{1}{2}[/tex] × (550/32.2) × (2)²

T = 34.16149 lb.ft

T =  ( 34.16149 × 12 ) lb.in  

T = 409.93788 lb.in

Now, the volume will be;

V = [tex]\frac{\pi }{4}[/tex]d²L

from the diagram; d = 0.5 ft and L = 1.5 ft

so we substitute

V =  [tex]\frac{\pi }{4}[/tex] × ( 0.5 × 12 in )² × ( 1.5 × 12 in )

V = 508.938 in³

So by conservation of energy;

Initial energy per unit volume = Strain energy per volume

⇒ T/V = σ²/2E

from the image; E = 6.48(10⁶) kip

so we substitute

⇒ 409.93788 / 508.938 = σ²/2[6.48(10⁶)]

508.938σ² =  5,312,794,924.8

σ² = 10,438,982.5967  

σ = √10,438,982.5967

σ = 3230.9414  

σ = 3.2309 ksi  ≈ 3.23 ksi    { three significant figures }

Therefore, the maximum stress in the cylinder is 3.23 ksi

In a certain company the cost of software depends on the license type which could be Individual or Enterprise. Write a program that reads  License Type wanted (just the first character of each type: I, i, E, e).  Number of Users to use the software. Type Price/User Minimum number of users Individual 500$ 1 Enterprise 300$ 5 Your program should:  Check if the number of users is greater than or equal than Minimum Number of Users allowed Compute the cost: (for example cost = Cost per user x Number of Users)

Answers

Solution :

import [tex]$\text{java}.$[/tex]util.*;

public [tex]$class$[/tex] currency{

  public static [tex]$\text{void}$[/tex] main(String[tex]$[]$[/tex] args) {

      Scanner input [tex]$=$[/tex] new Scanner(System[tex]$\text{.in}$[/tex]);

      System[tex]$\text{.out.}$[/tex]print("Enter [tex]$\text{number of}$[/tex] quarters:");

      int quarters = input.nextInt();

      System.out.print("Enter number of dimes:");

      int [tex]$\text{dimes =}$[/tex] input.nextInt();

      System[tex]$\text{.out.}$[/tex]print("Enter number of nickels:");

      int nickels = input.nextInt();

      System[tex]$\text{.out.}$[/tex]print("Enter number of pennies:");

      int [tex]$\text{pennies = }$[/tex] input.nextInt();

      // computing dollors

      double dollars = (double) ((quarters*0.25)+(dimes*0.10)+(nickels*0.05)+(pennies*0.01));

      System[tex]$\text{.out.}$[/tex]format("You have : $%.2f",dollars);

  }

}

Identify parts of the E-Cig that constitute voltage, current, and resistance. Discuss the role each plays in the E-Cig and typical values for each including units.
Discuss the electrical dangers of an E-Cig. Give specific examples.
There are many electrical safety rules. Pick one, and discuss its application on a small system, such as the E-Cig.

Answers

Answer: c

Explanation:

Explain what the engineering team should advise in the following scenario.

Situation: An engineering team is analyzing a problem with a dam near a large metropolitan area. The dam reservoir has a capacity of 300 billion cubic feet, and it produces hydroelectric power for local municipalities. The dam has developed a severe crack.

Answers

Answer: its c

Explanation:

The part in the photo below is

A. an evaporator.
B. an accumulator.
C. a condenser.
D. a compressor.

Answers

Answer: a accumulator

Explanation:

can someone please help me with this
I've an exams tomorrow ​

Answers

Answer:

I am in Eight Grade

Explanation:

state the parallelogram law of forces​

Answers

Answer:

The law of parallelogram of forces states that if two vectors acting on a particle at the same time be represented in magnitude and direction by the two adjacent sides of a parallelogram drawn from a point their resultant vector is represented in magnitude and direction by the diagonal of the parallelogram drawn from .

The law of parallelogram of forces states that if two vectors acting on a particle at the same time be represented in magnitude and direction by the two adjacent sides of a parallelogram drawn from a point their resultant vector is represented in magnitude

Three spheres are subjected to a hydraulic stress. The pressure on spheres 1 and 2 is the same, and they are made of the same material. The fractional compression on the third sphere is equal to the fractional compression on the first sphere times the reciprocal of the fractional compression on the second. If the pressure on the third sphere is 150000 N/m2 , what is the Bulk Modulus for the third sphere

Answers

Answer:

"150000 N/m²" is the right approach.

Explanation:

According to the question, the pressure on the two spheres 1 and 2 is same.

Sphere 1 and 2:

Then,

⇒  [tex]P_1=P_2[/tex]

⇒  [tex]\frac{\Delta V_1}{V_1}=\frac{\Delta V_2}{V_2}[/tex]

and the bulk modulus be,

⇒  [tex]B_1=B_2[/tex]

Sphere 3:

⇒  [tex]\frac{\Delta V_3}{V_3} =\frac{\frac{\Delta V_1}{V_1} }{\frac{\Delta V_2}{V_2} } =1[/tex]

then,

⇒  [tex]P_3=B\times \frac{\Delta V_3}{V_3}[/tex]

⇒       [tex]=B\times 1[/tex]

⇒       [tex]=150000\times 1[/tex]

⇒       [tex]=150000 \ N/m^2[/tex]

Refrigerant 134a enters the evaporator of a refrigeration system operating at steady state at -16oC and a quality of 20% at a velocity of 5 m/s. At the exit, the refrigerant is a saturated vapor at -16oC. The evaporator flow channel has constant diameter of 1.7 cm. Determine the mass flow rate of the refrigerant, in kg/s, and the velocity at the exit, in m/s.

Answers

Answer:

mass flow rate = 0.0534 kg/sec

velocity at exit = 29.34 m/sec

Explanation:

From the information given:

Inlet:

Temperature [tex]T_1 = -16^0\ C[/tex]

Quality [tex]x_1 = 0.2[/tex]

Outlet:

Temperature [tex]T_2 = -16^0 C[/tex]

Quality  [tex]x_2 = 1[/tex]

The following data were obtained at saturation properties of R134a at the temperature of -16° C

[tex]v_f= 0.7428 \times 10^{-3} \ m^3/kg \\ \\ v_g = 0.1247 \ m^3 /kg[/tex]

[tex]v_1 = v_f + x_1 ( vg - ( v_f)) \\ \\ v_1 = 0.7428 \times 10^{-3} + 0.2 (0.1247 -(0.7428 \times 10^{-3})) \\ \\ v_1 = 0.0255 \ m^3/kg \\ \\ \\ v_2 = v_g = 0.1247 \ m^3/kg[/tex]

[tex]m = \rho_1A_1v_1 = \rho_2A_2v_2 \\ \\ m = \dfrac{1}{0.0255} \times \dfrac{\pi}{4}\times (1.7 \times 10^{-2})^2\times 6 \\ \\ \mathbf{m = 0.0534 \ kg/sec}[/tex]

[tex]\rho_1A_1v_1 = \rho_2A_2v_2 \\ \\ A_1 =A_2 \\ \\ \rho_1v_1 = \rho_2v_2 \\ \\ \implies \dfrac{1}{0.0255} \times6 = \dfrac{1}{0.1247}\times (v_2)\\ \\ \\\mathbf{\\ v_2 = 29.34 \ m/sec}[/tex]

Knowing that the central portion of the link BD has a uniform cross-sectional area of 800 2, determine the magnitude of the load P for which the normal stress in that portion of BD is 50 .

Answers

Answer: 50

Explanation:

Which of these parts converts the spinning motion of the driveshaft 90° to turn the wheels?
A. Transmission
B. Axle
C. Differential
D. Engine

Answers

B. Axle by g hfffdggfrfgf
B. Axle aidgdkdidhskav

Define;
i) Voltage
ii) Current
iii) Electrical Power
iv) Electrical Energy​

Answers

Answer:

I) Voltage - is the pressure from an electrical circuit's power source that pushes charged electrons (current) through a conducting loop, enabling them to do work such as illuminating a light. In brief, voltage = pressure, and it is measured in volts (V).

II) Current - is the movement of electrons through a wire. Electric current is measured in amperes (amps) and refers to the number of charges that move through the wire per second. If we want current to flow directly from one point to another, we should use a wire that has as little resistance as possible.

III) Electrical Power - is the rate, per unit time, at which electrical energy is transferred by an electric circuit. The SI unit of power is the watt, one joule per second. Electric power is usually produced by electric generators, but can also be supplied by sources such as electric batteries.

IV) Electrical Energy - is a form of energy resulting from the flow of electric charge. Energy is the ability to do work or apply force to move an object. In the case of electrical energy, the force is electrical attraction or repulsion between charged particles.

Explanation:

I hope ot helps to you a lot! Correct me if I'm wrong.

Let xa(t)be an analog signal with bandwidth B=3kHz. We wishto use an ????=2m–pointDFT to compute the spectrum ofthe signal with a resolution less than or equal to 50 Hz.
Determine
(a) the minimum sampling rate,
(b) the minimum number of required samples, and
(c) the minimumlength of the analog signal record(in seconds).

Answers

Answer:

a) the minimum sampling rate is 6 kHz

b) the minimum numbers of required samples are 120

c) the minimum length of the analog signal is 0.02 s

Explanation:

Given the data in the question;

(a) the minimum sampling rate;

band width of analog signal xₐ(t) is;

bandwidth B = 3kHz

Now, according to sampling theorem, minimum sampling rate F[tex]_s[/tex] must be twice the bandwidth of the signal.

so

F[tex]_s[/tex] = 2B

F[tex]_s[/tex] = 2( 3 kHz )

F[tex]_s[/tex] = 6 kHz

Therefore, the minimum sampling rate is 6 kHz

(b) the minimum number of required samples;

Let L represent the minimum number of samples required,

given that; required resolution of the spectrum of the signal is less than or equal to 50 Hz

F[tex]_s[/tex]/L ≤ 50

L ≥ F[tex]_s[/tex]/50

L ≥ ( 6 × 1000 Hz ) / 50

L ≥ 6000 / 50

L ≥ 120

Therefore, the minimum numbers of required samples are 120

(c) the minimum length of the analog signal record(in seconds).

minimum number of samples required is 120

T = L /  F[tex]_s[/tex]

T = 120 / ( 6 × 1000 Hz )

T = 120 / 6000

T = 0.02 s

Therefore, the minimum length of the analog signal is 0.02 s

calculate the radius of a circular orbit for which the period is 1 day​

Answers

Answer:

(T²/D³)sys1 = (T²/D³)sys2

sys1 = earth-moon

sys2 = earth-sat

(27.33day)²/(3.8e8m)³ = (1day)²/D³

D = cbrt(7.3e22m³) = 4.2e7 m

Explanation:

exchange in capacity whilst a satellite tv for pc differences altitude How plenty paintings could be executed to flow the satellite tv for pc into yet another around orbit it is bigger above the outdoors of the Earth? satellite tv for pc exchange in capacity with top Assuming the satellite tv for pc is to be boosted to a clean top r? Gravitational capacity capacity (to center of earth) new orbit(2) has a greater robust PE than old one(a million), so exchange is helpful PE = G m?m?/r earth GM = 3.98e14 ?PE = (GM)(m)(a million/r? – a million/r?) KE additionally differences. Get speed at each and every top. New orbit(2) has decrease speed, so exchange is damaging v = ?(GM/R) V? = ?(GM/r?) V? = ?(GM/r?) ?KE = –½m(V?² – V?²) ?KE = –½mGM(a million/r? – a million/r?) including the two ?E = (GM)(m)(a million/r? – a million/r?)– ½mGM(a million/r? – a million/r?) ?E = ½mGM(a million/r? – a million/r?) ?E = ½(3.98e14)(7500) [a million/(0.5e7) –a million/(3.3e7) ] ?E = ½(3.98e14)(7500)(1e-7) [a million/(0.5) –a million/(3.3) ] ?E = ½(3.98e7)(7500) [2 – 0.303 ] ?E = ½(3.98e7)(7500)(a million.70) ?E = 2.04e11 Joules edit, corrected .

The radius of a circular orbit will be "[tex]\frac{V}{2 \pi} \ km[/tex]".

According to the question,

The orbit period of satellite,

Time = 1 day

Total distance will be equal to the orbit's circumference, then

Distance = [tex]2 \pi r[/tex]

Let,

The velocity be "V km/day".

As we know,

→ [tex]Distance = Velocity\times time[/tex]

By substituting the values, we get

→          [tex]2 \pi r = V\times 1[/tex]

→              [tex]r = \frac{V}{2 \pi} \ km[/tex]

Thus the above is the right answer.

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Fig_Q5
6. A steel rod is stressed by a tension force of 250 N. It is found that the rod has length of 45
m and diameter of 1.5 mm. If the modulus of elasticity of the steel rod is assumed as 2 x 105
MPa, determine the strain of the steel rod due to the applied force.

Answers

Answer:

The strain of the steel rod due to the applied force is 41.93

Explanation:

Modulus of elasticity is equal to stress divided by strain.

And stress is equal to force divided by area

Surface area of cylindrical rod

[tex]2\pi r (r+h)[/tex]

Substituting the given values we get -

[tex]2 *3.14 * \frac{1.5}{1000} * 45 (45 + \frac{1.5}{1000}) = 19.07[/tex]

[tex]2 * 10 ^5 = \frac{250}{19.07 * S=(\frac{\Delta L}{L} )}[/tex]

Hence, strain is equal to

Strain = 41.93

Combinations of velocity and acceleration

Answers

Answer:

acceleration=change in velocity/ time

Explanation:

The velocity of an object is its speed in a particular direction. Velocity is a vector quantity because it has both a magnitude and an associated direction. To calculate velocity, displacement is used in calculations, rather than distance.

A cylindrical rod of brass originally 10 mm in diameter is to be cold worked by drawing. The circular cross section will be maintained during deformation. A cold-worked tensile strength in excess of 380 MPa and a ductility of at least 15 %EL are desired. Furthermore, the final diameter must be 7.5 mm. Explain how this may be accomplished. Use the graphs given in previous question.

Answers

Answer:

Explanation:

From the information given:

original diameter [tex]d_o[/tex] = 10 mm

final diameter [tex]d_f =[/tex] 7.5 mm

Cold work tensile strength of brass = 380 MPa

Recall that;

[tex]\text {The percentage CW }= \dfrac{\pi (\dfrac{d_o}{2})^2 - \pi(\dfrac{d_f}{2})^2 }{\pi(\dfrac{d_o}{2})^2} \times 100[/tex]

[tex]\implies \dfrac{\pi (\dfrac{10}{2})^2 - \pi(\dfrac{7.5}{2})^2 }{\pi(\dfrac{10}{2})^2} \times 100[/tex]

[tex]\implies43.87\% \ CW[/tex]

→ At 43.87% CW, Brass has a tensile strength of around 550 MPa, which is greater than 380 MPa.

→ At 43.87% CW, the ductility is less than 5% EL, As a result, the conditions aren't met.

To achieve 15% EL, 28% CW is allowed at most

i.e

The lower bound cold work = 15%

The upper cold work = 28%

The average = [tex]\dfrac{15+28}{2}[/tex] = 21.5 CW

Now, after the first drawing, let the final diameter be [tex]d_o^'[/tex]; Then:

[tex]4.5\% \ CW = \dfrac{\pi (\dfrac{d_o^'}{2})^2 - (\dfrac{7.5}{2})^2}{\pi (\dfrac{d_o^'}{2})^2}\times 100[/tex]

By solving:

[tex]d_o^'} = 8.46 mm[/tex]

To meet all of the criteria raised by the question, we must first draw a wire with a diameter of 8.46 mm and then 21.5 percent CW on it.

In the production of soybean oil, dried and flaked soybeans are brought in contact with a solvent (often hexane) that extracts the oil and leaves behind the residual solids and a small amount of oil.

a. Draw flow diagram of the process, labeling the two feed streams (beans and solvent) and the leaving streams (solids and extract).
b. The soybeans contain 18.5 wt% oil and the remainder insoluble solids, and the hexane is fed at a rate corresponding to 2.0 kg hexane per kg beans. The residual solids leaving the extraction unit contain 35.0 wt% hexane, all of the non-oil solids that entered with beans, and 1.0% of the oil that entered the beans. For a feed rate of 1000 kg/h of dried flaked soybeans, calculate mass flow rates of extract and residual solids and the composition of extract.

Answers

Answer: its c

Explanation:

The evaporator:
A. directs airflow to the condenser.
B. absorbs heat from the passenger compartment.
C. removes moisture from the refrigerant.
D. restricts refrigerant flow.

Answers

Answer:

Option B

Explanation:

An evaporator along with  cold low pressure refrigerant absorbs heat from the air within the passenger compartment thereby supplying cool air for the occupants.

Hence, option B is correct

Barries of effective
communication?

Answers

Answer: barries

Explanation:

In 1951, a small approach embankment was constructed for a highway bridge over a river south of Los Angeles. The embankment was underlain by 5 ft of organic clay. Records of the settlement rate indicate that 90% of the consolidation settlements occurred in the first 4.5 years after construction. A new bridge over the river is now planned for a site a few hundred yards from the old bridge. The approach embankment to the new bridge will be underlain by 20 ft of the same organic clay found at the old bridge site. Estimate the time required to achieve an average degree of consolidation of 90% under the new embankment. Assume single drainage from the organic clay at both sites..

Answers

Answer:

72 years

Explanation:

The degree of consideration is the same for both bridges = 90%

Height  of first highway bridge( d1 ) = 5 ft

Time to consolidation ( t1 )= 4.5 years

Height of second bridge ( d2 ) = 20 ft

Time to consolidation ( t2 ) = ?

we will apply this relation below

Tv = Cv * t / d^v

Tv = constant

for a single drainage condition : t ∝ d^v hence; d = H

∴ [tex]\frac{t_{2} }{t_{1} } = (\frac{d_{2} }{d_{1} })[/tex]^2

t2 = t1 ( d2/d1 )^2

   = 4.5 ( 20 / 5 )^2

  = 72 years

In a parking garage, each level is 18 feet apart, Each ramp to a level is 120 feet long. Find the measurement of elevation for each ramp?

Answers

Answer:

Sin(angle)=0.1538

Angle=arcsine(0.1538)

Angle is 8.85 degrees

Explanation:

Firstly draw a diagram showing a triangle with the required angle, a hypotenuse of 130 ft and an opposite side of 20 ft.

The sine of the angle required uses the formula Opposite/Hypotenuse

(i am not sure sure)

Water from an upper tank is drained into a lower tank through a 5 cm diameter iron pipe with roughness 2 mm. The entrance to the pipe has minor loss coefficient 0.4 and the exit has minor loss coefficient of 1, both referenced to the velocity in the pipe. The water level of the upper tank is 4 m above the level of the lower tank, and the pipe is 5 m long. You will find the drainage volumetric flow rate. a) What is the relative roughness

Answers

Answer:

Relative roughness = 0.04

Explanation:

Given that:

Diameter = 5 cm

roughness = 2 mm

At inlet:

Minor coefficient loss [tex]k_{L1} = 0.4[/tex]

At exit:

Minor coefficient loss [tex]k_{L2} = 1[/tex]

Height h = 4m

Length = 5 m

To find the relative roughness:

Relative roughness is a term that is used to describe the set of irregularities that exist inside commercial pipes that transport fluids. The relative roughness can be evaluated by knowing the diameter of the pipe made with the absolute roughness in question. If we denote the absolute roughness as e and the diameter as D, the relative roughness is expressed as:

[tex]e_r = \dfrac{e}{D}[/tex]

[tex]e_r = \dfrac{0.2 }{5}[/tex]

[tex]\mathbf{e_r = 0.04}[/tex]

g Design of a spindle present in an existing design needs to be reviewed for use under new loading needs. It is currently designed to withstand combined torsional and compressive loading. The design team needs to determine what the maximum torque it can withstand before failing if a compressive, axial load of 15 kips is present. The ultimate compressive and tensile strengths of the material are 135 ksi and 40 ksi, respectively. Determine the torsional shear stress that will just cause failure using a non-conservative failure theory.

Answers

Answer: its c

Explanation:

For some metal alloy, a true stress of 345 MPa (50040 psi) produces a plastic true strain of 0.02. How much will a specimen of this material elongate when a true stress of 412 MPa (59760 psi) is applied if the original length is 480 mm (18.90 in.)? Assume a value of 0.22 for the strain-hardening exponent, n.

Answers

Answer:

the elongation of the metal alloy is 21.998 mm

Explanation:

Given the data in the question;

K = σT/ (εT)ⁿ

given that metal alloy true stress σT = 345 Mpa, plastic true strain εT = 0.02,

strain-hardening exponent n = 0.22

we substitute

K = 345 / [tex]0.02^{0.22[/tex]

K = 815.8165 Mpa

next, we determine the true strain

(εT) = (σT/ K)^1/n

given that σT = 412 MPa

we substitute

(εT) = (412 / 815.8165 )^(1/0.22)

(εT) = 0.04481 mm

Now, we calculate the instantaneous length

[tex]l_i[/tex] = [tex]l_0e^{ET[/tex]

given that [tex]l_0[/tex] = 480 mm

we substitute

[tex]l_i[/tex] =[tex]480mm[/tex] × [tex]e^{0.04481[/tex]

[tex]l_i[/tex] =  501.998 mm

Now we find the elongation;

Elongation = [tex]l_i - l_0[/tex]

we substitute

Elongation = 501.998 mm - 480 mm

Elongation = 21.998 mm

Therefore, the elongation of the metal alloy is 21.998 mm

a) For Well A, provide a cross-section sketch that shows (i) ground elevation, (ii) casing height, (iii) depth to
water table, (iv) sampling depth, (v) elevation of the well top of casing, (vi) water table elevation, (vii) elevation
head of the water sampled for bromide, and (viii) pressure head of the water sampled for bromide. Label each of
these distances with the above phrases, plus a unique variable.
b) Calculate the following for each well: (i) elevation of the well top of casing, (ii) water table elevation,
(iii) sampling port elevation, (iv) elevation head of the water sampled for bromide, and (v) pressure head of the
water sampled for bromide. Use sea level as your vertical datum. Write out all calculations (including equations
with variables) for Well A.

Answers

Don’t go on that file will give a virus! Sorry just looking out and I don’t know how to comment!

How may the desire for a perfect lawn be related to environmental pollution? A. Lawns have little to do with environmental pollution. B. Perfect lawns require excess use of manual labor. C. Manicured lawns are subject to increased runoff. D. Lawns absorb pesticides and fertilizers and these chemicals leach out of the lawn and run into creeks, streams and eventually, rivers, with rain or watering, eventually reaching the ocean.

Answers

Answer:  is b

Explanation:

Which statements describe the motion of car A and car B? Check all that apply. Car A and car B are both moving toward the origin. Car A and car B are moving in opposite directions. Car A is moving faster than car B. Car A and car B started at the same location. Car A and car B are moving toward each other until they cross over.

Answers

Answer:

car a is moving faster than the car b

Answer:

B: Car A and car B are moving in opposite directions.

C: Car A is moving faster than car B.

E: Car A and car B are moving toward each other until they cross over.

Explanation:

I just did the assignment on EDGE2020 and it's 200% correct!  

Also, heart and rate if you found this answer helpful!! :) (P.S It makes me feel good to know I helped someone today!!)  :)  

Define hermetic compressor

Answers

Answer:

Hermetic compressors are ideal for small refrigeration systems, where continuous maintenance cannot be ensured.

The distribution of SAT scores of all college-bound seniors taking the SAT in 2014 was approximately normal with mean μ=1497 and standard deviation σ=322. A certain test-retake preparation course is designed for students whose SAT scores are in the lower 25%, percent of those who take the test in a given year. What is the maximum SAT score in 2014 that meets the course requirements?

Answers

Answer:

1279

Explanation:

We have the mean u = 1497

Standard deviation sd = 322

We find the x distribution using 25%

P(Z<z) = 0.25

Z = -0.675

From here we use the formula for z score

X = z(sd) + u

X = -0.675*322 + 1497

X = -217.35 + 1497

X = 1279.6

Which is approximately 1279

So we conclude that the maximum sat scores in year 2014that meets with the requirements of this course is 1279

Answer:

1831

Explanation:

Other Questions
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