1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Musya8 [376]
3 years ago
15

Problem 4: The built-in function clock returns a row vector that contains 6 elements: the first three are the current date (year

, month, day) and the last three represent the current time in hours (24 hour clock), minutes, and seconds. The seconds is a real number, but all others are integers. Use function sprintf to accomplish the following formatting exercises. a) Get the current date and time and store them in p4a. The current date and time should be the date and time when the grader calls your script. b) Using the format 'YYYY:MM:DD', write the current date to string p4b. Here, YYYY, MM, and DD correspond to 4-digit year, 2-digit month, and 2-digit day, respectively. c) Using the format 'HH:MM:SS.SSSS', write the current time to string p4c. Here, HH, MM, and SS.SSSS correspond to 2-digit hour, 2-digit minute and 7-character second (2 digits before the decimal point and 4 digits after the decimal points), respectively. d) Remove the last 5 characters from the string in part (c) so that the format is now 'HH:MM:SS'. Put the answer into string p4d. e) Combine the strings in part (b) and part (d) together separated by a single space. Put theanswer in string p4e
Engineering
1 answer:
Vladimir [108]3 years ago
7 0
Answer is 77 because
You might be interested in
Design a PLC ladder logic program to control the operation of a conveyor-storage system using the following sequence: - 1. Progr
Phantasy [73]

Answer:

See explaination

Explanation:

Kindly check attachment for the step by step solution of the given problem.

8 0
4 years ago
Which of the following ranges depicts the 2% tolerance range to the full 9 digits provided?
Lyrx [107]

Answer:

the only one that meets the requirements is option C .

Explanation:

The tolerance of a quantity is the maximum limit of variation allowed for that quantity.

To find it we must have the value of the magnitude, its closest value is the average value, this value can be given or if it is not known it is calculated with the formula

         x_average = ∑ x_{i} / n

The tolerance or error is the current value over the mean value per 100

         Δx₁ = x₁ / x_average

         tolerance = | 100 -Δx₁  100 |

bars indicate absolute value

let's look for these values ​​for each case

a)

    x_average = (2.1700000+ 2.258571429) / 2

    x_average = 2.2142857145

fluctuation for x₁

        Δx₁ = 2.17000 / 2.2142857145

        Tolerance = 100 - 97.999999991

        Tolerance = 2.000000001%

fluctuation x₂

        Δx₂ = 2.258571429 / 2.2142857145

        Δx2 = 1.02

        tolerance = 100 - 102.000000009

        tolerance 2.000000001%

b)

    x_average = (2.2 + 2.29) / 2

    x_average = 2,245

fluctuation x₁

         Δx₁ = 2.2 / 2.245

         Δx₁ = 0.9799554

         tolerance = 100 - 97,999

         Tolerance = 2.00446%

fluctuation x₂

          Δx₂ = 2.29 / 2.245

          Δx₂ = 1.0200445

          Tolerance = 2.00445%

c)

   x_average = (2.211445 +2.3) / 2

   x_average = 2.2557225

       Δx₁ = 2.211445 / 2.2557225 = 0.9803710

       tolerance = 100 - 98.0371

       tolerance = 1.96%

       Δx₂ = 2.3 / 2.2557225 = 1.024624

       tolerance = 100 -101.962896

       tolerance = 1.96%

d)

   x_average = (2.20144927 + 2.29130435) / 2

   x_average = 2.24637681

       Δx₁ = 2.20144927 / 2.24637681 = 0.98000043

       tolerance = 100 - 98.000043

       tolerance = 2.000002%

       Δx₂ = 2.29130435 / 2.24637681 = 1.0200000017

       tolerance = 2.0000002%

e)

   x_average = (2 +2,3) / 2

   x_average = 2.15

   Δx₁ = 2 / 2.15 = 0.93023

   tolerance = 100 -93.023

   tolerance = 6.98%

   Δx₂ = 2.3 / 2.15 = 1.0698

   tolerance = 6.97%

Let's analyze these results, the result E is clearly not in the requested tolerance range, the other values ​​may be within the desired tolerance range depending on the required precision, for the high precision of this exercise the only one that meets the requirements is option C .

4 0
3 years ago
If angle A is a complement of angle B, angle B and angle C are vertical angles, and the supplement of C has a measure of 140°. F
Over [174]

Answer:

50°

Explanation:

Complementary angles add up to 90°.

Supplementary angles add up to 180°.

Vertical angles are equal.

A + B = 90°

B = C

C = 180° − 140°

C = 40°

B = 40°

A = 50°

7 0
4 years ago
Horizontal shear forces and, consequently, horizontal shear stresses are caused in a flexural member at those locations where th
jek_recluse [69]

Answer:

False

Explanation:

When the horizontal shear forces act on the surface there is transverse shear stress at a particular point which is equal in magnitude. Pure bending is less common than a non uniform bending because the beam is not in equilibrium.

5 0
3 years ago
The time factor for a doubly drained clay layer
Margarita [4]

Answer with Explanation:

Assuming that the degree of consolidation is less than 60% the relation between time factor and the degree of consolidation is

T_v=\frac{\pi }{4}(\frac{U}{100})^2

Solving for 'U' we get

\frac{\pi }{4}(\frac{U}{100})^2=0.2\\\\(\frac{U}{100})^2=\frac{4\times 0.2}{\pi }\\\\\therefore U=100\times \sqrt{\frac{4\times 0.2}{\pi }}=50.46%

Since our assumption is correct thus we conclude that degree of consolidation is 50.46%

The consolidation at different level's is obtained from the attached graph corresponding to Tv = 0.2

i)\frac{z}{H}=0.25=U=0.71 = 71% consolidation

ii)\frac{z}{H}=0.5=U=0.45 = 45% consolidation

iii)\frac{z}{H}=0.75U=0.3 = 30% consolidation

Part b)

The degree of consolidation is given by

\frac{\Delta H}{H_f}=U\\\\\frac{\Delta H}{1.0}=0.5046\\\\\therefore \Delta H=50.46cm

Thus a settlement of 50.46 centimeters has occurred

For time factor 0.7, U is given by

T_v=1.781-0.933log(100-U)\\\\0.7=1.781-0.933log(100-U)\\\\log(100-U)=\frac{1.780-.7}{0.933}=1.1586\\\\\therefore U=100-10^{1.1586}=85.59

thus consolidation of 85.59 % has occured if time factor is 0.7

The degree of consolidation is given by

\frac{\Delta H}{H_f}=U\\\\\frac{\Delta H}{1.0}=0.8559\\\\\therefore \Delta H=85.59cm

5 0
3 years ago
Other questions:
  • The idling engines of a landing turbojet produce forward thrust when operating in a normal manner, but they can produce reverse
    7·1 answer
  • The arm of the robot is extending at a constant rate = 1.5 ft/s when r = 3 ft, z = (4t2) ft, and  = (1.5 t) rad, where t is in
    14·1 answer
  • In addition to bandages and gauze, it's recommended that a first aid kit also contain ___________. A) A finger splintB) Portable
    9·1 answer
  • What are the important things to remember when arriving for an interview?
    15·1 answer
  • You’re going to write a program that models the Littleton City Lotto (not a real Lotto game). The program is going to allow to u
    6·1 answer
  • BIG POINTS AND WILL GIVE BRAINLIEST! Answer all 5 please or I can’t give brainliest and might report!
    10·1 answer
  • Give five examples of
    14·1 answer
  • Guyss I seriously and urgently need help what are the steps to build a headgear ??​
    5·2 answers
  • Propose any improvements if there are any in brake system
    7·1 answer
  • How does data mining help interactive marketing for a business?
    5·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!