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DochEvi [55]
3 years ago
14

In Lab 7, we worked through a program that displayed the homeless shelter occupancy over time. The same approach can be used for

displaying any dataset where the date and time are stored. For this program, use the lab as a starting point to display public school attendance from NYC OpenData. If you would like to test your program on other data, you can filter for an individual school by viewing the data and filtering on the school number ("School DBN"). Export the file as CSV and save. There is a sample file for the high school on campus on github Modify the program from Lab 7 that displays shelter population over time to: Ask the user to specify the input file, Ask the user to specify the output file, Convert the date column (which is stored as 'YYYYMMDD') to a datetime format recognized by pandas, for example if your dataframe is df, overwrite the 'Date' column to be: df["Date"] = pd.to_datetime(df["Date"].apply(str)) Make a plot of the percent of absent students over time from the data in the input file, and Store the plot in the output file the user specified. A sample run of the program: Enter name of input file:

Engineering
1 answer:
Bezzdna [24]3 years ago
6 0

Answer:

Explanation:

The python code to generate this is quite simple to run.

i hope you understand everything written here, you can as well try out other problems to understand better.

First to begin, we import the package;

Code:

import pandas as pd

import matplotlib.pyplot as plt

name = input('Enter name of the file: ')

op = input('Enter name of output file: ')

df = pd.read_csv(name)

df['Date'] = pd.to_datetime(df["Date"].apply(str))

plt.plot(df['Date'],df['Absent']/(df['Present']+df['Absent']+df['Released']),label="% Absent")

plt.legend(loc="upper right")

plt.xticks(rotation=20)

plt.savefig(op)

plt.show()

This should generate the data(plot) as seen in the uploaded screenshot.

thanks i hope this helps!!!

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An escalator handles a steady load of 26 people per minute in elevating them from the first to the second floor through a vertic
photoshop1234 [79]

Answer:

\eta = 70.711\,\%

Explanation:

The power needed to make the escalator working is obtained by means of the Work-Energy Theorem:

\dot W  = \dot U_{g}

\dot W = \dot n \cdot m_{p}\cdot g \cdot \Delta y

\dot W = \left(26\,\frac{persons}{min}\right)\cdot (124\,lbm)\cdot \left(32.174\,\frac{ft}{s^{2}}\right)\cdot \left(\frac{1\,lbf}{32.174\,\frac{lbm\cdot ft}{s^{2}} } \right)\cdot (27.5\,ft)

\dot W = 88660\,\frac{lbf\cdot ft}{min}\,\left(2.687\,hp\right)

The mechanical efficiency of the escalator is:

\eta = \frac{2.687\,hp}{3.8\,hp}\times 100\,\%

\eta = 70.711\,\%

3 0
3 years ago
Maximum iorsional shear siress.? Select one: a)- occurs at the center of a shaft. b)- occurs at the outer surface of a shaft c)-
pickupchik [31]

Answer:

b). Occurs at the outer surface of the shaft

Explanation:

We know from shear stress and torque relationship, we know that

\frac{T}{J}= \frac{\tau }{r}

where, T = torque

            J = polar moment of inertia of shaft

            τ = torsional shear stress

             r = raduis of the shaft

Therefore from the above relation we see that

            \tau = \frac{T.r}{J}

Thus torsional shear stress, τ is directly proportional to the radius,r of the shaft.

When r= 0, then τ = 0

and when r = R , τ is maximum

Thus, torsional shear stress is maximum at the outer surface of the shaft.

4 0
3 years ago
Do not answer pls thank you
astra-53 [7]

The answer is answered! Explanation:

4 0
3 years ago
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What are the advantages of studying a scheduled course over a more freeform one?
anyanavicka [17]

Answer:

Taking responsibility for your own learning makes it easier to identify your strengths and weaknesses. Once these have been identified you can work on a learning plan that focuses on the areas that you need most help with, increasing the speed of your learning, and build the skills you have been trying to perfect.

Explanation:

8 0
2 years ago
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Which of the following is correct regarding the principal stresses and maximum in-plane shear stresses? a. Principal stresses ca
ziro4ka [17]

Answer:

option B.

Explanation:

The correct answer is option B.

Principal stress is the maximum normal stress a body can have. In principal stress, there is purely normal stress. On principal plane shear stress is zero.

In-plane shear stress are the shear stress which is acting on the plane.

The statement which is correct regarding principle plane and shear stress is that The shear stress over principal stress planes is always zero.

4 0
3 years ago
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