Answer:
1.0MG
Explanation:
to solve this problem we use this formula
S₀-S/t = ksx --- (1)
the values have been given as
concentration = S₀ = 250mg
effluent concentration = S= 10mg
value of K = 0.04L/day
x = 3000 mg
when we put these values into this equation,
250-10/t = 0.04x10x3000
240/t = 1200
we cross multiply from this stage
240 = 1200t
t = 240/1200
t = 0.2
remember the question says that 5MGD is required to be treated
so the volume would be
v = 0.2x5
= 1.0 MG
Answer:
C = 292 Mbps
Explanation:
Given:
- Signal Transmitted Power P = 250mW
- The noise in channel N = 10 uW
- The signal bandwidth W = 20 MHz
Find:
what is the maximum capacity of the channel?
Solution:
-The capacity of the channel is given by Shannon's Formula:
C = W*log_2 ( 1 + P/N)
- Plug the values in:
C = (20*10^6)*log_2 ( 1 + 250*10^-3/10)
C = (20*10^6)*log_2 (25001)
C = (20*10^6)*14.6096
C = 292 Mbps
Solution :
Given :
Water have quality x = 0.7 (dryness fraction) at around pressure of 200 kPa
The phase diagram is provided below.
a). The phase is a standard mixture.
b). At pressure, p = 200 kPa, T = 
Temperature = 120.21°C
c). Specific volume
d). Specific energy (
)
e). Specific enthalpy 
At 
f). Enthalpy at m = 0.5 kg
= 1022.91 kJ
Answer:
import pandas pd
def read_prices(tickers):
price_dict = {}
# Read ingthe ticker data for all the tickers
for ticker in tickers:
# Read data for one ticker using pandas.read_csv
# We assume no column names in csv file
ticker_data = pd.read_csv("./" + ticker + ".csv", names=['date', 'price', 'volume'])
# ticker_data is now a panda data frame
# Creating dictionary
# for the ticker
price_dict[ticker] = {}
for i in range(len(ticker_data)):
# Use pandas.iloc to access data
date = ticker_data.iloc[i]['date']
price = ticker_data.iloc[i]['price']
price_dict[ticker][date] = price
return price_dict
Answer:
Just answered this to confirm my profile.
Explanation:
I dont have a clue, this is just to confirm my profile.
