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Stells [14]
4 years ago
5

Which distributions would you recommend be tested using Benford’s law? Select all choices that apply. (You may select more than

one answer. Single click the box with the question mark to produce a check mark for a correct answer and double click the box with the question mark to empty the box for a wrong answer. Any boxes left with a question mark will be automatically graded as incorrect.)
employee numbersunchecked

cash disbursementsunchecked

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accounts payable transactionsunchecked

Sales transactions

Engineering
1 answer:
Leni [432]4 years ago
6 0

Answer:

Please see the attached picture for the complete answer.

Explanation:

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Create a file named students containing the following data in your current directory. Each line in this file represents a studen
Ostrovityanka [42]

Answer:

#!/bin/bash

# Simple line count example, using bash

# Usage: ./line_count.sh file

# -----------------------------------------------------------------------------

# Link filedescriptor 10 with stdin

exec 10<&0

# stdin replaced with a file supplied as a first argument

exec < $1

# remember the name of the input file

in=$1

# init

file="current_line.txt"

let count=0

# this while loop iterates over all lines of the file

while read LINE

do

   # increase line counter  

   ((count++))

   # write current line to a tmp file with name $file (not needed for counting)

   echo $LINE > $file

   # this checks the return code of echo (not needed for writing; just for demo)

   if [ $? -ne 0 ]  

    then echo "Error in writing to file ${file}; check its permissions!"

   fi

done

echo "Number of lines: $count"

echo "The last line of the file is: `cat ${file}`"

# Note: You can achieve the same by just using the tool wc like this

echo "Expected number of lines: `wc -l $in`"

# restore stdin from filedescriptor 10

# and close filedescriptor 10

exec 0<&10 10<&-

Explanation:

4 0
4 years ago
You have accumulated several parking tickets while at school, but you are graduating later in the year and plan to return to you
Elena L [17]

Answer:

pay off the parking tickets

Explanation:

In the scenario being described, the best thing to do would be to pay off the parking tickets. The parking tickets stay under your name, and if they are not paid in time can cause problems down the road. For starters, if they are not paid in time the amount will increase largely which will be harder to pay. If that increased amount is also not paid, then the government will suspend your licence indefinitely which can later lead to higher insurance rates.

3 0
3 years ago
An alloy fin with a thermal conductivity of 200 W/ሺm ∙ Kሻ has a length of 2.5 cm and a thickness of 3.5 mm. The base of the fin
Svetllana [295]

Answer: heat flux into the fun is 21.714 mW/m^2

Explanation:

Heat flux Q = q/A

q = heat transfer rate W

A = area m^2

q = area * conductivity * temperature gradient

Temperature gradient = difference in temperature of the metal faces divided by the thickness.

Therefore Q = k * ( temp. gradient)

Q = 200 * ((400-20)/3.5*10^-2)

Q = 21714285.71 = 21.714 mW/m^2

Answer 2: convective heat transfer flux between fin and air

is 3800W/m^2

Explanation :

q = hA*(Ts-Ta)

h = convective heat transfer coefficient

Ts = temperature of fin

Ta = temperature of air

Q = q/A

Q = h(Ts-Ta)

Q = 10(400 - 20)

Q = 3800 W/m^2

5 0
4 years ago
A 25 kg iron block initially at 280°C is quenched in an insulated tank that contains 100 kg of water at 18°C. Assuming the water
defon

Answer: water at 18°C, w cw = 4.18 kJ/kg∙°C; at 27oC, Cfe = 0.554 kJ/kg∙°C. (Note that we could also have the following units for the heat capacities: kJ/kg∙K.) Assuming that the heat capacities are constant we have the following

The final temperature of the iron and the water will be the same: Tfe,2 = Tw,2 = T2. Substituting T2 for Tfe,2 and Tw,2, and solving for T2 gives the following result for the final temperature.

T2= ((Mw. Cw.T1w) + (Mfe.Cfe. T1fe))/((Mw.Cw) + (Mfe.Cfe))...equ 1

Where Me= mass of water= 100kg, Mfe =mass of iron = 25kg, T1fe = temp of iron before= 280°c,

Using substitution

T2= ((100*4.18*18) + (25*0.554*280))/ ((100*4.18) + (25*.554))

T2 = 26.4°c

So determine total entropy change

DStot = DSw + DSfe ...equat3

DStot = final entropy, DSw = entropy of water at T2, DSfe = final entropy of iron at T2 where,

DS = M.C. lin(T2/T1)...equ 5 temp is in Kelvin.

DSw = 100*4.18*lin(299.4/291) = 11.895

DSfe = 25*.554*lin(299.4/553) = -8.498.

Substituting answers into equa3

DStot = 11.895 - 8.498 = 3.397kj/kg*Kelvin

Explanation: the explanation is in the answers above..

3 0
3 years ago
1.The moist unit weights and degrees of saturation of a soil are given: moist unit weight (1) = 16.62 kN/m^3, degree of saturati
alexandr1967 [171]

Answer:

Gs = 2.647

e = 0.7986

Explanation:

We know that moist unit weight of soil is given as

\gamma_m \ or\ bulk\ density = \frac{(Gs+Se)\times \gamma_w}{(1+e)}

where,  \gamma_m = moist unit weight of the soil

Gs = specific gravity of the soil

S = degree of saturation

e = void ratio

\gamma_w = unit weight of water = 9.81 kN/m3

From data given we know that:

At 50% saturation,\gamma_m = 16.62 kN/m3

puttng all value to get Gs value;

16.62= \frac{(Gs+0.5*e)\timees 9.81}{(1+e)}

Gs - 1.194*e = 1.694 .........(1)

for saturaion 75%, unit weight = 17.71 KN/m3

17.71 = \frac{(Gs+0.75*e)\times 9.81}{(1+e)}

Gs - 1.055*e = 1.805 .........(2)

solving both  equations (1) and (2), we obtained;

Gs = 2.647

e = 0.7986

6 0
3 years ago
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