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Studentka2010 [4]
3 years ago
10

Which of the following statements best describes the relationship between availability of new green building products and custom

er demand for green building?
A. As more and more new green building products become available, customers will be more and more confused by the array of new choices.
B. The availability of new green building products needs to be accompanied by extensive marketing campaigns to make customers aware and drive demand.
C. New green building products can create new customer demand, which in turn can drive the development of more new products.
D. Customers only demand new green building products if, in addition to their environmental benefit, these products also lower costs and construction times.
Engineering
1 answer:
Fed [463]3 years ago
4 0

Answer:

c

Explanation:

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How much memory can a 32 -bit processor support ?
JulijaS [17]

It supporst only 4 Gigabytes of memory in the system

3 0
2 years ago
Two routes connect an origin and a destination. Routes 1 and 2 have performance functions t1 = 2 + X1 and t2 = 1 + X2, where the
Musya8 [376]

Solution :

Given

$t_1=2+x_1$

$t_2=1+x_2$

Now,

$P(h

$0.4=1-P(h \geq5)$

$0.6=P(h \geq5)$

$0.6= e^{\frac{-x_1 5}{3600}}$

Therefore,   $x_1=368 \ veh/h$

                        $=\frac{368}{1000} = 0.368$

Given,   $t_1=2+x_1$

                 = 2 + 0.368

                 = 2.368 min

At user equilibrium, $t_2=t_1$

∴  $t_2$ = 2.368 min

$t_2=1+x_2$

$2.368=1+x_2$

$x_2 = 1.368$

$x_2 = 1.368 \times 1000$

    = 1368 veh/h

7 0
3 years ago
The Ethernet (CSMA/CD) alternates between contention intervals and successful transmissions. Assume a 100 Mbps Ethernet over 1 k
Vesnalui [34]
<h3><u>CSMA/CD Protocol: </u></h3>

Carrier sensing can transmit the data at anytime only the condition is before sending the data sense carrier if the carrier is free then send the data.

But the problem is the standing at one end of channel, we can’t send the entire carrier. Because of this 2 stations can transmit the data (use the channel) at the same time resulting in collisions.

There are no acknowledgement to detect collisions, It's stations responsibility to detect whether its data is falling into collisions or not.

<u>Example: </u>

T_{P}=1 H r, at time t = 10.00 AM, A starts, 10:59:59 AM B starts at time 11:00 AM collision starts.

12:00 AM A will see collisions

Pocket Size to detect the collision.

\begin{aligned}&T_{t} \geq 2 T_{P}\\&\frac{L}{B} \geq 2 T_{P}\\&L \geq 2 \times T_{P} \times B\end{aligned}

CSMA/CD is widely used in Ethernet.

<u>Efficiency of CSMA/CD:</u>

  • In the previous example we have seen that in worst case 2 T_{P} time require to detect a collision.
  • There could be many collisions may happen before a successful completion of transmission of a packet.

We are given number of collisions (contentions slots)=4.

\text { Propagation day }=\frac{\text {distance}}{\text {speed}}

Distance = 1km = 1000m

\begin{aligned}&\text { Speed }=2 \times 10^{8} \mathrm{m} / \mathrm{sec}\\ &T_{P}=\frac{1000}{2 \times 10^{8}}=(0.5) \times 10^{-5}=5 \times 10^{-6}\\ &T_{t}=5 \mu \mathrm{sec}\end{aligned}

7 0
3 years ago
Write a C program that will update a bank balance. A user cannot withdraw an amount ofmoney that is more than the current balanc
GarryVolchara [31]

Answer:

Explanation:

Sample output:

BANK ACCOUT PROGRAM!

----------------------------------

Enter the old balance: 1234.50

Enter the transactions now.

Enter an F for the transaction type when you are finished.

Transaction Type (D=deposit, W=withdrawal, F=finished): D

Amount: 568.34

Transaction Type (D=deposit, W=withdrawal, F=finished): W

Amount: 25.68

Transaction Type (D=deposit, W=withdrawal, F=finished): W

Amount: 167.40

Transaction Type (D=deposit, W=withdrawal, F=finished): F

Your ending balance is $1609.76

Program is ending

Code to copy:

// include the necessary header files.

#include<stdio.h>

// Definition of the function

float withdraw(float account_balance, float withdraw_amount)

{

// Calculate the balace amount.

float balance_amount = account_balance - withdraw_amount;

// Check whether the withdraw amount

// is greater than 0 or not.

if (withdraw_amount > 0 && balance_amount >= 0)

{

// Assign value.

account_balance = balance_amount;

}

// return account_balance

return account_balance;

}

// Definition of the function deposit.

float deposit(float account_balance, float deposit_amount)

{

// Check whether the deposit amount is greater than zero

if (deposit_amount > 0)

{

// Update account balance.

account_balance = account_balance + deposit_amount;

}

// return account balance.

return account_balance;

}

int main()

{

// Declare the variables.

float account_balance;

float deposit_amount;

float withdrawl_amount;

char input;

// display the statement on console.

printf("BANK ACCOUT PROGRAM!\n");

printf("----------------------------------\n");

// prompt the user to enter the old balance.

printf("Enter the old balance: ");

// Input balance

scanf("%f", &account_balance);

// Display the statement on console.

printf("Enter the transactions now.\n");

printf("Enter an F for the transaction type when you are finished.\n");

// Start the do while loop

do

{

// prompt the user to enter transaction type.

printf("Transaction Type (D=deposit, W=withdrawal, F=finished): ");

// Input type.

scanf(" %c", &input);

// Check if the input is D

if (input == 'D')

{

// Prompt the user to input amount.

printf("Amount: ");

// input amount.

scanf("%f", &deposit_amount);

// Call to the function.

account_balance=deposit(account_balance,deposit_amount);

}

// Check if the input is W

if (input == 'W')

{

printf("Amount: ");

scanf("%f", &withdrawl_amount);

// Call to the function.

account_balance = withdraw(account_balance,withdrawl_amount);

}

// Check if the input is F

if (input == 'F')

{

// Dispplay the amount.

printf("Your ending balance is $%.2f\n", account_balance);

printf("Program is ending\n");

}

// End the while loop

} while(input != 'F');

return 0;

}

the picture uploaded below shows the program screenshot.

cheers, i hope this helps.

5 0
3 years ago
A 11.5 nC charge is at x = 0cm and a -1.2 nC charge is at x = 3 cm ..At what position or positions on the x-axis is the electric
diamong [38]

Answer:

Explanation:

Given

q_1=11.5\ nC charge is placed at x=0\ cm

another charge of q_2=-1.2\ nC is at x=3\ cm

We know that Electric field due to positive charge is away  from it and Electric field due to negative charge is towards it.

so net electric field is zero somewhere beyond negatively charged particle

Electric Field due to q_2 at some distance r from it

E_2=\frac{kq_2}{r^2}

Now Electric Field due to q_1 is

E_1=\frac{kq_1}{(3+r)^2}

Now E_1+E_2=0

\frac{k\times 11.5}{(r+3)^2}\frac{k\times (-1.2)}{r^2}=0

\frac{3+r}{r}=(\frac{11.5}{1.2})^{0.5}

\frac{3+r}{r}=3.095

thus r=1.43\ cm

Thus Electric field is zero at some distance r=1.43 cm right of q_2

3 0
3 years ago
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