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2 years ago

In which of the following phases of filmmaking would a production team be focused on the process of casting?

Computers and Technology

2 answers:

Mnenie [13.5K]2 years ago

8 0

Answer:

pre-production

Explanation:

edg2021

Once the film concept has been developed, the film company transitions to the production phases: pre-production, production, and post-production. In pre-production, the technical details are worked out, a schedule is set, and the resources to produce the film are assembled. This includes casting the film, filling the technical positions, and scheduling the production. The actual filming takes place during the production phase, followed by a post-production phase in which the footage is edited and the final film emerges.

Rzqust [24]2 years ago

3 0

Explanation:

There are five phases of film production and they include development, pre-production, production, post-production and distribution.

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1. How many bits would you need to address a 2M × 32 memory if:

Dominik [7]

Answer:

a) 23 b) 21
a) 43 b) 42
a) 0 b) 0

Explanation:

1) How many bits is needed to address a 2M * 32 memory

2M = 2^1*2^20, while item =32 bit long word

hence ; L = 2^21 ; w = 32

a) when the memory is byte addressable

w = 8; L = ( 2M * 32 ) / 8 = 2M * 4

hence number of bits = log2(2M * 4)= log2 ( 2 * 2^20 * 2^2 ) = 23 bits

b) when the memory is word addressable

W = 32 ; L = ( 2M * 32 )/ 32 = 2M

hence the number of bits = log2 ( 2M ) = Log2 (2 * 2^20 ) = 21 bits

2) How many bits are required to address a 4M × 16 main memory

4M = 4^1*4^20 while item = 16 bit long word

hence L ( length ) = 4^21 ; w = 16

a) when the memory is byte addressable

w = 8 ; L = ( 4M * 16 ) / 8 = 4M * 2

hence number of bits = log 2 ( 4M * 2 ) = log 2 ( 4^1*4^20*2^1 ) ≈ 43 bits

b) when the memory is word addressable

w = 16 ; L = ( 4M * 16 ) / 16 = 4M

hence number of bits = log 2 ( 4M ) = log2 ( 4^1*4^20 ) ≈ 42 bits

3) How many bits are required to address a 1M * 8 main memory

1M = 1^1 * 1^20 , item = 8

L = 1^21 ; w = 8

a) when the memory is byte addressable

w = 8 ; L = ( 1 M * 8 ) / 8 = 1M

hence number of bits = log 2 ( 1M ) = log2 ( 1^1 * 1^20 ) = 0 bit

b) when memory is word addressable

w = 8 ; L = ( 1 M * 8 ) / 8 = 1M

number of bits = 0

5 0

3 years ago

Which ipv6 header field is known as the priority field?

Yuki888 [10]

Traffic Class header field is known as the priority field.

4 0

3 years ago

Read 2 more answers

Given the function F (X, Y , Z)=Σm(0,1, 2 , 4 , 6)

Mnenie [13.5K]

Answer:

(1)Minterms complement = XYZ (2) Compliment of Minterms = Σm(0,1, 2 , 4 , 6) (3) (X+Y+Z) (4) Minimized SOP = Z + XY

Manterms = πM

Explanation:

Solution

Recall that:

Given the function F (X, Y , Z)=Σm(0,1, 2 , 4 , 6)

(1) Canonical Disjunctive Normal Form: In boolean algebra, the boolean function can be expressed as Canonical Disjunctive form known as minterms

In Minterm we assign 'I' to each uncomplimented variable and '0' to each complemented/complementary variable

For the given question stated we ave the following:

Minterms = XYZ, XYZ, XYZ, XYZ, XYZ.

(2) Canonical Conjunctive Normal Form: In boolean algebra, the boolean function can be expressed as Canonical Disjunctive form known as maxterms.

In Maxterms we assign '0' to each uncomplimented variable and '1' to each complemented/complementary variable

Compliment of Minterms = Σm(0,1, 2 , 4 , 6)

Maxterms = πM

Note: Kindly find an attached copy of the complete solution to this question below.

6 0

3 years ago

2. When a business practices offensive behavior, you have many options. The option with the loudest voice is

Andreas93 [3]

—-_-__-____- _—-_- -__-_-____-__

___-_-_ _- —|

4 0

2 years ago

Create an array to hold the rainfall values. Create a 2nd parallel array (as a constant) to hold the abbreviated names of the mo

Zarrin [17]

Answer:

#include <stdio.h>

int main()

{

//variable declaration

int low, high;

float lowRain, highRain, total, avg;

//array declaration

float rainfall[13];

char monthName[13][10] = {"", "Jan", "Feb", "Mar", "Apr", "May", "Jun", "Jul", "Aug", "Sep", "Oct", "Nov", "Dec"};

//get user input

for(int i=1; i<=12; i++)

{

printf("Enter the rainfall (in inches) for %s: ", monthName[i]);

scanf("%f", &rainfall[i]);

}

//display the monthly rainfall

printf("\nThe rainfall that was entered was:\n");

for(int i = 1; i<=6; i++)

printf("%s ", monthName[i]);

printf("\n");

for(int i = 1; i<=6; i++)

printf("%.1f ", rainfall[i]);

printf("\n");

for(int i = 7; i<=12; i++)

printf("%s ", monthName[i]);

printf("\n");

for(int i = 7; i<=12; i++)

printf("%.1f ", rainfall[i]);

//variable initialization

low = 1;

high = 1;

lowRain = rainfall[1];

highRain = rainfall[1];

total = 0;

//calculate the lowest, highest and averaage rainfall

for(int i=1; i<=12; i++)

{

if(lowRain>rainfall[i])

{

lowRain = rainfall[i];

low = i;

}

if(highRain<rainfall[i])

{

highRain = rainfall[i];

high = i;

}

total = total + rainfall[i];

}

avg = total / 12;

//display the result

printf("\n\nThe total rain that fell was %.1f inches", total);

printf("\nThe average monthly rainfall was %.1f inches.", avg);

printf("\nThe lowest monthly rainfall was %.1f inches in %s.", rainfall[low], monthName[low]);

printf("\nThe highest monthly rainfall was %.1f inches in %s.", rainfall[high], monthName[high]);

return 0;

}

4 0

3 years ago