Complete the sentence.

Instructions:Select the correct answer.

Gnesinka [82]

C) team leader is the answer

4 0

3 years ago

Host Y sends the first TCP ACK message for the transaction? a. true b. false

kupik [55]

Answer: True

Explanation:

TCP is the transmission control protocol which and TCP ACK is the transmission control handshake method which are used by TCP. It basically indicates the next sequence number in the method when the flag is set.

Firstly, the ACK (Acknowledge) send by the each end for the initial sequence number itself but it does not contained any type of data. In the TCP segment header ACK contain 32 bit field.

The acknowledgement is just a proof to clients that ACK is a specific to the SYN when the clients initiate.

8 0

3 years ago

The following equations estimate the calories burned when exercising (source): Men: Calories = ( (Age x 0.2017) — (Weight x 0.09

sammy [17]

In python:

age = float(input("How old are you? "))

weight = float(input("How much do you weigh? "))

heart_rate = float(input("What's your heart rate? "))

time = float(input("What's the time? "))

print("The calories burned for men is {}, and the calories burned for women is {}.".format(

((age * 0.2017) - (weight * 0.09036) + (heart_rate * 0.6309) - 55.0969) * (time / 4.184),

((age * 0.074) - (weight * 0.05741) + (heart_rate * 0.4472) - 20.4022) * (time / 4.184)))

This is the program.

When you enter 49 155 148 60, the output is:

The calories burned for men is 489.77724665391963, and the calories burned for women is 580.939531548757.

Round to whatever you desire.

6 0

3 years ago

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

2 years ago

Gina is upgrading your computer with a new processor. She installed the processor into your motherboard and adds the cooling sys

tigry1 [53]

Answer:

The answer is option d.

8 0

3 years ago

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