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aliina [53]
3 years ago
8

TRUE/FALSE

Computers and Technology
1 answer:
algol [13]3 years ago
5 0
<h2>Answer:</h2>

Following are given answers to each part with explanation:

<h2>Explanation:</h2>

1. lt takes multiple clock cycles to access data from memory.

The statement is TRUE.

Accessing data from memory acquires two steps. In first step the address is read and in second step we go to the address in order to access the data.

2.Each byte of memory has a unique address.

The statement is TRUE.

Each byte in the memory attains a unique address therefore we can access the data residing at each byte specifically by giving address.

3.The total memory used by all running programs can never be larger than the computer's physical memory.

The statement is TRUE.

The running parts are always a sub part o a system so combining them all cannot exceed the total size of computer's physical memory.

4.The lower half (or least significant half) of the EBX register is called BX.

The statement is TRUE.

BX is he least significant or lower half of the register EBX (extended BX).

5.The following data locations are in order of fastest access time to slowest access time: cache, registers, main memory.

The statement is TRUE.

Yes the order of data locations from faster to lower is given correctly:

Cache is accessed faster than registers and in the end comes main memory.

6.The ALU performs only addition, subtraction, multiplication, and division operation:s

The statement is FALSE.

ALU is an abbreviation of Arithmetic Logic unit. So it performs arithmetical operations such as addition, subtraction, multiplication, and division as well as the Logical operations that are AND OR NOT XOR.

7. The control unit (CU) coordinates the sequencing of execution steps in an instruction cycle.

The statement is TRUE.

As obvious from the name Control unit (CU) has assigned the coordination of execution of steps in sequence.

8.The EIP register is updated when an instruction is fetched.

The statement is TRUE.

Whenever the user retrieve any instruction from EIP or fetch the instruction from it. EIP gets updated.

9.The step to fetch an operand is always necessary in the instruction cycle.

The statement is TRUE.

Most of the Instructions demand an operand in the instruction cycle so that they could be operated on. While some instruction not demand the operand such as Exit. But majority demands the operand so it is true.

10.In an instruction cycle, the operands are fetched before the instruction is fetched.

The statement is FALSE.

The first part of instruction cycle is to fetch the instruction, whereas the operands are fetched after it. So the statement get FALSE.

<h3>I hope it will help you!</h3>
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Answer:

a. 1365 ways

b. Probability = 0.4096

c. Probability = 0.5904

Explanation:

Given

PCs = 15

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Solving (a): Ways to select 4 computers out of 15, we make use of Combination formula as follows;

^nC_r = \frac{n!}{(n-r)!r!}

Where n = 15\ and\ r = 4

^{15}C_4 = \frac{15!}{(15-4)!4!}

^{15}C_4 = \frac{15!}{11!4!}

^{15}C_4 = \frac{15 * 14 * 13 * 12 * 11!}{11! * 4 * 3 * 2 * 1}

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^{15}C_4 = \frac{32760}{24}

^{15}C_4 = 1365

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Solving (b): The probability that exactly 1 will be defective (from the selected 4)

First, we calculate the probability of a PC being defective (p) and probability of a PC not being defective (q)

<em>From the given parameters; 3 out of 15 is detective;</em>

So;

p = 3/15

p = 0.2

q = 1 - p

q = 1 - 0.2

q = 0.8

Solving further using binomial;

(p + q)^n = p^n + ^nC_1p^{n-1}q + ^nC_2p^{n-2}q^2 + .....+q^n

Where n = 4

For the probability that exactly 1 out of 4 will be defective, we make use of

Probability =  ^nC_3pq^3

Substitute 4 for n, 0.2 for p and 0.8 for q

Probability =  ^4C_3 * 0.2 * 0.8^3

Probability =  \frac{4!}{3!1!} * 0.2 * 0.8^3

Probability = 4 * 0.2 * 0.8^3

Probability = 0.4096

Solving (c): Probability that at least one is defective;

In probability, opposite probability sums to 1;

Hence;

<em>Probability that at least one is defective + Probability that at none is defective = 1</em>

Probability that none is defective is calculated as thus;

Probability =  q^n

Substitute 4 for n and 0.8 for q

Probability =  0.8^4

Probability = 0.4096

Substitute 0.4096 for Probability that at none is defective

Probability that at least one is defective + 0.4096= 1

Collect Like Terms

Probability = 1 - 0.4096

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