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Alenkasestr [34]
3 years ago
9

A certain microprocessor requires either 2, 3, 4, 8, or 12 machine cycles to perform various operations. Twenty-five percent of

its instructions require 2 machine cycles, 20% require 3 machine cycles, 17.5% require 4 machine cycles, 12.5% require 8 machine cycles, and 25% require 12 machine cycles. a) What is the average number of machine cycles per instruction for this microprocessor
Mathematics
1 answer:
myrzilka [38]3 years ago
3 0

Answer:

The average number of machine cycles per instruction for this microprocessor is 5.8.

Step-by-step explanation:

To find the average number of machine cycles per instruction for this microprocessor, we multiply each number of cycles for it's relative frequency.

We have that:

25% require 2 cycles.

20% require 3 cycles.

17.5% require 4 cycles.

12.5% require 8 cycles.

25% require 12 cycles.

So the average number is:

M = 0.25*2 + 0.2*3 + 0.175*4 + 0.125*8 + 0.25*12 = 5.8

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2/5 left of 1/4 gallon

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2 years ago
In a home theater system, the probability that the video components need repair within 1 year is 0.02, the probability that the
earnstyle [38]

Answer:

(a) The probability that at least one of these components will need repair within 1 year is 0.0278.

(b) The probability that exactly one of these component will need repair within 1 year is 0.0277.

Step-by-step explanation:

Denote the events as follows:

<em>A</em> = video components need repair within 1 year

<em>B</em> = electronic components need repair within 1 year

<em>C</em> = audio components need repair within 1 year

The information provided is:

P (A) = 0.02

P (B) = 0.007

P (C) = 0.001

The events <em>A</em>, <em>B</em> and <em>C</em> are independent.

(a)

Compute the probability that at least one of these components will need repair within 1 year as follows:

P (At least 1 component needs repair)

= 1 - P (No component needs repair)

=1-P(A^{c}\cap B^{c}\cap C^{c})\\=1-[P(A^{c})\times P(B^{c})\times P(C^{c})]\\=1-[(1-0.02)\times (1-0.007)\times (1-0.001)]\\=1-0.97216686\\=0.02783314\\\approx 0.0278

Thus, the probability that at least one of these components will need repair within 1 year is 0.0278.

(b)

Compute the probability that exactly one of these component will need repair within 1 year as follows:

P (Exactly 1 component needs repair)

= P (A or B or C)

=P(A\cap B^{c}\cap C^{c})+P(A^{c}\cap B\cap C^{c})+P(A^{c}\cap B^{c}\cap C)\\=[0.02\times (1-0.007)\times (1-0.001)]+[(1-0.02)\times 0.007\times (1-0.001)]\\+[(1-0.02)\times (1-0.007)\times 0.001]\\=0.02766642\\\approx 0.0277

Thus, the probability that exactly one of these component will need repair within 1 year is 0.0277.

7 0
3 years ago
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