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Tpy6a [65]
3 years ago
10

Ou have an application running on Oracle Cloud Infrastructure. You identified that the read and write operations are slowing you

r application down enough to impair user access. The application is currently using a VM.Standard1.2 compute without any block storage attached to it.
Which two options allow you to increase disk performance?

A. Terminate the compute instance preserving the boot volume. Create a new compute instance using a VM Dense IO shape using the boot volume preserved.
B. Terminate the compute instance preserving the boot volume. Create a new compute instance using a VM Standard shape and attach a new block volume to host your application.
C. Create a backup of the boot volume. Create a new compute instance using a VM Dense IO shape and restore the backup.
D. Terminate the compute instance and create a backup of the boot volume. Create a new compute instance using a VM Dense IO shape and restore the backup.
Computers and Technology
1 answer:
Otrada [13]3 years ago
3 0

Answer:

Options A and C.

Explanation:

In Oracle Cloud Infrastructure the two options which allows you to increase disk performance are;

1. Terminate the compute instance preserving the boot volume. Create a new compute instance using a VM Dense IO shape using the boot volume preserved.

2. Create a backup of the boot volume. Create a new compute instance a VM Dense IO shape and restore the backup.

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Which role is delegated to personnel of the IT department and is responsible for maintaining the integrity and security of the d
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Suppose that the format for license plates in a certain state is two letters followed by four numbers. (a) How many different pl
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(a) 6,760,000 plates

(b) 3,407,040 plates

(c) 6,084,000 plates

Explanation:

The very first thing to note about this question is the number of characters involved in the license plate format (6 characters in this case; 2 letters and 4 numbers). The letters come first and then the numbers follow.

There are a total of 26 possible letters (A-Z) and 10 possible numbers (0 - 9) that can be chosen. We can then proceed to the first question;

(a) Here, the total number of possible plates is to be determined. This is done as follows:

Character 1 (Letter): There are 26 possible letters

Character 2 (Letter): There are 26 possible letters

Character 3 (number): There are 10 possible numbers

Character 4 (number): There are 10 possible numbers

Character 5 (number): There are 10 possible numbers

Character 6 (number): There are 10 possible numbers

So, total number of different plates will be obtained by multiplying all the possibilities: 26 × 26 × 10 × 10 × 10 × 10 = 6,760,000 plates

(b) This second part puts a constraint on the usage of the numbers, unlike the question (a), where there was no constraint at all.

Since there is no constraint on the letters, we can write that:

Character 1 (Letter): There are 26 possible letters

Character 2 (Letter): There are 26 possible letters

For the first number as well, we can write:

Character 3 (number): There are 10 possible numbers

However, for the remaining characters, the possibilities will continually reduce by a value of 1, since we can not use a number that has been used before. So,

Character 4 (number): There are 9 possible numbers

Character 5 (number): There are 8 possible numbers

Character 6 (number): There are 7 possible numbers

So, total number of different plates will be: 26 × 26 × 10 × 9 × 8 × 7 = 3,407,040 plates

(c) Here, repetitions are allowed as in questions (a), but there can not be four zeros. This implies that the maximum number of zeros in any plate will be three. Thus, there will be maximum possibilities on all characters until the last one which will be constrained.

Character 1 (Letter): There are 26 possible letters

Character 2 (Letter): There are 26 possible letters

Character 3 (number): There are 10 possible numbers

Character 4 (number): There are 10 possible numbers

Character 5 (number): There are 10 possible numbers

Character 6 (number): There are 9 possible numbers

Total number of plates will therefore be: 26 × 26 × 10 × 10 × 10 × 9 = 6,084,000 plates.

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3 years ago
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