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borishaifa [10]

3 years ago

5

You are building a computer from spare parts in the office. You build the computer and realize you have several different types

of RAM chips. Which property of the memory modules does NOT apply when considering the ability to run at optimal performance

Computers and Technology

1 answer:

san4es73 [151]3 years ago

5 0

Answer:

The pin count is the same

Explanation:

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A computer retail store has 15 personal computers in stock. A buyer wants to purchase 3 of them. Unknown to either the retail st

Dima020 [189]

Answer:

a. 1365 ways

b. Probability = 0.4096

c. Probability = 0.5904

Explanation:

Given

PCs = 15

Purchase = 3

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 $n = 15\ and\ r = 4$

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

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

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

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

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

$^{15}C_4 = 1365$

<em>Hence, there are 1365 ways </em>

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

Probability = 0.5904

8 0

2 years ago

mojhsa [17]

Explanation:

The output of this program is 5 7, because the first time bruce is printed, his value is 5, and the second time, his value is 7. The comma at the end of the first print statement suppresses the newline after the output, which is why both outputs appear on the same line.

Here is what multiple assignment looks like in a state diagram:

With multiple assignment it is especially important to distinguish between an assignment operation and a statement of equality. Because Python uses the equal sign (=) for assignment, it is tempting to interpret a statement like a = b as a statement of equality. It is not!

First, equality is symmetric and assignment is not. For example, in mathematics, if a = 7 then 7 = a. But in Python, the statement a = 7 is legal and 7 = a is not.

Furthermore, in mathematics, a statement of equality is always true. If a = b now, then a will always equal b. In Python, an assignment statement can make two variables equal, but they don’t have to stay that way:

a = 5

4 0

2 years ago

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astra-53 [7]

send attached file with it

5 0

3 years ago

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earnstyle [38]

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3 years ago

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a_sh-v [17]

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3 years ago