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3 years ago

100 points!! What should you click on to relocate broken link files?

Computers and Technology

2 answers:

brilliants [131]3 years ago

6 0

Answer:

B. Locate!

let me know if im wrong pls :]

Explanation:

damaskus [11]3 years ago

6 0

Answer:

im pretty sure its browse

Explanation:

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Which phrase suggests feedback?

butalik [34]

Answer:

They also gave suggestions on some modules and said they would like to work with his team in the future.

Explanation:

This statement suggests feedback because, here the client gives Richard and his team some suggestions on what they can do with the modules. It implies that the client must have tested the software application and observed its functionality.

7 0

3 years ago

Who can give me answer for this kinda urgent pls

marissa [1.9K]

Answer:

true

Explanation:

it's TRUE the correct answer is true

6 0

3 years ago

Andrew has data in cell E14 and the cell should be blank. Andrew should _____.

denpristay [2]

Use the mouse to click on cell E14 and press delete

4 0

4 years ago

Read 2 more answers

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 $n = 15\ and\ r = 4$

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

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

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

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

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

$^{15}C_4 = 1365$

Hence, there are 1365 ways 

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)

From the given parameters; 3 out of 15 is detective;

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;

Probability that at least one is defective + Probability that at none is defective = 1

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

3 years ago

Write a little man program that accepts three values as input and produces the largest of the three as output.

ki77a [65]

Required: program to return the largest of three numbers.

pseudocode

input parameters, int A,B,C;
int T; // temporary storage
if (A>B) T=A;
else T=B;
if (T>C) print(T);
else print(C);

5 0

3 years ago