All categories

3 years ago

Change the case of letter by

Computers and Technology

2 answers:

Evgesh-ka [11]3 years ago

8 0

Answer:

writing it as a capital or lower case.

Explanation:

T - upper case

t - lower case

not sure if that is what you meant or not

Rasek [7]3 years ago

7 0

Answer:

Find the "Bloq Mayus" key at the left side of your keyboard and then type

Explanation:

Thats the only purpose of that key

You might be interested in

the outline view allows you to control how much information you can see in a document, because you can expand or collapse sectio

pochemuha

I believe its true because you can expand or collapse sections of it

5 0

3 years ago

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

Can you help please ill give branilist

konstantin123 [22]

HTML uses tags to help the computer know what different pieces of content in the web page actually are. Right now we've only learned how to tell the computer that some text is a paragraph, or that part of your website is the body. We've already seen how that affects the way our web pages look and are structured.

(I don't know how it should be organized, but hope this helped)

3 0

3 years ago

The 7-bit ASCII code for the character ‘&’ is: 0100110 An odd parity check bit is now added to this code so 8 bits are trans

Alex787 [66]

Answer

First part:

The transmitted 8-bit sequence for ASCII character '&' with odd parity will be 00100110. Here leftmost bit is odd parity bit.

Second part:

The invalid bit sequence are option a. 01001000 and d. 11100111

Explanation:

Explanation for first part:

In odd parity, check bit of either 0 or 1 is added to the binary number as leftmost bit for making the number of 1s in binary number odd.

If there are even number of 1s present in the original number then 1 is added as leftmost bit to make total number of 1s odd.

If there are odd number of 1s present in the original number then 0 is added as leftmost bit to keep the total number of 1s odd.

Explanation for second part:

A valid odd parity bit sequence will always have odd number of 1s.

Since in option a and d, total number of 1s are 2 and 6 i.e. even number. Therefore they are invalid odd parity check bit sequences.

And since in option b and c, total number of 1s are 5 and 7 i.e. odd numbers respectively. Therefore they are valid odd parity check bit sequences.

7 0

3 years ago

Sarah is working on a project in which she needs to record all the extracurricular activities in her college. Her college teache

EastWind [94]

It sounds like the answer is A because there are 5 types of sports and the question only asks for the sports folders

8 0

3 years ago

Read 2 more answers

Change the case of letter by​

Change the case of letter by